Is $mathbb Z[t]$ principal? [duplicate]
This question already has an answer here:
Is $mathbb{Z}[x]$ a principal ideal domain?
5 answers
I know that $mathbb Z[t]$ is factoriel, but is it principal ? For example, let consider the ideal $(5,t)$ it look that is not principal, but after all, $5$ and $t$ are co-prime, so by Bezout, there is $q(x),r(x)$ s.t. $$5q(t)+tk(t)=1$$ and thus $(5,t)=mathbb Z[t]$ ?
abstract-algebra
asked 15 hours ago
NewMath
3338
marked as duplicate by Pierre-Guy Plamondon, Dietrich Burde abstract-algebra
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14 hours ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Is $mathbb{Z}[x]$ a principal ideal domain?
5 answers
I know that $mathbb Z[t]$ is factoriel, but is it principal ? For example, let consider the ideal $(5,t)$ it look that is not principal, but after all, $5$ and $t$ are co-prime, so by Bezout, there is $q(x),r(x)$ s.t. $$5q(t)+tk(t)=1$$ and thus $(5,t)=mathbb Z[t]$ ?
abstract-algebra
asked 15 hours ago
NewMath
3338
marked as duplicate by Pierre-Guy Plamondon, Dietrich Burde abstract-algebra
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14 hours ago
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
This question already has an answer here:
Is $mathbb{Z}[x]$ a principal ideal domain?
5 answers
I know that $mathbb Z[t]$ is factoriel, but is it principal ? For example, let consider the ideal $(5,t)$ it look that is not principal, but after all, $5$ and $t$ are co-prime, so by Bezout, there is $q(x),r(x)$ s.t. $$5q(t)+tk(t)=1$$ and thus $(5,t)=mathbb Z[t]$ ?
abstract-algebra
asked 15 hours ago
NewMath
3338
This question already has an answer here:
Is $mathbb{Z}[x]$ a principal ideal domain?
5 answers
I know that $mathbb Z[t]$ is factoriel, but is it principal ? For example, let consider the ideal $(5,t)$ it look that is not principal, but after all, $5$ and $t$ are co-prime, so by Bezout, there is $q(x),r(x)$ s.t. $$5q(t)+tk(t)=1$$ and thus $(5,t)=mathbb Z[t]$ ?
This question already has an answer here:
Is $mathbb{Z}[x]$ a principal ideal domain?
5 answers
abstract-algebra
abstract-algebra
asked 15 hours ago
NewMath
3338
asked 15 hours ago
NewMath
3338
asked 15 hours ago
NewMath
3338
asked 15 hours ago
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3338
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14 hours ago
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14 hours ago
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Bezout's doesn't apply here. We can clearly see that in the polynomial $5q(t) + tk(t)$, the constant term is divisible by $5$. More formally, consider the evaluation homomorphism $v_0:Bbb Z[t]to Bbb Z$ at $t = 0$. We see that $5mid v_0(5q(t) + tk(t))$, but $5nmid v_0(1)$, so we cannot have $5q(t) + tk(t) = 1$.
And no, it's not a PID. The only principal ideal which contains both $5$ and $t$ is $(1)$, and as we have seen, $1notin (5, t)$.
answered 15 hours ago
Arthur
111k7105186
Sorry, I don't get why Bezout doesn't apply there. Don't we have that $gcd(f(t),g(t))=1iff x(t)f(t)+y(t)g(t)=1$ for some polynomial $x(t),y(t)$ in $mathbb Z[t]$ ? I thought that Bezout equality work in all factorial ring.
– NewMath
14 hours ago
@NewMath I showed you why Bezout doesn't apply. I demonstrated that no linear combination of $5$ and $t$ can give $1$. There isn't any fundamental reason other than "it just doesn't, because see here, it clearly fails in at least this one case". One can start to point at properties that $Bbb Z$ has and $Bbb Z[t]$ doesn't which are essential to Bezout, but that's not really important.
– Arthur
14 hours ago
yes indeed. I was confuse between with $(a,b)$ and $(a)+(b)$. Could you tel me which operation between $I=(a)$ and $J=(b)$ gives the ideal $(a,b)$ ? For example, $I+J$ is the $gcd$, $Icap J$ is the $lcm$, but which operation between $I$ and $J$ gives $(a,b)$ ? It look that $IJsubset (a,b)$. Could it be $(a,b)=I+J+IJ$ ?
– NewMath
13 hours ago
1
@NewMath No, no confusion there. $(a, b) = (a) + (b)$. Plain and simple. The elements in $(a, b)$ are all the linear combinations of $a$ and $b$, while the elements in $(a) + (b)$ are all possible sums of one element in $(a)$ with one element in $(b)$. Same thing. Alternatively, $(a, b)$ is the smallest ideal which contains both $a$ and $b$, while $(a)+(b)$ is the smallest ideal which contains both $(a)$ and $(b)$. Again, same thing.
– Arthur
13 hours ago
Thank you very much dear @arthur. It's perfectly clear.
– NewMath
13 hours ago
add a comment |
In the ring ${Bbb Z}[t]$, we have
$gcd(2, t) = 1$. But there is no representation of the form $2cdot f + tcdot g =1$,
since the constant term on the left hand side is even.
answered 15 hours ago
Wuestenfux
3,6041411
add a comment |
2 Answers
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votes
2 Answers
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active
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active
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votes
Bezout's doesn't apply here. We can clearly see that in the polynomial $5q(t) + tk(t)$, the constant term is divisible by $5$. More formally, consider the evaluation homomorphism $v_0:Bbb Z[t]to Bbb Z$ at $t = 0$. We see that $5mid v_0(5q(t) + tk(t))$, but $5nmid v_0(1)$, so we cannot have $5q(t) + tk(t) = 1$.
And no, it's not a PID. The only principal ideal which contains both $5$ and $t$ is $(1)$, and as we have seen, $1notin (5, t)$.
answered 15 hours ago
Arthur
111k7105186
Sorry, I don't get why Bezout doesn't apply there. Don't we have that $gcd(f(t),g(t))=1iff x(t)f(t)+y(t)g(t)=1$ for some polynomial $x(t),y(t)$ in $mathbb Z[t]$ ? I thought that Bezout equality work in all factorial ring.
– NewMath
14 hours ago
@NewMath I showed you why Bezout doesn't apply. I demonstrated that no linear combination of $5$ and $t$ can give $1$. There isn't any fundamental reason other than "it just doesn't, because see here, it clearly fails in at least this one case". One can start to point at properties that $Bbb Z$ has and $Bbb Z[t]$ doesn't which are essential to Bezout, but that's not really important.
– Arthur
14 hours ago
yes indeed. I was confuse between with $(a,b)$ and $(a)+(b)$. Could you tel me which operation between $I=(a)$ and $J=(b)$ gives the ideal $(a,b)$ ? For example, $I+J$ is the $gcd$, $Icap J$ is the $lcm$, but which operation between $I$ and $J$ gives $(a,b)$ ? It look that $IJsubset (a,b)$. Could it be $(a,b)=I+J+IJ$ ?
– NewMath
13 hours ago
1
@NewMath No, no confusion there. $(a, b) = (a) + (b)$. Plain and simple. The elements in $(a, b)$ are all the linear combinations of $a$ and $b$, while the elements in $(a) + (b)$ are all possible sums of one element in $(a)$ with one element in $(b)$. Same thing. Alternatively, $(a, b)$ is the smallest ideal which contains both $a$ and $b$, while $(a)+(b)$ is the smallest ideal which contains both $(a)$ and $(b)$. Again, same thing.
– Arthur
13 hours ago
Thank you very much dear @arthur. It's perfectly clear.
– NewMath
13 hours ago
add a comment |
Bezout's doesn't apply here. We can clearly see that in the polynomial $5q(t) + tk(t)$, the constant term is divisible by $5$. More formally, consider the evaluation homomorphism $v_0:Bbb Z[t]to Bbb Z$ at $t = 0$. We see that $5mid v_0(5q(t) + tk(t))$, but $5nmid v_0(1)$, so we cannot have $5q(t) + tk(t) = 1$.
And no, it's not a PID. The only principal ideal which contains both $5$ and $t$ is $(1)$, and as we have seen, $1notin (5, t)$.
answered 15 hours ago
Arthur
111k7105186
Sorry, I don't get why Bezout doesn't apply there. Don't we have that $gcd(f(t),g(t))=1iff x(t)f(t)+y(t)g(t)=1$ for some polynomial $x(t),y(t)$ in $mathbb Z[t]$ ? I thought that Bezout equality work in all factorial ring.
– NewMath
14 hours ago
@NewMath I showed you why Bezout doesn't apply. I demonstrated that no linear combination of $5$ and $t$ can give $1$. There isn't any fundamental reason other than "it just doesn't, because see here, it clearly fails in at least this one case". One can start to point at properties that $Bbb Z$ has and $Bbb Z[t]$ doesn't which are essential to Bezout, but that's not really important.
– Arthur
14 hours ago
yes indeed. I was confuse between with $(a,b)$ and $(a)+(b)$. Could you tel me which operation between $I=(a)$ and $J=(b)$ gives the ideal $(a,b)$ ? For example, $I+J$ is the $gcd$, $Icap J$ is the $lcm$, but which operation between $I$ and $J$ gives $(a,b)$ ? It look that $IJsubset (a,b)$. Could it be $(a,b)=I+J+IJ$ ?
– NewMath
13 hours ago
1
@NewMath No, no confusion there. $(a, b) = (a) + (b)$. Plain and simple. The elements in $(a, b)$ are all the linear combinations of $a$ and $b$, while the elements in $(a) + (b)$ are all possible sums of one element in $(a)$ with one element in $(b)$. Same thing. Alternatively, $(a, b)$ is the smallest ideal which contains both $a$ and $b$, while $(a)+(b)$ is the smallest ideal which contains both $(a)$ and $(b)$. Again, same thing.
– Arthur
13 hours ago
Thank you very much dear @arthur. It's perfectly clear.
– NewMath
13 hours ago
add a comment |
Bezout's doesn't apply here. We can clearly see that in the polynomial $5q(t) + tk(t)$, the constant term is divisible by $5$. More formally, consider the evaluation homomorphism $v_0:Bbb Z[t]to Bbb Z$ at $t = 0$. We see that $5mid v_0(5q(t) + tk(t))$, but $5nmid v_0(1)$, so we cannot have $5q(t) + tk(t) = 1$.
And no, it's not a PID. The only principal ideal which contains both $5$ and $t$ is $(1)$, and as we have seen, $1notin (5, t)$.
answered 15 hours ago
Arthur
111k7105186
Bezout's doesn't apply here. We can clearly see that in the polynomial $5q(t) + tk(t)$, the constant term is divisible by $5$. More formally, consider the evaluation homomorphism $v_0:Bbb Z[t]to Bbb Z$ at $t = 0$. We see that $5mid v_0(5q(t) + tk(t))$, but $5nmid v_0(1)$, so we cannot have $5q(t) + tk(t) = 1$.
And no, it's not a PID. The only principal ideal which contains both $5$ and $t$ is $(1)$, and as we have seen, $1notin (5, t)$.
answered 15 hours ago
Arthur
111k7105186
answered 15 hours ago
Arthur
111k7105186
answered 15 hours ago
Arthur
111k7105186
answered 15 hours ago
Arthur
111k7105186
111k7105186
Sorry, I don't get why Bezout doesn't apply there. Don't we have that $gcd(f(t),g(t))=1iff x(t)f(t)+y(t)g(t)=1$ for some polynomial $x(t),y(t)$ in $mathbb Z[t]$ ? I thought that Bezout equality work in all factorial ring.
– NewMath
14 hours ago
@NewMath I showed you why Bezout doesn't apply. I demonstrated that no linear combination of $5$ and $t$ can give $1$. There isn't any fundamental reason other than "it just doesn't, because see here, it clearly fails in at least this one case". One can start to point at properties that $Bbb Z$ has and $Bbb Z[t]$ doesn't which are essential to Bezout, but that's not really important.
– Arthur
14 hours ago
yes indeed. I was confuse between with $(a,b)$ and $(a)+(b)$. Could you tel me which operation between $I=(a)$ and $J=(b)$ gives the ideal $(a,b)$ ? For example, $I+J$ is the $gcd$, $Icap J$ is the $lcm$, but which operation between $I$ and $J$ gives $(a,b)$ ? It look that $IJsubset (a,b)$. Could it be $(a,b)=I+J+IJ$ ?
– NewMath
13 hours ago
1
@NewMath No, no confusion there. $(a, b) = (a) + (b)$. Plain and simple. The elements in $(a, b)$ are all the linear combinations of $a$ and $b$, while the elements in $(a) + (b)$ are all possible sums of one element in $(a)$ with one element in $(b)$. Same thing. Alternatively, $(a, b)$ is the smallest ideal which contains both $a$ and $b$, while $(a)+(b)$ is the smallest ideal which contains both $(a)$ and $(b)$. Again, same thing.
– Arthur
13 hours ago
Thank you very much dear @arthur. It's perfectly clear.
– NewMath
13 hours ago
add a comment |
Sorry, I don't get why Bezout doesn't apply there. Don't we have that $gcd(f(t),g(t))=1iff x(t)f(t)+y(t)g(t)=1$ for some polynomial $x(t),y(t)$ in $mathbb Z[t]$ ? I thought that Bezout equality work in all factorial ring.
– NewMath
14 hours ago
@NewMath I showed you why Bezout doesn't apply. I demonstrated that no linear combination of $5$ and $t$ can give $1$. There isn't any fundamental reason other than "it just doesn't, because see here, it clearly fails in at least this one case". One can start to point at properties that $Bbb Z$ has and $Bbb Z[t]$ doesn't which are essential to Bezout, but that's not really important.
– Arthur
14 hours ago
yes indeed. I was confuse between with $(a,b)$ and $(a)+(b)$. Could you tel me which operation between $I=(a)$ and $J=(b)$ gives the ideal $(a,b)$ ? For example, $I+J$ is the $gcd$, $Icap J$ is the $lcm$, but which operation between $I$ and $J$ gives $(a,b)$ ? It look that $IJsubset (a,b)$. Could it be $(a,b)=I+J+IJ$ ?
– NewMath
13 hours ago
1
@NewMath No, no confusion there. $(a, b) = (a) + (b)$. Plain and simple. The elements in $(a, b)$ are all the linear combinations of $a$ and $b$, while the elements in $(a) + (b)$ are all possible sums of one element in $(a)$ with one element in $(b)$. Same thing. Alternatively, $(a, b)$ is the smallest ideal which contains both $a$ and $b$, while $(a)+(b)$ is the smallest ideal which contains both $(a)$ and $(b)$. Again, same thing.
– Arthur
13 hours ago
Thank you very much dear @arthur. It's perfectly clear.
– NewMath
13 hours ago
Sorry, I don't get why Bezout doesn't apply there. Don't we have that $gcd(f(t),g(t))=1iff x(t)f(t)+y(t)g(t)=1$ for some polynomial $x(t),y(t)$ in $mathbb Z[t]$ ? I thought that Bezout equality work in all factorial ring.
– NewMath
14 hours ago
Sorry, I don't get why Bezout doesn't apply there. Don't we have that $gcd(f(t),g(t))=1iff x(t)f(t)+y(t)g(t)=1$ for some polynomial $x(t),y(t)$ in $mathbb Z[t]$ ? I thought that Bezout equality work in all factorial ring.
– NewMath
14 hours ago
@NewMath I showed you why Bezout doesn't apply. I demonstrated that no linear combination of $5$ and $t$ can give $1$. There isn't any fundamental reason other than "it just doesn't, because see here, it clearly fails in at least this one case". One can start to point at properties that $Bbb Z$ has and $Bbb Z[t]$ doesn't which are essential to Bezout, but that's not really important.
– Arthur
14 hours ago
@NewMath I showed you why Bezout doesn't apply. I demonstrated that no linear combination of $5$ and $t$ can give $1$. There isn't any fundamental reason other than "it just doesn't, because see here, it clearly fails in at least this one case". One can start to point at properties that $Bbb Z$ has and $Bbb Z[t]$ doesn't which are essential to Bezout, but that's not really important.
– Arthur
14 hours ago
yes indeed. I was confuse between with $(a,b)$ and $(a)+(b)$. Could you tel me which operation between $I=(a)$ and $J=(b)$ gives the ideal $(a,b)$ ? For example, $I+J$ is the $gcd$, $Icap J$ is the $lcm$, but which operation between $I$ and $J$ gives $(a,b)$ ? It look that $IJsubset (a,b)$. Could it be $(a,b)=I+J+IJ$ ?
– NewMath
13 hours ago
yes indeed. I was confuse between with $(a,b)$ and $(a)+(b)$. Could you tel me which operation between $I=(a)$ and $J=(b)$ gives the ideal $(a,b)$ ? For example, $I+J$ is the $gcd$, $Icap J$ is the $lcm$, but which operation between $I$ and $J$ gives $(a,b)$ ? It look that $IJsubset (a,b)$. Could it be $(a,b)=I+J+IJ$ ?
– NewMath
13 hours ago
1
1
@NewMath No, no confusion there. $(a, b) = (a) + (b)$. Plain and simple. The elements in $(a, b)$ are all the linear combinations of $a$ and $b$, while the elements in $(a) + (b)$ are all possible sums of one element in $(a)$ with one element in $(b)$. Same thing. Alternatively, $(a, b)$ is the smallest ideal which contains both $a$ and $b$, while $(a)+(b)$ is the smallest ideal which contains both $(a)$ and $(b)$. Again, same thing.
– Arthur
13 hours ago
@NewMath No, no confusion there. $(a, b) = (a) + (b)$. Plain and simple. The elements in $(a, b)$ are all the linear combinations of $a$ and $b$, while the elements in $(a) + (b)$ are all possible sums of one element in $(a)$ with one element in $(b)$. Same thing. Alternatively, $(a, b)$ is the smallest ideal which contains both $a$ and $b$, while $(a)+(b)$ is the smallest ideal which contains both $(a)$ and $(b)$. Again, same thing.
– Arthur
13 hours ago
Thank you very much dear @arthur. It's perfectly clear.
– NewMath
13 hours ago
Thank you very much dear @arthur. It's perfectly clear.
– NewMath
13 hours ago
add a comment |
In the ring ${Bbb Z}[t]$, we have
$gcd(2, t) = 1$. But there is no representation of the form $2cdot f + tcdot g =1$,
since the constant term on the left hand side is even.
answered 15 hours ago
Wuestenfux
3,6041411
add a comment |
In the ring ${Bbb Z}[t]$, we have
$gcd(2, t) = 1$. But there is no representation of the form $2cdot f + tcdot g =1$,
since the constant term on the left hand side is even.
answered 15 hours ago
Wuestenfux
3,6041411
add a comment |
In the ring ${Bbb Z}[t]$, we have
$gcd(2, t) = 1$. But there is no representation of the form $2cdot f + tcdot g =1$,
since the constant term on the left hand side is even.
answered 15 hours ago
Wuestenfux
3,6041411
In the ring ${Bbb Z}[t]$, we have
$gcd(2, t) = 1$. But there is no representation of the form $2cdot f + tcdot g =1$,
since the constant term on the left hand side is even.
answered 15 hours ago
Wuestenfux
3,6041411
edited 15 hours ago
answered 15 hours ago
Wuestenfux
3,6041411
answered 15 hours ago
Wuestenfux
3,6041411
answered 15 hours ago
Wuestenfux
3,6041411
3,6041411
add a comment |
add a comment |
