Why doesn't the definition “$p$ is called 'prime' if $pmid abimplies pmid a,text{ or },pmid b$” hold up...












7














So, I've been given the actual definition of a prime for the first time, as opposed to the definition of an irreducible which I was previously taught (as is customary), it goes as follows:




An element $p$ of a ring $R$ is called "prime" if $a,bin R$ and $p|abrightarrow p|a$ or $p|b$




$4^2 = 16$. So inversely $4|16$. Now let's look at all the combinations of two integers $a,bin R$ where $ab = 16$ and whether $p|a$ or $p|b$:



$1 * 16$ ($4|16$)



$2 * 8$ ($4|8$)



$4 * 4$ ($4|4$)



By my logic, $4$ is prime.



However, as we all know $4 = 2^2$. But $2$ is not a unit so $4$ is an irreducible. Since the ring of integers is an integral domain $4$ must, therefore, be prime as well. Why does this happen?



Edit



Thanks for your help everyone. I understand your answers logically and experimentally but not really conceptually. The reason this definition works (from my perspective) because your saying that




A number is prime if it necessary to represent its multiples in
factorisations.




Whereby factorisations, I mean $8 = 2^3, 6 = 2*3$ and relevantly $16 = 2^4$.



The answer I had kind-of expected from this question (and was going to include in my original query until I forgot) was that more generally if there exists the product of any amount of numbers that is equivalent to $ab$ and for each of those numbers $n$, $n$ suffices $p|n$ then $p$ is composite.



This makes much more sense to me logically. In the case of $16$ you saying that you don't need $4$ to express $16$ since you can instead use $2*2*2*2$ or $2^4$ to represent the same thing. For this reason, $4$ is composite.



By this logic, you don't need to prove it for every multiple of 4, only 1. Have I completely confused myself? Can someone provide a counter-example? Also, could you provide an explanation of why my logic doesn't work?



Edit 2



Yes everyone, the correct definition of includes checking the factors of every $ab$ such $p|ab$. However, as @Vincent and I showed in our respective answers (the former much better) the definition I used is equivalent to the definition everyone else is using in integral domains. I highly suggest you read @Vincent answer as he makes this very clear.



Regardless, I'm not accepting any answer that simply says that my definition is wrong when it is actually equivalent.










share|cite|improve this question




















  • 35




    For an element to be prime this must hold for all a,b, not just one example.
    – Erik Parkinson
    2 days ago






  • 11




    You have shown that there exists $c$ such that $4|c$ and $forall a,b$ such that $ab=c, 4|a$ or $4|b$. Your definition demands this to be true for every $c$, not just 16.
    – RcnSc
    2 days ago






  • 13




    Your definition of prime is wrong. An element $p$ is prime if for all $a,bin R$, $p|abto p|atext{ or }p|b$. You left out the all-important "for all" and inserted an "and".
    – bof
    2 days ago








  • 3




    Where did you get your definition? It is valid only if we interpret it as $(a,b in R text{ and } p|ab) rightarrow (p|a text{ or } p|b)$, while you seem to be interpreting it as $a,b in R text{ and } (p|ab rightarrow (p|a text{ or } p|b))$.
    – Acccumulation
    2 days ago








  • 4




    $2cdot6$ is divisible by $4$ but neither $2$ nor $6$ is. Therefore $4$ is not a prime.
    – Jyrki Lahtonen
    yesterday
















7














So, I've been given the actual definition of a prime for the first time, as opposed to the definition of an irreducible which I was previously taught (as is customary), it goes as follows:




An element $p$ of a ring $R$ is called "prime" if $a,bin R$ and $p|abrightarrow p|a$ or $p|b$




$4^2 = 16$. So inversely $4|16$. Now let's look at all the combinations of two integers $a,bin R$ where $ab = 16$ and whether $p|a$ or $p|b$:



$1 * 16$ ($4|16$)



$2 * 8$ ($4|8$)



$4 * 4$ ($4|4$)



By my logic, $4$ is prime.



However, as we all know $4 = 2^2$. But $2$ is not a unit so $4$ is an irreducible. Since the ring of integers is an integral domain $4$ must, therefore, be prime as well. Why does this happen?



Edit



Thanks for your help everyone. I understand your answers logically and experimentally but not really conceptually. The reason this definition works (from my perspective) because your saying that




A number is prime if it necessary to represent its multiples in
factorisations.




Whereby factorisations, I mean $8 = 2^3, 6 = 2*3$ and relevantly $16 = 2^4$.



The answer I had kind-of expected from this question (and was going to include in my original query until I forgot) was that more generally if there exists the product of any amount of numbers that is equivalent to $ab$ and for each of those numbers $n$, $n$ suffices $p|n$ then $p$ is composite.



This makes much more sense to me logically. In the case of $16$ you saying that you don't need $4$ to express $16$ since you can instead use $2*2*2*2$ or $2^4$ to represent the same thing. For this reason, $4$ is composite.



By this logic, you don't need to prove it for every multiple of 4, only 1. Have I completely confused myself? Can someone provide a counter-example? Also, could you provide an explanation of why my logic doesn't work?



Edit 2



Yes everyone, the correct definition of includes checking the factors of every $ab$ such $p|ab$. However, as @Vincent and I showed in our respective answers (the former much better) the definition I used is equivalent to the definition everyone else is using in integral domains. I highly suggest you read @Vincent answer as he makes this very clear.



Regardless, I'm not accepting any answer that simply says that my definition is wrong when it is actually equivalent.










share|cite|improve this question




















  • 35




    For an element to be prime this must hold for all a,b, not just one example.
    – Erik Parkinson
    2 days ago






  • 11




    You have shown that there exists $c$ such that $4|c$ and $forall a,b$ such that $ab=c, 4|a$ or $4|b$. Your definition demands this to be true for every $c$, not just 16.
    – RcnSc
    2 days ago






  • 13




    Your definition of prime is wrong. An element $p$ is prime if for all $a,bin R$, $p|abto p|atext{ or }p|b$. You left out the all-important "for all" and inserted an "and".
    – bof
    2 days ago








  • 3




    Where did you get your definition? It is valid only if we interpret it as $(a,b in R text{ and } p|ab) rightarrow (p|a text{ or } p|b)$, while you seem to be interpreting it as $a,b in R text{ and } (p|ab rightarrow (p|a text{ or } p|b))$.
    – Acccumulation
    2 days ago








  • 4




    $2cdot6$ is divisible by $4$ but neither $2$ nor $6$ is. Therefore $4$ is not a prime.
    – Jyrki Lahtonen
    yesterday














7












7








7


1





So, I've been given the actual definition of a prime for the first time, as opposed to the definition of an irreducible which I was previously taught (as is customary), it goes as follows:




An element $p$ of a ring $R$ is called "prime" if $a,bin R$ and $p|abrightarrow p|a$ or $p|b$




$4^2 = 16$. So inversely $4|16$. Now let's look at all the combinations of two integers $a,bin R$ where $ab = 16$ and whether $p|a$ or $p|b$:



$1 * 16$ ($4|16$)



$2 * 8$ ($4|8$)



$4 * 4$ ($4|4$)



By my logic, $4$ is prime.



However, as we all know $4 = 2^2$. But $2$ is not a unit so $4$ is an irreducible. Since the ring of integers is an integral domain $4$ must, therefore, be prime as well. Why does this happen?



Edit



Thanks for your help everyone. I understand your answers logically and experimentally but not really conceptually. The reason this definition works (from my perspective) because your saying that




A number is prime if it necessary to represent its multiples in
factorisations.




Whereby factorisations, I mean $8 = 2^3, 6 = 2*3$ and relevantly $16 = 2^4$.



The answer I had kind-of expected from this question (and was going to include in my original query until I forgot) was that more generally if there exists the product of any amount of numbers that is equivalent to $ab$ and for each of those numbers $n$, $n$ suffices $p|n$ then $p$ is composite.



This makes much more sense to me logically. In the case of $16$ you saying that you don't need $4$ to express $16$ since you can instead use $2*2*2*2$ or $2^4$ to represent the same thing. For this reason, $4$ is composite.



By this logic, you don't need to prove it for every multiple of 4, only 1. Have I completely confused myself? Can someone provide a counter-example? Also, could you provide an explanation of why my logic doesn't work?



Edit 2



Yes everyone, the correct definition of includes checking the factors of every $ab$ such $p|ab$. However, as @Vincent and I showed in our respective answers (the former much better) the definition I used is equivalent to the definition everyone else is using in integral domains. I highly suggest you read @Vincent answer as he makes this very clear.



Regardless, I'm not accepting any answer that simply says that my definition is wrong when it is actually equivalent.










share|cite|improve this question















So, I've been given the actual definition of a prime for the first time, as opposed to the definition of an irreducible which I was previously taught (as is customary), it goes as follows:




An element $p$ of a ring $R$ is called "prime" if $a,bin R$ and $p|abrightarrow p|a$ or $p|b$




$4^2 = 16$. So inversely $4|16$. Now let's look at all the combinations of two integers $a,bin R$ where $ab = 16$ and whether $p|a$ or $p|b$:



$1 * 16$ ($4|16$)



$2 * 8$ ($4|8$)



$4 * 4$ ($4|4$)



By my logic, $4$ is prime.



However, as we all know $4 = 2^2$. But $2$ is not a unit so $4$ is an irreducible. Since the ring of integers is an integral domain $4$ must, therefore, be prime as well. Why does this happen?



Edit



Thanks for your help everyone. I understand your answers logically and experimentally but not really conceptually. The reason this definition works (from my perspective) because your saying that




A number is prime if it necessary to represent its multiples in
factorisations.




Whereby factorisations, I mean $8 = 2^3, 6 = 2*3$ and relevantly $16 = 2^4$.



The answer I had kind-of expected from this question (and was going to include in my original query until I forgot) was that more generally if there exists the product of any amount of numbers that is equivalent to $ab$ and for each of those numbers $n$, $n$ suffices $p|n$ then $p$ is composite.



This makes much more sense to me logically. In the case of $16$ you saying that you don't need $4$ to express $16$ since you can instead use $2*2*2*2$ or $2^4$ to represent the same thing. For this reason, $4$ is composite.



By this logic, you don't need to prove it for every multiple of 4, only 1. Have I completely confused myself? Can someone provide a counter-example? Also, could you provide an explanation of why my logic doesn't work?



Edit 2



Yes everyone, the correct definition of includes checking the factors of every $ab$ such $p|ab$. However, as @Vincent and I showed in our respective answers (the former much better) the definition I used is equivalent to the definition everyone else is using in integral domains. I highly suggest you read @Vincent answer as he makes this very clear.



Regardless, I'm not accepting any answer that simply says that my definition is wrong when it is actually equivalent.







abstract-algebra group-theory ring-theory prime-numbers






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 33 mins ago

























asked 2 days ago









PolymorphismPrince

1546




1546








  • 35




    For an element to be prime this must hold for all a,b, not just one example.
    – Erik Parkinson
    2 days ago






  • 11




    You have shown that there exists $c$ such that $4|c$ and $forall a,b$ such that $ab=c, 4|a$ or $4|b$. Your definition demands this to be true for every $c$, not just 16.
    – RcnSc
    2 days ago






  • 13




    Your definition of prime is wrong. An element $p$ is prime if for all $a,bin R$, $p|abto p|atext{ or }p|b$. You left out the all-important "for all" and inserted an "and".
    – bof
    2 days ago








  • 3




    Where did you get your definition? It is valid only if we interpret it as $(a,b in R text{ and } p|ab) rightarrow (p|a text{ or } p|b)$, while you seem to be interpreting it as $a,b in R text{ and } (p|ab rightarrow (p|a text{ or } p|b))$.
    – Acccumulation
    2 days ago








  • 4




    $2cdot6$ is divisible by $4$ but neither $2$ nor $6$ is. Therefore $4$ is not a prime.
    – Jyrki Lahtonen
    yesterday














  • 35




    For an element to be prime this must hold for all a,b, not just one example.
    – Erik Parkinson
    2 days ago






  • 11




    You have shown that there exists $c$ such that $4|c$ and $forall a,b$ such that $ab=c, 4|a$ or $4|b$. Your definition demands this to be true for every $c$, not just 16.
    – RcnSc
    2 days ago






  • 13




    Your definition of prime is wrong. An element $p$ is prime if for all $a,bin R$, $p|abto p|atext{ or }p|b$. You left out the all-important "for all" and inserted an "and".
    – bof
    2 days ago








  • 3




    Where did you get your definition? It is valid only if we interpret it as $(a,b in R text{ and } p|ab) rightarrow (p|a text{ or } p|b)$, while you seem to be interpreting it as $a,b in R text{ and } (p|ab rightarrow (p|a text{ or } p|b))$.
    – Acccumulation
    2 days ago








  • 4




    $2cdot6$ is divisible by $4$ but neither $2$ nor $6$ is. Therefore $4$ is not a prime.
    – Jyrki Lahtonen
    yesterday








35




35




For an element to be prime this must hold for all a,b, not just one example.
– Erik Parkinson
2 days ago




For an element to be prime this must hold for all a,b, not just one example.
– Erik Parkinson
2 days ago




11




11




You have shown that there exists $c$ such that $4|c$ and $forall a,b$ such that $ab=c, 4|a$ or $4|b$. Your definition demands this to be true for every $c$, not just 16.
– RcnSc
2 days ago




You have shown that there exists $c$ such that $4|c$ and $forall a,b$ such that $ab=c, 4|a$ or $4|b$. Your definition demands this to be true for every $c$, not just 16.
– RcnSc
2 days ago




13




13




Your definition of prime is wrong. An element $p$ is prime if for all $a,bin R$, $p|abto p|atext{ or }p|b$. You left out the all-important "for all" and inserted an "and".
– bof
2 days ago






Your definition of prime is wrong. An element $p$ is prime if for all $a,bin R$, $p|abto p|atext{ or }p|b$. You left out the all-important "for all" and inserted an "and".
– bof
2 days ago






3




3




Where did you get your definition? It is valid only if we interpret it as $(a,b in R text{ and } p|ab) rightarrow (p|a text{ or } p|b)$, while you seem to be interpreting it as $a,b in R text{ and } (p|ab rightarrow (p|a text{ or } p|b))$.
– Acccumulation
2 days ago






Where did you get your definition? It is valid only if we interpret it as $(a,b in R text{ and } p|ab) rightarrow (p|a text{ or } p|b)$, while you seem to be interpreting it as $a,b in R text{ and } (p|ab rightarrow (p|a text{ or } p|b))$.
– Acccumulation
2 days ago






4




4




$2cdot6$ is divisible by $4$ but neither $2$ nor $6$ is. Therefore $4$ is not a prime.
– Jyrki Lahtonen
yesterday




$2cdot6$ is divisible by $4$ but neither $2$ nor $6$ is. Therefore $4$ is not a prime.
– Jyrki Lahtonen
yesterday










6 Answers
6






active

oldest

votes


















23














Take $mathbb Z$ and $12$.



$12$ is divided by $4$.



$12=6cdot2$, but $6$ is not divided by $4$ and $2$ is not divided by $4$, which says that $4$ is not prime.






share|cite|improve this answer































    11














    Let's take a step further from your last line:



    $4 mid 4 implies 4 mid 2cdot2$, but $4 mid 2 $ does not hold, so $4$ is not a prime. (We have $a = b = 2$, $p=4$)





    Going back to your definition:




    An element $p$ of a ring $R$ is called "prime" if $a,bin R$ and $p|abrightarrow p|a$ or $p|b$




    As was mentioned in the comments, the implied requirement is that this applies for all $a$ and $b$.



    You're trying the case of $4$ being prime, so $p=4$.
    Letting $a=2$ and $b=2$ is a counterexample,
    which shows that $4$ is not a prime.



    The definition really just says that a multiplication can't make new prime factors appear, but instead all prime factors must be present in the composition of (at least one of) the multiplied items.





    If it were enough to find just any one pair $a,b$ satisfying $p|a$ and $p|b$ to make $p$ a prime, then we could easily prove that any number is prime. It would be simplest to let $a = b = p$, but that's the squaring you already did. So let's try with e.g. $a = 12$, $b=18$, $p=6$:



    $6mid 216$ and $216 = 12cdot 18$, so since $6mid 12$ and $6mid 18$, then $6$ is prime. This is of course nonsensical.






    share|cite|improve this answer























    • I'm sorry, I don't understand. $ab$ should multiply to $16$?
      – PolymorphismPrince
      2 days ago










    • @PolymorphismPrince It should be true for all $ab$. For $ab=4$ it's not true.
      – Michael Rozenberg
      2 days ago










    • Responding to your edit: please reread my question, I did show it for every combination of $a$ & $b$ that multiply to $16$: $4$ and $4$, $2$ and $8$ and $1$ and $16$. All of them contain at least one multiple of $4$. In contrast, with your example, the factors of $216$ include $27$ and $8$, neither are divisible by $6$. That's why $6$ is composite and it doesn't contradict what I asked in my questsion.
      – PolymorphismPrince
      16 hours ago












    • @PolymorphismPrince, right, so you meant that, then. Ok, you're right, you did show that for any $a$ and $b$ where $ab =16$, it does look like $4$ is prime. But the definition you quoted doesn't mention anything about $ab$ being a particular number, so $p=4, a=b=2$ is still a counterexample that proves $4$ is composite. (or $a=2, b=6$, or $a=b=6$, etc.)
      – ilkkachu
      9 hours ago





















    2














    Going back to your example, with $p=16$.



    If I choose $a=12$ and $b=4$, then I have $p | ab$, but neither $p|a$ nor $p|b$. Hence, $p$ is not prime.



    The definition requires that you consider any $a, b$ such that $p|ab$, not only those such that $p=ab$.






    share|cite|improve this answer








    New contributor




    user2233709 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.


























      2














      This is an answer to the accepted answer more than to the question, but I believe that that is ok, since both are posed by the same poster.



      NOTE: I CHANGED SOME THINGS AFTER READING DARIJ'S COMMENT



      The question seems to be which of four definitions of prime element is the 'correct' one. I restate the definitions here, but let them define four different types of elements (prime1, prime1', prime2, prime3). The reason for this somewhat awkward notation is that it enables us to say things like 'The element $4 in mathbb{Z}$ is prime2 but not prime1' and know exactly what we are talking about.



      So here are the definitions.




      An element $p$ in a ring $R$ is called prime1 if it is not a unit and $forall a, b in R$ such that $p|ab$, we have that $p|a$ or $p|b$. This is the definition used in most of the other answers.



      An element $p$ in a ring $R$ is called prime1' if it is not a unit and $forall c in R$ such that $p|c$, and for all $a, b in R$ such that $ab = c$, we have that $p|a$ or $p|b$. This definition appears in the accepted answer, although at that point the OP seems to have stopped believing that it is the correct one.




      I use the names prime1 and prime1' to emphasis that these definitions are really identical. Just try to think for 2 seconds what you would need to do to show that an element $p$ is not prime1 or not prime1'. In both cases it amounts to the same thing: find two elements ($a, b$) that are not divisible by $p$ while their product ($c$) is.



      Things get more interesting when we invoke the other two definitions.




      An element $p$ in a ring $R$ is called prime2 if it is not a unit and $exists c in R$ such that $p|c$ and $forall a, b in R$ such that $ab = c$ we have that $p|a$ or $p|c$. This definition is nowhere written out explicitly but it is very strongly implicitly present in the original question.



      An element $p$ in a ring $R$ is called prime3 if it is not a unit and $forall c in R$ such that $p|c$, and for all finite $S subset R$ such that $prod_{s in S} = c$, there is an $s in S$ such that $p|s$. This definition is proposed by the OP in the accepted answer as the "correct" one.




      Now the content of the original question is that the number $4 in mathbb{Z}$ is prime2, showing that 'prime2-ness' is not a good way of generalizing 'primeness' (primality) of integers to arbitrary rings. Very nice, I never thought of that before. However, for now that means that prime2 is out as a candidate definition of "prime" and the race is between prime1 and prime3. (I ignore prime1' for the moment as it is identical to prime1).



      The counterexample the OP is asking for seems to be a natural number that is prime3 but not prime in the ordinary sense. In fact it would already be interesting to see any element of any integral domain that is either prime3 but not prime1 or prime1 but not prime3. The point of my answer is: these elements do not exist. Concretely we have:




      Theorem 1: let $R$ be an integral domain. Then every element that is prime3 is also prime1



      Theorem 2: let $R$ be an integral domain. Then every element that is prime1 is also prime3.




      Together the theorems say that the two definitions are equivalent and hence that choosing the one over the other is a matter of taste (or pedagogy) and doesn't change anything mathematically.



      I will prove the theorems below.



      Proving Theorem 1 is really easy, there is hardly anything to do. We have an element $p$ that is prime3 and elements $a, b$ such that $p|ab$. Taking $c = ab$ and $S = {a, b}$ we see that from prime3-ness of $p$ that $p|a$ or $p|b$ which in turn implies that $p$ is prime1. End of proof.



      Proving Theorem 2 is a bit more involved.



      Let $p$ be a prime1 element of $R$. Let $c$ be any element of $R$ such that $p|c$ and let $S subset R$ be any finite set such that $prod_{sin S} s = c$. We need to show that there is an element $s in S$ such that $p|s$ in order to show that $p$ is prime3. We do this by induction on the number $n$ of elements of $S$.



      The first hurdle we have to tackle is the fact that if $n = 0$ then our goal is unreachable: clearly in an empty set there is no element $s$ such that $p|s$. Luckily this case does not occur! The point is: the product of all elements of an empty set is 1 by definition. But if $c = 1$ and $p|c$ then $p$ is a unit, which contradicts its prime1-ness. So this case cannot happen.



      We move on to the case $n = 1$. If $S$ contains only a single element then clearly this single element must be the element $c$, which was the product of all elements. But we already know that $p|c$ and so, indeed, $p$ divides a (or rather: the) element of $S$.



      Now let $n geq 2$ and we suppose that ('Induction Hypothesis') that for all $a in R$ with $p|a$ and all $S' subset R$ with $n-1$ elements such that $prod_{s in S'} s = a$ we have that $p$ divides some element of $S'$.



      We are still working with the $n$-element set $S$ and we label the the elements $s_1, s_2, ldots, s_{n-1}, s_n$. Let $a = s_1 cdot s_2 cdots s_{n-1}$ and $b = s_n$. Then $ab = c$ and hence $p|ab$ and hence, since $p$ is prime1 we have that $p|a$ or $p|b$. In the second case we see that $p|s_n$ and hence we see that $p$ divides an element of $S$ as we wanted to show. In the first case we set $S' = {s_1, ldots, s_{n-1}}$. Since $p|a$ we know from the Induction Hypothesis that $p$ divides some element of $S'$. But every element of $S'$ is also an element of $S$, so in this case we have that $p$ divides an element of $S$ as well, as we wanted to show.



      End of proof.



      In summary: the new definition from your answer and the old definition from the other answers are equivalent and single out the same elements of $R$.






      share|cite|improve this answer























      • I don't get why $4$ is prime2. Set $a=2$, $b=2$ and $c=4$. Did you misstate the definition or did I misread it?
        – darij grinberg
        2 hours ago










      • No you are right, 4 is not prime2 as stated... I have to think about how to edit my post. So the point of the original question was to show that 4 is prime4 defined as $p$ is prime 4 if there exist a $c$ such that $p|c$ and for all $a, b$ such that $ab = c$ we have that $p|a$ or $p|c$. However the reason I defined prime2 as I did and not as prime4 above was that the definition of prime2 appears in the answer by the OP that inspired my answer. So now I am a bit torn what to do. The best is probably to include both prime2 and prime4 into the post and say something about both...
        – Vincent
        2 hours ago










      • Ok, what do you think of this version? For future readers: the old prime2 darij is refering to is the new prime1' and I edited in (what in above comment is) prime4 as prime2 to keep the flow of the original answer.
        – Vincent
        1 hour ago










      • The new version is fine (though "prime1'" looks really strange with quotation marks).
        – darij grinberg
        1 hour ago












      • because of the prime (pun not intended) followed by quotes? Hmm yes. Ok, while I'm at it I might as well edit out all the quotes and replace them with italics...
        – Vincent
        1 hour ago



















      0














      Here's how I would reword it:




      An element $p$ of a ring $R$ is called prime if for any combination of $a, b in R$ such that $p mid ab$, at least one of $p mid a$ or $p mid b$ also holds true.




      So if $p = a = b = 4$ we see that $p mid ab$ and then $p mid a$ and $p mid b$ are trivially true. But that's not the only possible combination of $a$ and $b$ that we can choose so that $p mid ab$. You have already discovered two more. But there are in fact infinitely many more.



      It suffices to choose one in which $a = 2$ and $b neq pm 2$ though it is singly even (that is, not divisible by $4$). So, for example, $b = 14$, then we have that $4 mid 28$ and $28 = 2 times 14$, but $2$ is not divisible by $4$ and neither is $14$. Hence $4$ is not prime.



      I'm sure you can generalize this to the squares of odd primes. Just my two cents.






      share|cite|improve this answer





















      • Please read mine or @Vincent's answer as to why you can use just one multiple of $p$ ($ab$) as long as you check all the factors.
        – PolymorphismPrince
        29 mins ago










      • I might or I might not, I'm writing an answer to a more difficult question and that might not leave enough time before my social obligations tonight. In the meantime, I suggest that you accept Vincent's answer instead of your own. I would appreciate an upvote for you, but my answer is really more for the benefit of others who come across your question.
        – David R.
        23 mins ago



















      -2














      Please read this answer before taking any of the others as fact (I'm OP)



      Short version



      I believe the correct definition is




      An element $p$ in a ring $R$ is called "prime" if for some $cin R$ such $p|c$, $forall Ssubset R$ such the product of elements in $S$ is $c$, $exists nin S$ such $p|n$.




      The only real difference is that we can make factorisations of the multiple of $p$ that are more than just two coefficients.



      Under this definition, $4$ can be composite since there exists one $S$ that multiplies to $16$ which is $left {2,2,2 right }$ and $4|2$ is false.



      Long version explaining why I believe other answers are wrong



      I'd like to firstly acknowledge that I'm slightly disappointed that countless people have copped countless amounts of karma for saying the following:




      The definition should be interpreted as




      An element $p$ in a ring $R$ is called "prime" if $forall a,bin R$ such $p|ab$, $p|a$ or $p|b$.




      Rather than (the way OP interpreted it)




      An element $p$ in a ring $R$ is called "prime" if $forall cin R$ such $p|c$, $forall a,bin R$ such $ab = c$, $p|a$ or $p|b$.





      In the edit of my original question, I said that this answer didn't make sense to me intuitively, I suggested that a correct definition would have been:




      An element $p$ in a ring $R$ is called "prime" if for some $cin R$ such $p|c$, $forall Ssubset R$ such the product of elements in $S$ is $c$, $exists nin S$ such $n|$.




      I asked for a counter-example, meaning, a number that couldn't be correctly proved prime or composite using this definition. At the time of writing, I'm yet to receive one.



      I am now convinced in the legitimacy of my definition for the following reason:



      The standard way of proving if an integer, $n$, is prime or composite is by checking whether it is divisible by any number from $2$ to $sqrt{n}$.



      If we assume my definition is correct we can check if $n$ is prime in the following way:




      1. Make $c = n$

      2. See if $c$ has any factors that are not divisible by $n$. All of them will be since $c = n$ and a positive integer will never be divisible by an integer less than it.

      3. If we find one, $n$ is not prime.


      This process is exactly the same as the standard process I previously described. However, if you take other people's definitions I have to check the factors of every possible $nk$ where $k in mathbb{Z}$. That's something we never have to do.



      I only showed this works in the set of integers. I'm unsure of non-integral domains. But it was the set of integers my question regarded.






      share|cite|improve this answer























      • A minor edit: instead of '$p|a$ or $p|b$ you need to write something like '$p|s$ of one of the elements $s in S$ since you explicitly keep open the possibility of $S$ having more than one element (also as written it is not clear that $a$ and $b$ are elements of $S$.
        – Vincent
        4 hours ago










      • Anyway, see my answer below. I hope this is what you were looking for.
        – Vincent
        2 hours ago










      • @Vincent Thanks, editing now.
        – PolymorphismPrince
        45 mins ago










      • I downvoted for the following reasons. (1) Answers are not obligated to update themselves in response to your edits (it might be convenient if they were, but it'd be thoroughly impractical), and the fact that the answers don't address your edit therefore doesn't make them wrong. They may have been unhelpful, and you are not obligated to accept them, however, that doesn't make them wrong. (2) The definition you gave in your question originally also isn't wrong. It may have confused you, but again, that doesn't make it incorrect. (cont)
        – jgon
        18 mins ago










      • (3) Your definition is incorrect. It implies that all irreducibles are prime, which is false. It only works in a UFD, which makes it unhelpful for defining primes. (If $pi$ is irreducible, then taking $c=pi$, any product of elements of the ring that ends up being $pi$ will have all except one factor units, and $pi$ divides the remaining factor.) Your definition would work if we changed $exists c$ to $forall c$.
        – jgon
        15 mins ago













      Your Answer





      StackExchange.ifUsing("editor", function () {
      return StackExchange.using("mathjaxEditing", function () {
      StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
      StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
      });
      });
      }, "mathjax-editing");

      StackExchange.ready(function() {
      var channelOptions = {
      tags: "".split(" "),
      id: "69"
      };
      initTagRenderer("".split(" "), "".split(" "), channelOptions);

      StackExchange.using("externalEditor", function() {
      // Have to fire editor after snippets, if snippets enabled
      if (StackExchange.settings.snippets.snippetsEnabled) {
      StackExchange.using("snippets", function() {
      createEditor();
      });
      }
      else {
      createEditor();
      }
      });

      function createEditor() {
      StackExchange.prepareEditor({
      heartbeatType: 'answer',
      autoActivateHeartbeat: false,
      convertImagesToLinks: true,
      noModals: true,
      showLowRepImageUploadWarning: true,
      reputationToPostImages: 10,
      bindNavPrevention: true,
      postfix: "",
      imageUploader: {
      brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
      contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
      allowUrls: true
      },
      noCode: true, onDemand: true,
      discardSelector: ".discard-answer"
      ,immediatelyShowMarkdownHelp:true
      });


      }
      });














      draft saved

      draft discarded


















      StackExchange.ready(
      function () {
      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059225%2fwhy-doesnt-the-definition-p-is-called-prime-if-p-mid-ab-implies-p-mid-a%23new-answer', 'question_page');
      }
      );

      Post as a guest















      Required, but never shown

























      6 Answers
      6






      active

      oldest

      votes








      6 Answers
      6






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      23














      Take $mathbb Z$ and $12$.



      $12$ is divided by $4$.



      $12=6cdot2$, but $6$ is not divided by $4$ and $2$ is not divided by $4$, which says that $4$ is not prime.






      share|cite|improve this answer




























        23














        Take $mathbb Z$ and $12$.



        $12$ is divided by $4$.



        $12=6cdot2$, but $6$ is not divided by $4$ and $2$ is not divided by $4$, which says that $4$ is not prime.






        share|cite|improve this answer


























          23












          23








          23






          Take $mathbb Z$ and $12$.



          $12$ is divided by $4$.



          $12=6cdot2$, but $6$ is not divided by $4$ and $2$ is not divided by $4$, which says that $4$ is not prime.






          share|cite|improve this answer














          Take $mathbb Z$ and $12$.



          $12$ is divided by $4$.



          $12=6cdot2$, but $6$ is not divided by $4$ and $2$ is not divided by $4$, which says that $4$ is not prime.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 days ago

























          answered 2 days ago









          Michael Rozenberg

          96.8k1589188




          96.8k1589188























              11














              Let's take a step further from your last line:



              $4 mid 4 implies 4 mid 2cdot2$, but $4 mid 2 $ does not hold, so $4$ is not a prime. (We have $a = b = 2$, $p=4$)





              Going back to your definition:




              An element $p$ of a ring $R$ is called "prime" if $a,bin R$ and $p|abrightarrow p|a$ or $p|b$




              As was mentioned in the comments, the implied requirement is that this applies for all $a$ and $b$.



              You're trying the case of $4$ being prime, so $p=4$.
              Letting $a=2$ and $b=2$ is a counterexample,
              which shows that $4$ is not a prime.



              The definition really just says that a multiplication can't make new prime factors appear, but instead all prime factors must be present in the composition of (at least one of) the multiplied items.





              If it were enough to find just any one pair $a,b$ satisfying $p|a$ and $p|b$ to make $p$ a prime, then we could easily prove that any number is prime. It would be simplest to let $a = b = p$, but that's the squaring you already did. So let's try with e.g. $a = 12$, $b=18$, $p=6$:



              $6mid 216$ and $216 = 12cdot 18$, so since $6mid 12$ and $6mid 18$, then $6$ is prime. This is of course nonsensical.






              share|cite|improve this answer























              • I'm sorry, I don't understand. $ab$ should multiply to $16$?
                – PolymorphismPrince
                2 days ago










              • @PolymorphismPrince It should be true for all $ab$. For $ab=4$ it's not true.
                – Michael Rozenberg
                2 days ago










              • Responding to your edit: please reread my question, I did show it for every combination of $a$ & $b$ that multiply to $16$: $4$ and $4$, $2$ and $8$ and $1$ and $16$. All of them contain at least one multiple of $4$. In contrast, with your example, the factors of $216$ include $27$ and $8$, neither are divisible by $6$. That's why $6$ is composite and it doesn't contradict what I asked in my questsion.
                – PolymorphismPrince
                16 hours ago












              • @PolymorphismPrince, right, so you meant that, then. Ok, you're right, you did show that for any $a$ and $b$ where $ab =16$, it does look like $4$ is prime. But the definition you quoted doesn't mention anything about $ab$ being a particular number, so $p=4, a=b=2$ is still a counterexample that proves $4$ is composite. (or $a=2, b=6$, or $a=b=6$, etc.)
                – ilkkachu
                9 hours ago


















              11














              Let's take a step further from your last line:



              $4 mid 4 implies 4 mid 2cdot2$, but $4 mid 2 $ does not hold, so $4$ is not a prime. (We have $a = b = 2$, $p=4$)





              Going back to your definition:




              An element $p$ of a ring $R$ is called "prime" if $a,bin R$ and $p|abrightarrow p|a$ or $p|b$




              As was mentioned in the comments, the implied requirement is that this applies for all $a$ and $b$.



              You're trying the case of $4$ being prime, so $p=4$.
              Letting $a=2$ and $b=2$ is a counterexample,
              which shows that $4$ is not a prime.



              The definition really just says that a multiplication can't make new prime factors appear, but instead all prime factors must be present in the composition of (at least one of) the multiplied items.





              If it were enough to find just any one pair $a,b$ satisfying $p|a$ and $p|b$ to make $p$ a prime, then we could easily prove that any number is prime. It would be simplest to let $a = b = p$, but that's the squaring you already did. So let's try with e.g. $a = 12$, $b=18$, $p=6$:



              $6mid 216$ and $216 = 12cdot 18$, so since $6mid 12$ and $6mid 18$, then $6$ is prime. This is of course nonsensical.






              share|cite|improve this answer























              • I'm sorry, I don't understand. $ab$ should multiply to $16$?
                – PolymorphismPrince
                2 days ago










              • @PolymorphismPrince It should be true for all $ab$. For $ab=4$ it's not true.
                – Michael Rozenberg
                2 days ago










              • Responding to your edit: please reread my question, I did show it for every combination of $a$ & $b$ that multiply to $16$: $4$ and $4$, $2$ and $8$ and $1$ and $16$. All of them contain at least one multiple of $4$. In contrast, with your example, the factors of $216$ include $27$ and $8$, neither are divisible by $6$. That's why $6$ is composite and it doesn't contradict what I asked in my questsion.
                – PolymorphismPrince
                16 hours ago












              • @PolymorphismPrince, right, so you meant that, then. Ok, you're right, you did show that for any $a$ and $b$ where $ab =16$, it does look like $4$ is prime. But the definition you quoted doesn't mention anything about $ab$ being a particular number, so $p=4, a=b=2$ is still a counterexample that proves $4$ is composite. (or $a=2, b=6$, or $a=b=6$, etc.)
                – ilkkachu
                9 hours ago
















              11












              11








              11






              Let's take a step further from your last line:



              $4 mid 4 implies 4 mid 2cdot2$, but $4 mid 2 $ does not hold, so $4$ is not a prime. (We have $a = b = 2$, $p=4$)





              Going back to your definition:




              An element $p$ of a ring $R$ is called "prime" if $a,bin R$ and $p|abrightarrow p|a$ or $p|b$




              As was mentioned in the comments, the implied requirement is that this applies for all $a$ and $b$.



              You're trying the case of $4$ being prime, so $p=4$.
              Letting $a=2$ and $b=2$ is a counterexample,
              which shows that $4$ is not a prime.



              The definition really just says that a multiplication can't make new prime factors appear, but instead all prime factors must be present in the composition of (at least one of) the multiplied items.





              If it were enough to find just any one pair $a,b$ satisfying $p|a$ and $p|b$ to make $p$ a prime, then we could easily prove that any number is prime. It would be simplest to let $a = b = p$, but that's the squaring you already did. So let's try with e.g. $a = 12$, $b=18$, $p=6$:



              $6mid 216$ and $216 = 12cdot 18$, so since $6mid 12$ and $6mid 18$, then $6$ is prime. This is of course nonsensical.






              share|cite|improve this answer














              Let's take a step further from your last line:



              $4 mid 4 implies 4 mid 2cdot2$, but $4 mid 2 $ does not hold, so $4$ is not a prime. (We have $a = b = 2$, $p=4$)





              Going back to your definition:




              An element $p$ of a ring $R$ is called "prime" if $a,bin R$ and $p|abrightarrow p|a$ or $p|b$




              As was mentioned in the comments, the implied requirement is that this applies for all $a$ and $b$.



              You're trying the case of $4$ being prime, so $p=4$.
              Letting $a=2$ and $b=2$ is a counterexample,
              which shows that $4$ is not a prime.



              The definition really just says that a multiplication can't make new prime factors appear, but instead all prime factors must be present in the composition of (at least one of) the multiplied items.





              If it were enough to find just any one pair $a,b$ satisfying $p|a$ and $p|b$ to make $p$ a prime, then we could easily prove that any number is prime. It would be simplest to let $a = b = p$, but that's the squaring you already did. So let's try with e.g. $a = 12$, $b=18$, $p=6$:



              $6mid 216$ and $216 = 12cdot 18$, so since $6mid 12$ and $6mid 18$, then $6$ is prime. This is of course nonsensical.







              share|cite|improve this answer














              share|cite|improve this answer



              share|cite|improve this answer








              edited yesterday

























              answered 2 days ago









              ilkkachu

              50139




              50139












              • I'm sorry, I don't understand. $ab$ should multiply to $16$?
                – PolymorphismPrince
                2 days ago










              • @PolymorphismPrince It should be true for all $ab$. For $ab=4$ it's not true.
                – Michael Rozenberg
                2 days ago










              • Responding to your edit: please reread my question, I did show it for every combination of $a$ & $b$ that multiply to $16$: $4$ and $4$, $2$ and $8$ and $1$ and $16$. All of them contain at least one multiple of $4$. In contrast, with your example, the factors of $216$ include $27$ and $8$, neither are divisible by $6$. That's why $6$ is composite and it doesn't contradict what I asked in my questsion.
                – PolymorphismPrince
                16 hours ago












              • @PolymorphismPrince, right, so you meant that, then. Ok, you're right, you did show that for any $a$ and $b$ where $ab =16$, it does look like $4$ is prime. But the definition you quoted doesn't mention anything about $ab$ being a particular number, so $p=4, a=b=2$ is still a counterexample that proves $4$ is composite. (or $a=2, b=6$, or $a=b=6$, etc.)
                – ilkkachu
                9 hours ago




















              • I'm sorry, I don't understand. $ab$ should multiply to $16$?
                – PolymorphismPrince
                2 days ago










              • @PolymorphismPrince It should be true for all $ab$. For $ab=4$ it's not true.
                – Michael Rozenberg
                2 days ago










              • Responding to your edit: please reread my question, I did show it for every combination of $a$ & $b$ that multiply to $16$: $4$ and $4$, $2$ and $8$ and $1$ and $16$. All of them contain at least one multiple of $4$. In contrast, with your example, the factors of $216$ include $27$ and $8$, neither are divisible by $6$. That's why $6$ is composite and it doesn't contradict what I asked in my questsion.
                – PolymorphismPrince
                16 hours ago












              • @PolymorphismPrince, right, so you meant that, then. Ok, you're right, you did show that for any $a$ and $b$ where $ab =16$, it does look like $4$ is prime. But the definition you quoted doesn't mention anything about $ab$ being a particular number, so $p=4, a=b=2$ is still a counterexample that proves $4$ is composite. (or $a=2, b=6$, or $a=b=6$, etc.)
                – ilkkachu
                9 hours ago


















              I'm sorry, I don't understand. $ab$ should multiply to $16$?
              – PolymorphismPrince
              2 days ago




              I'm sorry, I don't understand. $ab$ should multiply to $16$?
              – PolymorphismPrince
              2 days ago












              @PolymorphismPrince It should be true for all $ab$. For $ab=4$ it's not true.
              – Michael Rozenberg
              2 days ago




              @PolymorphismPrince It should be true for all $ab$. For $ab=4$ it's not true.
              – Michael Rozenberg
              2 days ago












              Responding to your edit: please reread my question, I did show it for every combination of $a$ & $b$ that multiply to $16$: $4$ and $4$, $2$ and $8$ and $1$ and $16$. All of them contain at least one multiple of $4$. In contrast, with your example, the factors of $216$ include $27$ and $8$, neither are divisible by $6$. That's why $6$ is composite and it doesn't contradict what I asked in my questsion.
              – PolymorphismPrince
              16 hours ago






              Responding to your edit: please reread my question, I did show it for every combination of $a$ & $b$ that multiply to $16$: $4$ and $4$, $2$ and $8$ and $1$ and $16$. All of them contain at least one multiple of $4$. In contrast, with your example, the factors of $216$ include $27$ and $8$, neither are divisible by $6$. That's why $6$ is composite and it doesn't contradict what I asked in my questsion.
              – PolymorphismPrince
              16 hours ago














              @PolymorphismPrince, right, so you meant that, then. Ok, you're right, you did show that for any $a$ and $b$ where $ab =16$, it does look like $4$ is prime. But the definition you quoted doesn't mention anything about $ab$ being a particular number, so $p=4, a=b=2$ is still a counterexample that proves $4$ is composite. (or $a=2, b=6$, or $a=b=6$, etc.)
              – ilkkachu
              9 hours ago






              @PolymorphismPrince, right, so you meant that, then. Ok, you're right, you did show that for any $a$ and $b$ where $ab =16$, it does look like $4$ is prime. But the definition you quoted doesn't mention anything about $ab$ being a particular number, so $p=4, a=b=2$ is still a counterexample that proves $4$ is composite. (or $a=2, b=6$, or $a=b=6$, etc.)
              – ilkkachu
              9 hours ago













              2














              Going back to your example, with $p=16$.



              If I choose $a=12$ and $b=4$, then I have $p | ab$, but neither $p|a$ nor $p|b$. Hence, $p$ is not prime.



              The definition requires that you consider any $a, b$ such that $p|ab$, not only those such that $p=ab$.






              share|cite|improve this answer








              New contributor




              user2233709 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
              Check out our Code of Conduct.























                2














                Going back to your example, with $p=16$.



                If I choose $a=12$ and $b=4$, then I have $p | ab$, but neither $p|a$ nor $p|b$. Hence, $p$ is not prime.



                The definition requires that you consider any $a, b$ such that $p|ab$, not only those such that $p=ab$.






                share|cite|improve this answer








                New contributor




                user2233709 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                Check out our Code of Conduct.





















                  2












                  2








                  2






                  Going back to your example, with $p=16$.



                  If I choose $a=12$ and $b=4$, then I have $p | ab$, but neither $p|a$ nor $p|b$. Hence, $p$ is not prime.



                  The definition requires that you consider any $a, b$ such that $p|ab$, not only those such that $p=ab$.






                  share|cite|improve this answer








                  New contributor




                  user2233709 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  Going back to your example, with $p=16$.



                  If I choose $a=12$ and $b=4$, then I have $p | ab$, but neither $p|a$ nor $p|b$. Hence, $p$ is not prime.



                  The definition requires that you consider any $a, b$ such that $p|ab$, not only those such that $p=ab$.







                  share|cite|improve this answer








                  New contributor




                  user2233709 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  share|cite|improve this answer



                  share|cite|improve this answer






                  New contributor




                  user2233709 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.









                  answered yesterday









                  user2233709

                  1212




                  1212




                  New contributor




                  user2233709 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.





                  New contributor





                  user2233709 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.






                  user2233709 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
                  Check out our Code of Conduct.























                      2














                      This is an answer to the accepted answer more than to the question, but I believe that that is ok, since both are posed by the same poster.



                      NOTE: I CHANGED SOME THINGS AFTER READING DARIJ'S COMMENT



                      The question seems to be which of four definitions of prime element is the 'correct' one. I restate the definitions here, but let them define four different types of elements (prime1, prime1', prime2, prime3). The reason for this somewhat awkward notation is that it enables us to say things like 'The element $4 in mathbb{Z}$ is prime2 but not prime1' and know exactly what we are talking about.



                      So here are the definitions.




                      An element $p$ in a ring $R$ is called prime1 if it is not a unit and $forall a, b in R$ such that $p|ab$, we have that $p|a$ or $p|b$. This is the definition used in most of the other answers.



                      An element $p$ in a ring $R$ is called prime1' if it is not a unit and $forall c in R$ such that $p|c$, and for all $a, b in R$ such that $ab = c$, we have that $p|a$ or $p|b$. This definition appears in the accepted answer, although at that point the OP seems to have stopped believing that it is the correct one.




                      I use the names prime1 and prime1' to emphasis that these definitions are really identical. Just try to think for 2 seconds what you would need to do to show that an element $p$ is not prime1 or not prime1'. In both cases it amounts to the same thing: find two elements ($a, b$) that are not divisible by $p$ while their product ($c$) is.



                      Things get more interesting when we invoke the other two definitions.




                      An element $p$ in a ring $R$ is called prime2 if it is not a unit and $exists c in R$ such that $p|c$ and $forall a, b in R$ such that $ab = c$ we have that $p|a$ or $p|c$. This definition is nowhere written out explicitly but it is very strongly implicitly present in the original question.



                      An element $p$ in a ring $R$ is called prime3 if it is not a unit and $forall c in R$ such that $p|c$, and for all finite $S subset R$ such that $prod_{s in S} = c$, there is an $s in S$ such that $p|s$. This definition is proposed by the OP in the accepted answer as the "correct" one.




                      Now the content of the original question is that the number $4 in mathbb{Z}$ is prime2, showing that 'prime2-ness' is not a good way of generalizing 'primeness' (primality) of integers to arbitrary rings. Very nice, I never thought of that before. However, for now that means that prime2 is out as a candidate definition of "prime" and the race is between prime1 and prime3. (I ignore prime1' for the moment as it is identical to prime1).



                      The counterexample the OP is asking for seems to be a natural number that is prime3 but not prime in the ordinary sense. In fact it would already be interesting to see any element of any integral domain that is either prime3 but not prime1 or prime1 but not prime3. The point of my answer is: these elements do not exist. Concretely we have:




                      Theorem 1: let $R$ be an integral domain. Then every element that is prime3 is also prime1



                      Theorem 2: let $R$ be an integral domain. Then every element that is prime1 is also prime3.




                      Together the theorems say that the two definitions are equivalent and hence that choosing the one over the other is a matter of taste (or pedagogy) and doesn't change anything mathematically.



                      I will prove the theorems below.



                      Proving Theorem 1 is really easy, there is hardly anything to do. We have an element $p$ that is prime3 and elements $a, b$ such that $p|ab$. Taking $c = ab$ and $S = {a, b}$ we see that from prime3-ness of $p$ that $p|a$ or $p|b$ which in turn implies that $p$ is prime1. End of proof.



                      Proving Theorem 2 is a bit more involved.



                      Let $p$ be a prime1 element of $R$. Let $c$ be any element of $R$ such that $p|c$ and let $S subset R$ be any finite set such that $prod_{sin S} s = c$. We need to show that there is an element $s in S$ such that $p|s$ in order to show that $p$ is prime3. We do this by induction on the number $n$ of elements of $S$.



                      The first hurdle we have to tackle is the fact that if $n = 0$ then our goal is unreachable: clearly in an empty set there is no element $s$ such that $p|s$. Luckily this case does not occur! The point is: the product of all elements of an empty set is 1 by definition. But if $c = 1$ and $p|c$ then $p$ is a unit, which contradicts its prime1-ness. So this case cannot happen.



                      We move on to the case $n = 1$. If $S$ contains only a single element then clearly this single element must be the element $c$, which was the product of all elements. But we already know that $p|c$ and so, indeed, $p$ divides a (or rather: the) element of $S$.



                      Now let $n geq 2$ and we suppose that ('Induction Hypothesis') that for all $a in R$ with $p|a$ and all $S' subset R$ with $n-1$ elements such that $prod_{s in S'} s = a$ we have that $p$ divides some element of $S'$.



                      We are still working with the $n$-element set $S$ and we label the the elements $s_1, s_2, ldots, s_{n-1}, s_n$. Let $a = s_1 cdot s_2 cdots s_{n-1}$ and $b = s_n$. Then $ab = c$ and hence $p|ab$ and hence, since $p$ is prime1 we have that $p|a$ or $p|b$. In the second case we see that $p|s_n$ and hence we see that $p$ divides an element of $S$ as we wanted to show. In the first case we set $S' = {s_1, ldots, s_{n-1}}$. Since $p|a$ we know from the Induction Hypothesis that $p$ divides some element of $S'$. But every element of $S'$ is also an element of $S$, so in this case we have that $p$ divides an element of $S$ as well, as we wanted to show.



                      End of proof.



                      In summary: the new definition from your answer and the old definition from the other answers are equivalent and single out the same elements of $R$.






                      share|cite|improve this answer























                      • I don't get why $4$ is prime2. Set $a=2$, $b=2$ and $c=4$. Did you misstate the definition or did I misread it?
                        – darij grinberg
                        2 hours ago










                      • No you are right, 4 is not prime2 as stated... I have to think about how to edit my post. So the point of the original question was to show that 4 is prime4 defined as $p$ is prime 4 if there exist a $c$ such that $p|c$ and for all $a, b$ such that $ab = c$ we have that $p|a$ or $p|c$. However the reason I defined prime2 as I did and not as prime4 above was that the definition of prime2 appears in the answer by the OP that inspired my answer. So now I am a bit torn what to do. The best is probably to include both prime2 and prime4 into the post and say something about both...
                        – Vincent
                        2 hours ago










                      • Ok, what do you think of this version? For future readers: the old prime2 darij is refering to is the new prime1' and I edited in (what in above comment is) prime4 as prime2 to keep the flow of the original answer.
                        – Vincent
                        1 hour ago










                      • The new version is fine (though "prime1'" looks really strange with quotation marks).
                        – darij grinberg
                        1 hour ago












                      • because of the prime (pun not intended) followed by quotes? Hmm yes. Ok, while I'm at it I might as well edit out all the quotes and replace them with italics...
                        – Vincent
                        1 hour ago
















                      2














                      This is an answer to the accepted answer more than to the question, but I believe that that is ok, since both are posed by the same poster.



                      NOTE: I CHANGED SOME THINGS AFTER READING DARIJ'S COMMENT



                      The question seems to be which of four definitions of prime element is the 'correct' one. I restate the definitions here, but let them define four different types of elements (prime1, prime1', prime2, prime3). The reason for this somewhat awkward notation is that it enables us to say things like 'The element $4 in mathbb{Z}$ is prime2 but not prime1' and know exactly what we are talking about.



                      So here are the definitions.




                      An element $p$ in a ring $R$ is called prime1 if it is not a unit and $forall a, b in R$ such that $p|ab$, we have that $p|a$ or $p|b$. This is the definition used in most of the other answers.



                      An element $p$ in a ring $R$ is called prime1' if it is not a unit and $forall c in R$ such that $p|c$, and for all $a, b in R$ such that $ab = c$, we have that $p|a$ or $p|b$. This definition appears in the accepted answer, although at that point the OP seems to have stopped believing that it is the correct one.




                      I use the names prime1 and prime1' to emphasis that these definitions are really identical. Just try to think for 2 seconds what you would need to do to show that an element $p$ is not prime1 or not prime1'. In both cases it amounts to the same thing: find two elements ($a, b$) that are not divisible by $p$ while their product ($c$) is.



                      Things get more interesting when we invoke the other two definitions.




                      An element $p$ in a ring $R$ is called prime2 if it is not a unit and $exists c in R$ such that $p|c$ and $forall a, b in R$ such that $ab = c$ we have that $p|a$ or $p|c$. This definition is nowhere written out explicitly but it is very strongly implicitly present in the original question.



                      An element $p$ in a ring $R$ is called prime3 if it is not a unit and $forall c in R$ such that $p|c$, and for all finite $S subset R$ such that $prod_{s in S} = c$, there is an $s in S$ such that $p|s$. This definition is proposed by the OP in the accepted answer as the "correct" one.




                      Now the content of the original question is that the number $4 in mathbb{Z}$ is prime2, showing that 'prime2-ness' is not a good way of generalizing 'primeness' (primality) of integers to arbitrary rings. Very nice, I never thought of that before. However, for now that means that prime2 is out as a candidate definition of "prime" and the race is between prime1 and prime3. (I ignore prime1' for the moment as it is identical to prime1).



                      The counterexample the OP is asking for seems to be a natural number that is prime3 but not prime in the ordinary sense. In fact it would already be interesting to see any element of any integral domain that is either prime3 but not prime1 or prime1 but not prime3. The point of my answer is: these elements do not exist. Concretely we have:




                      Theorem 1: let $R$ be an integral domain. Then every element that is prime3 is also prime1



                      Theorem 2: let $R$ be an integral domain. Then every element that is prime1 is also prime3.




                      Together the theorems say that the two definitions are equivalent and hence that choosing the one over the other is a matter of taste (or pedagogy) and doesn't change anything mathematically.



                      I will prove the theorems below.



                      Proving Theorem 1 is really easy, there is hardly anything to do. We have an element $p$ that is prime3 and elements $a, b$ such that $p|ab$. Taking $c = ab$ and $S = {a, b}$ we see that from prime3-ness of $p$ that $p|a$ or $p|b$ which in turn implies that $p$ is prime1. End of proof.



                      Proving Theorem 2 is a bit more involved.



                      Let $p$ be a prime1 element of $R$. Let $c$ be any element of $R$ such that $p|c$ and let $S subset R$ be any finite set such that $prod_{sin S} s = c$. We need to show that there is an element $s in S$ such that $p|s$ in order to show that $p$ is prime3. We do this by induction on the number $n$ of elements of $S$.



                      The first hurdle we have to tackle is the fact that if $n = 0$ then our goal is unreachable: clearly in an empty set there is no element $s$ such that $p|s$. Luckily this case does not occur! The point is: the product of all elements of an empty set is 1 by definition. But if $c = 1$ and $p|c$ then $p$ is a unit, which contradicts its prime1-ness. So this case cannot happen.



                      We move on to the case $n = 1$. If $S$ contains only a single element then clearly this single element must be the element $c$, which was the product of all elements. But we already know that $p|c$ and so, indeed, $p$ divides a (or rather: the) element of $S$.



                      Now let $n geq 2$ and we suppose that ('Induction Hypothesis') that for all $a in R$ with $p|a$ and all $S' subset R$ with $n-1$ elements such that $prod_{s in S'} s = a$ we have that $p$ divides some element of $S'$.



                      We are still working with the $n$-element set $S$ and we label the the elements $s_1, s_2, ldots, s_{n-1}, s_n$. Let $a = s_1 cdot s_2 cdots s_{n-1}$ and $b = s_n$. Then $ab = c$ and hence $p|ab$ and hence, since $p$ is prime1 we have that $p|a$ or $p|b$. In the second case we see that $p|s_n$ and hence we see that $p$ divides an element of $S$ as we wanted to show. In the first case we set $S' = {s_1, ldots, s_{n-1}}$. Since $p|a$ we know from the Induction Hypothesis that $p$ divides some element of $S'$. But every element of $S'$ is also an element of $S$, so in this case we have that $p$ divides an element of $S$ as well, as we wanted to show.



                      End of proof.



                      In summary: the new definition from your answer and the old definition from the other answers are equivalent and single out the same elements of $R$.






                      share|cite|improve this answer























                      • I don't get why $4$ is prime2. Set $a=2$, $b=2$ and $c=4$. Did you misstate the definition or did I misread it?
                        – darij grinberg
                        2 hours ago










                      • No you are right, 4 is not prime2 as stated... I have to think about how to edit my post. So the point of the original question was to show that 4 is prime4 defined as $p$ is prime 4 if there exist a $c$ such that $p|c$ and for all $a, b$ such that $ab = c$ we have that $p|a$ or $p|c$. However the reason I defined prime2 as I did and not as prime4 above was that the definition of prime2 appears in the answer by the OP that inspired my answer. So now I am a bit torn what to do. The best is probably to include both prime2 and prime4 into the post and say something about both...
                        – Vincent
                        2 hours ago










                      • Ok, what do you think of this version? For future readers: the old prime2 darij is refering to is the new prime1' and I edited in (what in above comment is) prime4 as prime2 to keep the flow of the original answer.
                        – Vincent
                        1 hour ago










                      • The new version is fine (though "prime1'" looks really strange with quotation marks).
                        – darij grinberg
                        1 hour ago












                      • because of the prime (pun not intended) followed by quotes? Hmm yes. Ok, while I'm at it I might as well edit out all the quotes and replace them with italics...
                        – Vincent
                        1 hour ago














                      2












                      2








                      2






                      This is an answer to the accepted answer more than to the question, but I believe that that is ok, since both are posed by the same poster.



                      NOTE: I CHANGED SOME THINGS AFTER READING DARIJ'S COMMENT



                      The question seems to be which of four definitions of prime element is the 'correct' one. I restate the definitions here, but let them define four different types of elements (prime1, prime1', prime2, prime3). The reason for this somewhat awkward notation is that it enables us to say things like 'The element $4 in mathbb{Z}$ is prime2 but not prime1' and know exactly what we are talking about.



                      So here are the definitions.




                      An element $p$ in a ring $R$ is called prime1 if it is not a unit and $forall a, b in R$ such that $p|ab$, we have that $p|a$ or $p|b$. This is the definition used in most of the other answers.



                      An element $p$ in a ring $R$ is called prime1' if it is not a unit and $forall c in R$ such that $p|c$, and for all $a, b in R$ such that $ab = c$, we have that $p|a$ or $p|b$. This definition appears in the accepted answer, although at that point the OP seems to have stopped believing that it is the correct one.




                      I use the names prime1 and prime1' to emphasis that these definitions are really identical. Just try to think for 2 seconds what you would need to do to show that an element $p$ is not prime1 or not prime1'. In both cases it amounts to the same thing: find two elements ($a, b$) that are not divisible by $p$ while their product ($c$) is.



                      Things get more interesting when we invoke the other two definitions.




                      An element $p$ in a ring $R$ is called prime2 if it is not a unit and $exists c in R$ such that $p|c$ and $forall a, b in R$ such that $ab = c$ we have that $p|a$ or $p|c$. This definition is nowhere written out explicitly but it is very strongly implicitly present in the original question.



                      An element $p$ in a ring $R$ is called prime3 if it is not a unit and $forall c in R$ such that $p|c$, and for all finite $S subset R$ such that $prod_{s in S} = c$, there is an $s in S$ such that $p|s$. This definition is proposed by the OP in the accepted answer as the "correct" one.




                      Now the content of the original question is that the number $4 in mathbb{Z}$ is prime2, showing that 'prime2-ness' is not a good way of generalizing 'primeness' (primality) of integers to arbitrary rings. Very nice, I never thought of that before. However, for now that means that prime2 is out as a candidate definition of "prime" and the race is between prime1 and prime3. (I ignore prime1' for the moment as it is identical to prime1).



                      The counterexample the OP is asking for seems to be a natural number that is prime3 but not prime in the ordinary sense. In fact it would already be interesting to see any element of any integral domain that is either prime3 but not prime1 or prime1 but not prime3. The point of my answer is: these elements do not exist. Concretely we have:




                      Theorem 1: let $R$ be an integral domain. Then every element that is prime3 is also prime1



                      Theorem 2: let $R$ be an integral domain. Then every element that is prime1 is also prime3.




                      Together the theorems say that the two definitions are equivalent and hence that choosing the one over the other is a matter of taste (or pedagogy) and doesn't change anything mathematically.



                      I will prove the theorems below.



                      Proving Theorem 1 is really easy, there is hardly anything to do. We have an element $p$ that is prime3 and elements $a, b$ such that $p|ab$. Taking $c = ab$ and $S = {a, b}$ we see that from prime3-ness of $p$ that $p|a$ or $p|b$ which in turn implies that $p$ is prime1. End of proof.



                      Proving Theorem 2 is a bit more involved.



                      Let $p$ be a prime1 element of $R$. Let $c$ be any element of $R$ such that $p|c$ and let $S subset R$ be any finite set such that $prod_{sin S} s = c$. We need to show that there is an element $s in S$ such that $p|s$ in order to show that $p$ is prime3. We do this by induction on the number $n$ of elements of $S$.



                      The first hurdle we have to tackle is the fact that if $n = 0$ then our goal is unreachable: clearly in an empty set there is no element $s$ such that $p|s$. Luckily this case does not occur! The point is: the product of all elements of an empty set is 1 by definition. But if $c = 1$ and $p|c$ then $p$ is a unit, which contradicts its prime1-ness. So this case cannot happen.



                      We move on to the case $n = 1$. If $S$ contains only a single element then clearly this single element must be the element $c$, which was the product of all elements. But we already know that $p|c$ and so, indeed, $p$ divides a (or rather: the) element of $S$.



                      Now let $n geq 2$ and we suppose that ('Induction Hypothesis') that for all $a in R$ with $p|a$ and all $S' subset R$ with $n-1$ elements such that $prod_{s in S'} s = a$ we have that $p$ divides some element of $S'$.



                      We are still working with the $n$-element set $S$ and we label the the elements $s_1, s_2, ldots, s_{n-1}, s_n$. Let $a = s_1 cdot s_2 cdots s_{n-1}$ and $b = s_n$. Then $ab = c$ and hence $p|ab$ and hence, since $p$ is prime1 we have that $p|a$ or $p|b$. In the second case we see that $p|s_n$ and hence we see that $p$ divides an element of $S$ as we wanted to show. In the first case we set $S' = {s_1, ldots, s_{n-1}}$. Since $p|a$ we know from the Induction Hypothesis that $p$ divides some element of $S'$. But every element of $S'$ is also an element of $S$, so in this case we have that $p$ divides an element of $S$ as well, as we wanted to show.



                      End of proof.



                      In summary: the new definition from your answer and the old definition from the other answers are equivalent and single out the same elements of $R$.






                      share|cite|improve this answer














                      This is an answer to the accepted answer more than to the question, but I believe that that is ok, since both are posed by the same poster.



                      NOTE: I CHANGED SOME THINGS AFTER READING DARIJ'S COMMENT



                      The question seems to be which of four definitions of prime element is the 'correct' one. I restate the definitions here, but let them define four different types of elements (prime1, prime1', prime2, prime3). The reason for this somewhat awkward notation is that it enables us to say things like 'The element $4 in mathbb{Z}$ is prime2 but not prime1' and know exactly what we are talking about.



                      So here are the definitions.




                      An element $p$ in a ring $R$ is called prime1 if it is not a unit and $forall a, b in R$ such that $p|ab$, we have that $p|a$ or $p|b$. This is the definition used in most of the other answers.



                      An element $p$ in a ring $R$ is called prime1' if it is not a unit and $forall c in R$ such that $p|c$, and for all $a, b in R$ such that $ab = c$, we have that $p|a$ or $p|b$. This definition appears in the accepted answer, although at that point the OP seems to have stopped believing that it is the correct one.




                      I use the names prime1 and prime1' to emphasis that these definitions are really identical. Just try to think for 2 seconds what you would need to do to show that an element $p$ is not prime1 or not prime1'. In both cases it amounts to the same thing: find two elements ($a, b$) that are not divisible by $p$ while their product ($c$) is.



                      Things get more interesting when we invoke the other two definitions.




                      An element $p$ in a ring $R$ is called prime2 if it is not a unit and $exists c in R$ such that $p|c$ and $forall a, b in R$ such that $ab = c$ we have that $p|a$ or $p|c$. This definition is nowhere written out explicitly but it is very strongly implicitly present in the original question.



                      An element $p$ in a ring $R$ is called prime3 if it is not a unit and $forall c in R$ such that $p|c$, and for all finite $S subset R$ such that $prod_{s in S} = c$, there is an $s in S$ such that $p|s$. This definition is proposed by the OP in the accepted answer as the "correct" one.




                      Now the content of the original question is that the number $4 in mathbb{Z}$ is prime2, showing that 'prime2-ness' is not a good way of generalizing 'primeness' (primality) of integers to arbitrary rings. Very nice, I never thought of that before. However, for now that means that prime2 is out as a candidate definition of "prime" and the race is between prime1 and prime3. (I ignore prime1' for the moment as it is identical to prime1).



                      The counterexample the OP is asking for seems to be a natural number that is prime3 but not prime in the ordinary sense. In fact it would already be interesting to see any element of any integral domain that is either prime3 but not prime1 or prime1 but not prime3. The point of my answer is: these elements do not exist. Concretely we have:




                      Theorem 1: let $R$ be an integral domain. Then every element that is prime3 is also prime1



                      Theorem 2: let $R$ be an integral domain. Then every element that is prime1 is also prime3.




                      Together the theorems say that the two definitions are equivalent and hence that choosing the one over the other is a matter of taste (or pedagogy) and doesn't change anything mathematically.



                      I will prove the theorems below.



                      Proving Theorem 1 is really easy, there is hardly anything to do. We have an element $p$ that is prime3 and elements $a, b$ such that $p|ab$. Taking $c = ab$ and $S = {a, b}$ we see that from prime3-ness of $p$ that $p|a$ or $p|b$ which in turn implies that $p$ is prime1. End of proof.



                      Proving Theorem 2 is a bit more involved.



                      Let $p$ be a prime1 element of $R$. Let $c$ be any element of $R$ such that $p|c$ and let $S subset R$ be any finite set such that $prod_{sin S} s = c$. We need to show that there is an element $s in S$ such that $p|s$ in order to show that $p$ is prime3. We do this by induction on the number $n$ of elements of $S$.



                      The first hurdle we have to tackle is the fact that if $n = 0$ then our goal is unreachable: clearly in an empty set there is no element $s$ such that $p|s$. Luckily this case does not occur! The point is: the product of all elements of an empty set is 1 by definition. But if $c = 1$ and $p|c$ then $p$ is a unit, which contradicts its prime1-ness. So this case cannot happen.



                      We move on to the case $n = 1$. If $S$ contains only a single element then clearly this single element must be the element $c$, which was the product of all elements. But we already know that $p|c$ and so, indeed, $p$ divides a (or rather: the) element of $S$.



                      Now let $n geq 2$ and we suppose that ('Induction Hypothesis') that for all $a in R$ with $p|a$ and all $S' subset R$ with $n-1$ elements such that $prod_{s in S'} s = a$ we have that $p$ divides some element of $S'$.



                      We are still working with the $n$-element set $S$ and we label the the elements $s_1, s_2, ldots, s_{n-1}, s_n$. Let $a = s_1 cdot s_2 cdots s_{n-1}$ and $b = s_n$. Then $ab = c$ and hence $p|ab$ and hence, since $p$ is prime1 we have that $p|a$ or $p|b$. In the second case we see that $p|s_n$ and hence we see that $p$ divides an element of $S$ as we wanted to show. In the first case we set $S' = {s_1, ldots, s_{n-1}}$. Since $p|a$ we know from the Induction Hypothesis that $p$ divides some element of $S'$. But every element of $S'$ is also an element of $S$, so in this case we have that $p$ divides an element of $S$ as well, as we wanted to show.



                      End of proof.



                      In summary: the new definition from your answer and the old definition from the other answers are equivalent and single out the same elements of $R$.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited 1 hour ago

























                      answered 2 hours ago









                      Vincent

                      3,03611228




                      3,03611228












                      • I don't get why $4$ is prime2. Set $a=2$, $b=2$ and $c=4$. Did you misstate the definition or did I misread it?
                        – darij grinberg
                        2 hours ago










                      • No you are right, 4 is not prime2 as stated... I have to think about how to edit my post. So the point of the original question was to show that 4 is prime4 defined as $p$ is prime 4 if there exist a $c$ such that $p|c$ and for all $a, b$ such that $ab = c$ we have that $p|a$ or $p|c$. However the reason I defined prime2 as I did and not as prime4 above was that the definition of prime2 appears in the answer by the OP that inspired my answer. So now I am a bit torn what to do. The best is probably to include both prime2 and prime4 into the post and say something about both...
                        – Vincent
                        2 hours ago










                      • Ok, what do you think of this version? For future readers: the old prime2 darij is refering to is the new prime1' and I edited in (what in above comment is) prime4 as prime2 to keep the flow of the original answer.
                        – Vincent
                        1 hour ago










                      • The new version is fine (though "prime1'" looks really strange with quotation marks).
                        – darij grinberg
                        1 hour ago












                      • because of the prime (pun not intended) followed by quotes? Hmm yes. Ok, while I'm at it I might as well edit out all the quotes and replace them with italics...
                        – Vincent
                        1 hour ago


















                      • I don't get why $4$ is prime2. Set $a=2$, $b=2$ and $c=4$. Did you misstate the definition or did I misread it?
                        – darij grinberg
                        2 hours ago










                      • No you are right, 4 is not prime2 as stated... I have to think about how to edit my post. So the point of the original question was to show that 4 is prime4 defined as $p$ is prime 4 if there exist a $c$ such that $p|c$ and for all $a, b$ such that $ab = c$ we have that $p|a$ or $p|c$. However the reason I defined prime2 as I did and not as prime4 above was that the definition of prime2 appears in the answer by the OP that inspired my answer. So now I am a bit torn what to do. The best is probably to include both prime2 and prime4 into the post and say something about both...
                        – Vincent
                        2 hours ago










                      • Ok, what do you think of this version? For future readers: the old prime2 darij is refering to is the new prime1' and I edited in (what in above comment is) prime4 as prime2 to keep the flow of the original answer.
                        – Vincent
                        1 hour ago










                      • The new version is fine (though "prime1'" looks really strange with quotation marks).
                        – darij grinberg
                        1 hour ago












                      • because of the prime (pun not intended) followed by quotes? Hmm yes. Ok, while I'm at it I might as well edit out all the quotes and replace them with italics...
                        – Vincent
                        1 hour ago
















                      I don't get why $4$ is prime2. Set $a=2$, $b=2$ and $c=4$. Did you misstate the definition or did I misread it?
                      – darij grinberg
                      2 hours ago




                      I don't get why $4$ is prime2. Set $a=2$, $b=2$ and $c=4$. Did you misstate the definition or did I misread it?
                      – darij grinberg
                      2 hours ago












                      No you are right, 4 is not prime2 as stated... I have to think about how to edit my post. So the point of the original question was to show that 4 is prime4 defined as $p$ is prime 4 if there exist a $c$ such that $p|c$ and for all $a, b$ such that $ab = c$ we have that $p|a$ or $p|c$. However the reason I defined prime2 as I did and not as prime4 above was that the definition of prime2 appears in the answer by the OP that inspired my answer. So now I am a bit torn what to do. The best is probably to include both prime2 and prime4 into the post and say something about both...
                      – Vincent
                      2 hours ago




                      No you are right, 4 is not prime2 as stated... I have to think about how to edit my post. So the point of the original question was to show that 4 is prime4 defined as $p$ is prime 4 if there exist a $c$ such that $p|c$ and for all $a, b$ such that $ab = c$ we have that $p|a$ or $p|c$. However the reason I defined prime2 as I did and not as prime4 above was that the definition of prime2 appears in the answer by the OP that inspired my answer. So now I am a bit torn what to do. The best is probably to include both prime2 and prime4 into the post and say something about both...
                      – Vincent
                      2 hours ago












                      Ok, what do you think of this version? For future readers: the old prime2 darij is refering to is the new prime1' and I edited in (what in above comment is) prime4 as prime2 to keep the flow of the original answer.
                      – Vincent
                      1 hour ago




                      Ok, what do you think of this version? For future readers: the old prime2 darij is refering to is the new prime1' and I edited in (what in above comment is) prime4 as prime2 to keep the flow of the original answer.
                      – Vincent
                      1 hour ago












                      The new version is fine (though "prime1'" looks really strange with quotation marks).
                      – darij grinberg
                      1 hour ago






                      The new version is fine (though "prime1'" looks really strange with quotation marks).
                      – darij grinberg
                      1 hour ago














                      because of the prime (pun not intended) followed by quotes? Hmm yes. Ok, while I'm at it I might as well edit out all the quotes and replace them with italics...
                      – Vincent
                      1 hour ago




                      because of the prime (pun not intended) followed by quotes? Hmm yes. Ok, while I'm at it I might as well edit out all the quotes and replace them with italics...
                      – Vincent
                      1 hour ago











                      0














                      Here's how I would reword it:




                      An element $p$ of a ring $R$ is called prime if for any combination of $a, b in R$ such that $p mid ab$, at least one of $p mid a$ or $p mid b$ also holds true.




                      So if $p = a = b = 4$ we see that $p mid ab$ and then $p mid a$ and $p mid b$ are trivially true. But that's not the only possible combination of $a$ and $b$ that we can choose so that $p mid ab$. You have already discovered two more. But there are in fact infinitely many more.



                      It suffices to choose one in which $a = 2$ and $b neq pm 2$ though it is singly even (that is, not divisible by $4$). So, for example, $b = 14$, then we have that $4 mid 28$ and $28 = 2 times 14$, but $2$ is not divisible by $4$ and neither is $14$. Hence $4$ is not prime.



                      I'm sure you can generalize this to the squares of odd primes. Just my two cents.






                      share|cite|improve this answer





















                      • Please read mine or @Vincent's answer as to why you can use just one multiple of $p$ ($ab$) as long as you check all the factors.
                        – PolymorphismPrince
                        29 mins ago










                      • I might or I might not, I'm writing an answer to a more difficult question and that might not leave enough time before my social obligations tonight. In the meantime, I suggest that you accept Vincent's answer instead of your own. I would appreciate an upvote for you, but my answer is really more for the benefit of others who come across your question.
                        – David R.
                        23 mins ago
















                      0














                      Here's how I would reword it:




                      An element $p$ of a ring $R$ is called prime if for any combination of $a, b in R$ such that $p mid ab$, at least one of $p mid a$ or $p mid b$ also holds true.




                      So if $p = a = b = 4$ we see that $p mid ab$ and then $p mid a$ and $p mid b$ are trivially true. But that's not the only possible combination of $a$ and $b$ that we can choose so that $p mid ab$. You have already discovered two more. But there are in fact infinitely many more.



                      It suffices to choose one in which $a = 2$ and $b neq pm 2$ though it is singly even (that is, not divisible by $4$). So, for example, $b = 14$, then we have that $4 mid 28$ and $28 = 2 times 14$, but $2$ is not divisible by $4$ and neither is $14$. Hence $4$ is not prime.



                      I'm sure you can generalize this to the squares of odd primes. Just my two cents.






                      share|cite|improve this answer





















                      • Please read mine or @Vincent's answer as to why you can use just one multiple of $p$ ($ab$) as long as you check all the factors.
                        – PolymorphismPrince
                        29 mins ago










                      • I might or I might not, I'm writing an answer to a more difficult question and that might not leave enough time before my social obligations tonight. In the meantime, I suggest that you accept Vincent's answer instead of your own. I would appreciate an upvote for you, but my answer is really more for the benefit of others who come across your question.
                        – David R.
                        23 mins ago














                      0












                      0








                      0






                      Here's how I would reword it:




                      An element $p$ of a ring $R$ is called prime if for any combination of $a, b in R$ such that $p mid ab$, at least one of $p mid a$ or $p mid b$ also holds true.




                      So if $p = a = b = 4$ we see that $p mid ab$ and then $p mid a$ and $p mid b$ are trivially true. But that's not the only possible combination of $a$ and $b$ that we can choose so that $p mid ab$. You have already discovered two more. But there are in fact infinitely many more.



                      It suffices to choose one in which $a = 2$ and $b neq pm 2$ though it is singly even (that is, not divisible by $4$). So, for example, $b = 14$, then we have that $4 mid 28$ and $28 = 2 times 14$, but $2$ is not divisible by $4$ and neither is $14$. Hence $4$ is not prime.



                      I'm sure you can generalize this to the squares of odd primes. Just my two cents.






                      share|cite|improve this answer












                      Here's how I would reword it:




                      An element $p$ of a ring $R$ is called prime if for any combination of $a, b in R$ such that $p mid ab$, at least one of $p mid a$ or $p mid b$ also holds true.




                      So if $p = a = b = 4$ we see that $p mid ab$ and then $p mid a$ and $p mid b$ are trivially true. But that's not the only possible combination of $a$ and $b$ that we can choose so that $p mid ab$. You have already discovered two more. But there are in fact infinitely many more.



                      It suffices to choose one in which $a = 2$ and $b neq pm 2$ though it is singly even (that is, not divisible by $4$). So, for example, $b = 14$, then we have that $4 mid 28$ and $28 = 2 times 14$, but $2$ is not divisible by $4$ and neither is $14$. Hence $4$ is not prime.



                      I'm sure you can generalize this to the squares of odd primes. Just my two cents.







                      share|cite|improve this answer












                      share|cite|improve this answer



                      share|cite|improve this answer










                      answered 33 mins ago









                      David R.

                      2211728




                      2211728












                      • Please read mine or @Vincent's answer as to why you can use just one multiple of $p$ ($ab$) as long as you check all the factors.
                        – PolymorphismPrince
                        29 mins ago










                      • I might or I might not, I'm writing an answer to a more difficult question and that might not leave enough time before my social obligations tonight. In the meantime, I suggest that you accept Vincent's answer instead of your own. I would appreciate an upvote for you, but my answer is really more for the benefit of others who come across your question.
                        – David R.
                        23 mins ago


















                      • Please read mine or @Vincent's answer as to why you can use just one multiple of $p$ ($ab$) as long as you check all the factors.
                        – PolymorphismPrince
                        29 mins ago










                      • I might or I might not, I'm writing an answer to a more difficult question and that might not leave enough time before my social obligations tonight. In the meantime, I suggest that you accept Vincent's answer instead of your own. I would appreciate an upvote for you, but my answer is really more for the benefit of others who come across your question.
                        – David R.
                        23 mins ago
















                      Please read mine or @Vincent's answer as to why you can use just one multiple of $p$ ($ab$) as long as you check all the factors.
                      – PolymorphismPrince
                      29 mins ago




                      Please read mine or @Vincent's answer as to why you can use just one multiple of $p$ ($ab$) as long as you check all the factors.
                      – PolymorphismPrince
                      29 mins ago












                      I might or I might not, I'm writing an answer to a more difficult question and that might not leave enough time before my social obligations tonight. In the meantime, I suggest that you accept Vincent's answer instead of your own. I would appreciate an upvote for you, but my answer is really more for the benefit of others who come across your question.
                      – David R.
                      23 mins ago




                      I might or I might not, I'm writing an answer to a more difficult question and that might not leave enough time before my social obligations tonight. In the meantime, I suggest that you accept Vincent's answer instead of your own. I would appreciate an upvote for you, but my answer is really more for the benefit of others who come across your question.
                      – David R.
                      23 mins ago











                      -2














                      Please read this answer before taking any of the others as fact (I'm OP)



                      Short version



                      I believe the correct definition is




                      An element $p$ in a ring $R$ is called "prime" if for some $cin R$ such $p|c$, $forall Ssubset R$ such the product of elements in $S$ is $c$, $exists nin S$ such $p|n$.




                      The only real difference is that we can make factorisations of the multiple of $p$ that are more than just two coefficients.



                      Under this definition, $4$ can be composite since there exists one $S$ that multiplies to $16$ which is $left {2,2,2 right }$ and $4|2$ is false.



                      Long version explaining why I believe other answers are wrong



                      I'd like to firstly acknowledge that I'm slightly disappointed that countless people have copped countless amounts of karma for saying the following:




                      The definition should be interpreted as




                      An element $p$ in a ring $R$ is called "prime" if $forall a,bin R$ such $p|ab$, $p|a$ or $p|b$.




                      Rather than (the way OP interpreted it)




                      An element $p$ in a ring $R$ is called "prime" if $forall cin R$ such $p|c$, $forall a,bin R$ such $ab = c$, $p|a$ or $p|b$.





                      In the edit of my original question, I said that this answer didn't make sense to me intuitively, I suggested that a correct definition would have been:




                      An element $p$ in a ring $R$ is called "prime" if for some $cin R$ such $p|c$, $forall Ssubset R$ such the product of elements in $S$ is $c$, $exists nin S$ such $n|$.




                      I asked for a counter-example, meaning, a number that couldn't be correctly proved prime or composite using this definition. At the time of writing, I'm yet to receive one.



                      I am now convinced in the legitimacy of my definition for the following reason:



                      The standard way of proving if an integer, $n$, is prime or composite is by checking whether it is divisible by any number from $2$ to $sqrt{n}$.



                      If we assume my definition is correct we can check if $n$ is prime in the following way:




                      1. Make $c = n$

                      2. See if $c$ has any factors that are not divisible by $n$. All of them will be since $c = n$ and a positive integer will never be divisible by an integer less than it.

                      3. If we find one, $n$ is not prime.


                      This process is exactly the same as the standard process I previously described. However, if you take other people's definitions I have to check the factors of every possible $nk$ where $k in mathbb{Z}$. That's something we never have to do.



                      I only showed this works in the set of integers. I'm unsure of non-integral domains. But it was the set of integers my question regarded.






                      share|cite|improve this answer























                      • A minor edit: instead of '$p|a$ or $p|b$ you need to write something like '$p|s$ of one of the elements $s in S$ since you explicitly keep open the possibility of $S$ having more than one element (also as written it is not clear that $a$ and $b$ are elements of $S$.
                        – Vincent
                        4 hours ago










                      • Anyway, see my answer below. I hope this is what you were looking for.
                        – Vincent
                        2 hours ago










                      • @Vincent Thanks, editing now.
                        – PolymorphismPrince
                        45 mins ago










                      • I downvoted for the following reasons. (1) Answers are not obligated to update themselves in response to your edits (it might be convenient if they were, but it'd be thoroughly impractical), and the fact that the answers don't address your edit therefore doesn't make them wrong. They may have been unhelpful, and you are not obligated to accept them, however, that doesn't make them wrong. (2) The definition you gave in your question originally also isn't wrong. It may have confused you, but again, that doesn't make it incorrect. (cont)
                        – jgon
                        18 mins ago










                      • (3) Your definition is incorrect. It implies that all irreducibles are prime, which is false. It only works in a UFD, which makes it unhelpful for defining primes. (If $pi$ is irreducible, then taking $c=pi$, any product of elements of the ring that ends up being $pi$ will have all except one factor units, and $pi$ divides the remaining factor.) Your definition would work if we changed $exists c$ to $forall c$.
                        – jgon
                        15 mins ago


















                      -2














                      Please read this answer before taking any of the others as fact (I'm OP)



                      Short version



                      I believe the correct definition is




                      An element $p$ in a ring $R$ is called "prime" if for some $cin R$ such $p|c$, $forall Ssubset R$ such the product of elements in $S$ is $c$, $exists nin S$ such $p|n$.




                      The only real difference is that we can make factorisations of the multiple of $p$ that are more than just two coefficients.



                      Under this definition, $4$ can be composite since there exists one $S$ that multiplies to $16$ which is $left {2,2,2 right }$ and $4|2$ is false.



                      Long version explaining why I believe other answers are wrong



                      I'd like to firstly acknowledge that I'm slightly disappointed that countless people have copped countless amounts of karma for saying the following:




                      The definition should be interpreted as




                      An element $p$ in a ring $R$ is called "prime" if $forall a,bin R$ such $p|ab$, $p|a$ or $p|b$.




                      Rather than (the way OP interpreted it)




                      An element $p$ in a ring $R$ is called "prime" if $forall cin R$ such $p|c$, $forall a,bin R$ such $ab = c$, $p|a$ or $p|b$.





                      In the edit of my original question, I said that this answer didn't make sense to me intuitively, I suggested that a correct definition would have been:




                      An element $p$ in a ring $R$ is called "prime" if for some $cin R$ such $p|c$, $forall Ssubset R$ such the product of elements in $S$ is $c$, $exists nin S$ such $n|$.




                      I asked for a counter-example, meaning, a number that couldn't be correctly proved prime or composite using this definition. At the time of writing, I'm yet to receive one.



                      I am now convinced in the legitimacy of my definition for the following reason:



                      The standard way of proving if an integer, $n$, is prime or composite is by checking whether it is divisible by any number from $2$ to $sqrt{n}$.



                      If we assume my definition is correct we can check if $n$ is prime in the following way:




                      1. Make $c = n$

                      2. See if $c$ has any factors that are not divisible by $n$. All of them will be since $c = n$ and a positive integer will never be divisible by an integer less than it.

                      3. If we find one, $n$ is not prime.


                      This process is exactly the same as the standard process I previously described. However, if you take other people's definitions I have to check the factors of every possible $nk$ where $k in mathbb{Z}$. That's something we never have to do.



                      I only showed this works in the set of integers. I'm unsure of non-integral domains. But it was the set of integers my question regarded.






                      share|cite|improve this answer























                      • A minor edit: instead of '$p|a$ or $p|b$ you need to write something like '$p|s$ of one of the elements $s in S$ since you explicitly keep open the possibility of $S$ having more than one element (also as written it is not clear that $a$ and $b$ are elements of $S$.
                        – Vincent
                        4 hours ago










                      • Anyway, see my answer below. I hope this is what you were looking for.
                        – Vincent
                        2 hours ago










                      • @Vincent Thanks, editing now.
                        – PolymorphismPrince
                        45 mins ago










                      • I downvoted for the following reasons. (1) Answers are not obligated to update themselves in response to your edits (it might be convenient if they were, but it'd be thoroughly impractical), and the fact that the answers don't address your edit therefore doesn't make them wrong. They may have been unhelpful, and you are not obligated to accept them, however, that doesn't make them wrong. (2) The definition you gave in your question originally also isn't wrong. It may have confused you, but again, that doesn't make it incorrect. (cont)
                        – jgon
                        18 mins ago










                      • (3) Your definition is incorrect. It implies that all irreducibles are prime, which is false. It only works in a UFD, which makes it unhelpful for defining primes. (If $pi$ is irreducible, then taking $c=pi$, any product of elements of the ring that ends up being $pi$ will have all except one factor units, and $pi$ divides the remaining factor.) Your definition would work if we changed $exists c$ to $forall c$.
                        – jgon
                        15 mins ago
















                      -2












                      -2








                      -2






                      Please read this answer before taking any of the others as fact (I'm OP)



                      Short version



                      I believe the correct definition is




                      An element $p$ in a ring $R$ is called "prime" if for some $cin R$ such $p|c$, $forall Ssubset R$ such the product of elements in $S$ is $c$, $exists nin S$ such $p|n$.




                      The only real difference is that we can make factorisations of the multiple of $p$ that are more than just two coefficients.



                      Under this definition, $4$ can be composite since there exists one $S$ that multiplies to $16$ which is $left {2,2,2 right }$ and $4|2$ is false.



                      Long version explaining why I believe other answers are wrong



                      I'd like to firstly acknowledge that I'm slightly disappointed that countless people have copped countless amounts of karma for saying the following:




                      The definition should be interpreted as




                      An element $p$ in a ring $R$ is called "prime" if $forall a,bin R$ such $p|ab$, $p|a$ or $p|b$.




                      Rather than (the way OP interpreted it)




                      An element $p$ in a ring $R$ is called "prime" if $forall cin R$ such $p|c$, $forall a,bin R$ such $ab = c$, $p|a$ or $p|b$.





                      In the edit of my original question, I said that this answer didn't make sense to me intuitively, I suggested that a correct definition would have been:




                      An element $p$ in a ring $R$ is called "prime" if for some $cin R$ such $p|c$, $forall Ssubset R$ such the product of elements in $S$ is $c$, $exists nin S$ such $n|$.




                      I asked for a counter-example, meaning, a number that couldn't be correctly proved prime or composite using this definition. At the time of writing, I'm yet to receive one.



                      I am now convinced in the legitimacy of my definition for the following reason:



                      The standard way of proving if an integer, $n$, is prime or composite is by checking whether it is divisible by any number from $2$ to $sqrt{n}$.



                      If we assume my definition is correct we can check if $n$ is prime in the following way:




                      1. Make $c = n$

                      2. See if $c$ has any factors that are not divisible by $n$. All of them will be since $c = n$ and a positive integer will never be divisible by an integer less than it.

                      3. If we find one, $n$ is not prime.


                      This process is exactly the same as the standard process I previously described. However, if you take other people's definitions I have to check the factors of every possible $nk$ where $k in mathbb{Z}$. That's something we never have to do.



                      I only showed this works in the set of integers. I'm unsure of non-integral domains. But it was the set of integers my question regarded.






                      share|cite|improve this answer














                      Please read this answer before taking any of the others as fact (I'm OP)



                      Short version



                      I believe the correct definition is




                      An element $p$ in a ring $R$ is called "prime" if for some $cin R$ such $p|c$, $forall Ssubset R$ such the product of elements in $S$ is $c$, $exists nin S$ such $p|n$.




                      The only real difference is that we can make factorisations of the multiple of $p$ that are more than just two coefficients.



                      Under this definition, $4$ can be composite since there exists one $S$ that multiplies to $16$ which is $left {2,2,2 right }$ and $4|2$ is false.



                      Long version explaining why I believe other answers are wrong



                      I'd like to firstly acknowledge that I'm slightly disappointed that countless people have copped countless amounts of karma for saying the following:




                      The definition should be interpreted as




                      An element $p$ in a ring $R$ is called "prime" if $forall a,bin R$ such $p|ab$, $p|a$ or $p|b$.




                      Rather than (the way OP interpreted it)




                      An element $p$ in a ring $R$ is called "prime" if $forall cin R$ such $p|c$, $forall a,bin R$ such $ab = c$, $p|a$ or $p|b$.





                      In the edit of my original question, I said that this answer didn't make sense to me intuitively, I suggested that a correct definition would have been:




                      An element $p$ in a ring $R$ is called "prime" if for some $cin R$ such $p|c$, $forall Ssubset R$ such the product of elements in $S$ is $c$, $exists nin S$ such $n|$.




                      I asked for a counter-example, meaning, a number that couldn't be correctly proved prime or composite using this definition. At the time of writing, I'm yet to receive one.



                      I am now convinced in the legitimacy of my definition for the following reason:



                      The standard way of proving if an integer, $n$, is prime or composite is by checking whether it is divisible by any number from $2$ to $sqrt{n}$.



                      If we assume my definition is correct we can check if $n$ is prime in the following way:




                      1. Make $c = n$

                      2. See if $c$ has any factors that are not divisible by $n$. All of them will be since $c = n$ and a positive integer will never be divisible by an integer less than it.

                      3. If we find one, $n$ is not prime.


                      This process is exactly the same as the standard process I previously described. However, if you take other people's definitions I have to check the factors of every possible $nk$ where $k in mathbb{Z}$. That's something we never have to do.



                      I only showed this works in the set of integers. I'm unsure of non-integral domains. But it was the set of integers my question regarded.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited 40 mins ago

























                      answered 15 hours ago









                      PolymorphismPrince

                      1546




                      1546












                      • A minor edit: instead of '$p|a$ or $p|b$ you need to write something like '$p|s$ of one of the elements $s in S$ since you explicitly keep open the possibility of $S$ having more than one element (also as written it is not clear that $a$ and $b$ are elements of $S$.
                        – Vincent
                        4 hours ago










                      • Anyway, see my answer below. I hope this is what you were looking for.
                        – Vincent
                        2 hours ago










                      • @Vincent Thanks, editing now.
                        – PolymorphismPrince
                        45 mins ago










                      • I downvoted for the following reasons. (1) Answers are not obligated to update themselves in response to your edits (it might be convenient if they were, but it'd be thoroughly impractical), and the fact that the answers don't address your edit therefore doesn't make them wrong. They may have been unhelpful, and you are not obligated to accept them, however, that doesn't make them wrong. (2) The definition you gave in your question originally also isn't wrong. It may have confused you, but again, that doesn't make it incorrect. (cont)
                        – jgon
                        18 mins ago










                      • (3) Your definition is incorrect. It implies that all irreducibles are prime, which is false. It only works in a UFD, which makes it unhelpful for defining primes. (If $pi$ is irreducible, then taking $c=pi$, any product of elements of the ring that ends up being $pi$ will have all except one factor units, and $pi$ divides the remaining factor.) Your definition would work if we changed $exists c$ to $forall c$.
                        – jgon
                        15 mins ago




















                      • A minor edit: instead of '$p|a$ or $p|b$ you need to write something like '$p|s$ of one of the elements $s in S$ since you explicitly keep open the possibility of $S$ having more than one element (also as written it is not clear that $a$ and $b$ are elements of $S$.
                        – Vincent
                        4 hours ago










                      • Anyway, see my answer below. I hope this is what you were looking for.
                        – Vincent
                        2 hours ago










                      • @Vincent Thanks, editing now.
                        – PolymorphismPrince
                        45 mins ago










                      • I downvoted for the following reasons. (1) Answers are not obligated to update themselves in response to your edits (it might be convenient if they were, but it'd be thoroughly impractical), and the fact that the answers don't address your edit therefore doesn't make them wrong. They may have been unhelpful, and you are not obligated to accept them, however, that doesn't make them wrong. (2) The definition you gave in your question originally also isn't wrong. It may have confused you, but again, that doesn't make it incorrect. (cont)
                        – jgon
                        18 mins ago










                      • (3) Your definition is incorrect. It implies that all irreducibles are prime, which is false. It only works in a UFD, which makes it unhelpful for defining primes. (If $pi$ is irreducible, then taking $c=pi$, any product of elements of the ring that ends up being $pi$ will have all except one factor units, and $pi$ divides the remaining factor.) Your definition would work if we changed $exists c$ to $forall c$.
                        – jgon
                        15 mins ago


















                      A minor edit: instead of '$p|a$ or $p|b$ you need to write something like '$p|s$ of one of the elements $s in S$ since you explicitly keep open the possibility of $S$ having more than one element (also as written it is not clear that $a$ and $b$ are elements of $S$.
                      – Vincent
                      4 hours ago




                      A minor edit: instead of '$p|a$ or $p|b$ you need to write something like '$p|s$ of one of the elements $s in S$ since you explicitly keep open the possibility of $S$ having more than one element (also as written it is not clear that $a$ and $b$ are elements of $S$.
                      – Vincent
                      4 hours ago












                      Anyway, see my answer below. I hope this is what you were looking for.
                      – Vincent
                      2 hours ago




                      Anyway, see my answer below. I hope this is what you were looking for.
                      – Vincent
                      2 hours ago












                      @Vincent Thanks, editing now.
                      – PolymorphismPrince
                      45 mins ago




                      @Vincent Thanks, editing now.
                      – PolymorphismPrince
                      45 mins ago












                      I downvoted for the following reasons. (1) Answers are not obligated to update themselves in response to your edits (it might be convenient if they were, but it'd be thoroughly impractical), and the fact that the answers don't address your edit therefore doesn't make them wrong. They may have been unhelpful, and you are not obligated to accept them, however, that doesn't make them wrong. (2) The definition you gave in your question originally also isn't wrong. It may have confused you, but again, that doesn't make it incorrect. (cont)
                      – jgon
                      18 mins ago




                      I downvoted for the following reasons. (1) Answers are not obligated to update themselves in response to your edits (it might be convenient if they were, but it'd be thoroughly impractical), and the fact that the answers don't address your edit therefore doesn't make them wrong. They may have been unhelpful, and you are not obligated to accept them, however, that doesn't make them wrong. (2) The definition you gave in your question originally also isn't wrong. It may have confused you, but again, that doesn't make it incorrect. (cont)
                      – jgon
                      18 mins ago












                      (3) Your definition is incorrect. It implies that all irreducibles are prime, which is false. It only works in a UFD, which makes it unhelpful for defining primes. (If $pi$ is irreducible, then taking $c=pi$, any product of elements of the ring that ends up being $pi$ will have all except one factor units, and $pi$ divides the remaining factor.) Your definition would work if we changed $exists c$ to $forall c$.
                      – jgon
                      15 mins ago






                      (3) Your definition is incorrect. It implies that all irreducibles are prime, which is false. It only works in a UFD, which makes it unhelpful for defining primes. (If $pi$ is irreducible, then taking $c=pi$, any product of elements of the ring that ends up being $pi$ will have all except one factor units, and $pi$ divides the remaining factor.) Your definition would work if we changed $exists c$ to $forall c$.
                      – jgon
                      15 mins ago




















                      draft saved

                      draft discarded




















































                      Thanks for contributing an answer to Mathematics Stack Exchange!


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      Use MathJax to format equations. MathJax reference.


                      To learn more, see our tips on writing great answers.





                      Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


                      Please pay close attention to the following guidance:


                      • Please be sure to answer the question. Provide details and share your research!

                      But avoid



                      • Asking for help, clarification, or responding to other answers.

                      • Making statements based on opinion; back them up with references or personal experience.


                      To learn more, see our tips on writing great answers.




                      draft saved


                      draft discarded














                      StackExchange.ready(
                      function () {
                      StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3059225%2fwhy-doesnt-the-definition-p-is-called-prime-if-p-mid-ab-implies-p-mid-a%23new-answer', 'question_page');
                      }
                      );

                      Post as a guest















                      Required, but never shown





















































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown

































                      Required, but never shown














                      Required, but never shown












                      Required, but never shown







                      Required, but never shown







                      Popular posts from this blog

                      Mario Kart Wii

                      The Binding of Isaac: Rebirth/Afterbirth

                      What does “Dominus providebit” mean?