Proving a polynomial is irreducible over a finite field












2












$begingroup$


Question: $mathbb{F}_5={0,1,2,3,4,5}$ be the field with $5$ elements, let $mathbb{F}_5[X]$ be the polynomial ring over $mathbb{F}_5$. Let $m(X) = X^2+X+1$. Prove that $m(X)$ is irreducible over $mathbb{F}_5$.



Proof.



Suppose $m(X)$ is reducible then there exist $r(X),q(X)$ where $deg r(X)=deg q(X)= 1$



So by the quadratic formula we have



$$m(X)= left(X-left(frac{-1+3i}{2}right)right)left(X-left(frac{-1-3i}{2}right)right)$$



However $$frac{-1+3i}{2},frac{-1-3i}{2}notin mathbb{F}_5.$$



Therefore $m(X)$ is irreducible.



Would this be correct?






abstract-algebra polynomials proof-verification finite-fields irreducible-polynomials







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Notice that $Bbb F_5={0,1,2,3,4}$
    $endgroup$
    – Masacroso
    Oct 7 '16 at 10:52






  • 1




    $begingroup$
    You can show (by simply plugging in all values) that $m(X)$ has no roots over $mathbb{F}_5$.
    $endgroup$
    – Janik
    Oct 7 '16 at 10:59






  • 2




    $begingroup$
    Not sure how you applied the quadratic formula. $i$, as in $sqrt {-1}$ is an element of $mathbb F_{,5}$ as $2^2=-1pmod 5$. But you should have had $sqrt {-3}$ and $-3$ is not a square $pmod 5$.
    $endgroup$
    – lulu
    Oct 7 '16 at 11:03












  • $begingroup$
    I'm sorry, I don't quite understand how modulus is involved? Isn't this a polynomial ring?
    $endgroup$
    – aoshdosi
    Oct 7 '16 at 11:12










  • $begingroup$
    Working in $mathbb F_{,5}$ is the same as working in the integers $pmod 5$.
    $endgroup$
    – lulu
    Oct 7 '16 at 11:20
















2












$begingroup$


Question: $mathbb{F}_5={0,1,2,3,4,5}$ be the field with $5$ elements, let $mathbb{F}_5[X]$ be the polynomial ring over $mathbb{F}_5$. Let $m(X) = X^2+X+1$. Prove that $m(X)$ is irreducible over $mathbb{F}_5$.



Proof.



Suppose $m(X)$ is reducible then there exist $r(X),q(X)$ where $deg r(X)=deg q(X)= 1$



So by the quadratic formula we have



$$m(X)= left(X-left(frac{-1+3i}{2}right)right)left(X-left(frac{-1-3i}{2}right)right)$$



However $$frac{-1+3i}{2},frac{-1-3i}{2}notin mathbb{F}_5.$$



Therefore $m(X)$ is irreducible.



Would this be correct?






abstract-algebra polynomials proof-verification finite-fields irreducible-polynomials







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    Notice that $Bbb F_5={0,1,2,3,4}$
    $endgroup$
    – Masacroso
    Oct 7 '16 at 10:52






  • 1




    $begingroup$
    You can show (by simply plugging in all values) that $m(X)$ has no roots over $mathbb{F}_5$.
    $endgroup$
    – Janik
    Oct 7 '16 at 10:59






  • 2




    $begingroup$
    Not sure how you applied the quadratic formula. $i$, as in $sqrt {-1}$ is an element of $mathbb F_{,5}$ as $2^2=-1pmod 5$. But you should have had $sqrt {-3}$ and $-3$ is not a square $pmod 5$.
    $endgroup$
    – lulu
    Oct 7 '16 at 11:03












  • $begingroup$
    I'm sorry, I don't quite understand how modulus is involved? Isn't this a polynomial ring?
    $endgroup$
    – aoshdosi
    Oct 7 '16 at 11:12










  • $begingroup$
    Working in $mathbb F_{,5}$ is the same as working in the integers $pmod 5$.
    $endgroup$
    – lulu
    Oct 7 '16 at 11:20














2












2








2


2



$begingroup$


Question: $mathbb{F}_5={0,1,2,3,4,5}$ be the field with $5$ elements, let $mathbb{F}_5[X]$ be the polynomial ring over $mathbb{F}_5$. Let $m(X) = X^2+X+1$. Prove that $m(X)$ is irreducible over $mathbb{F}_5$.



Proof.



Suppose $m(X)$ is reducible then there exist $r(X),q(X)$ where $deg r(X)=deg q(X)= 1$



So by the quadratic formula we have



$$m(X)= left(X-left(frac{-1+3i}{2}right)right)left(X-left(frac{-1-3i}{2}right)right)$$



However $$frac{-1+3i}{2},frac{-1-3i}{2}notin mathbb{F}_5.$$



Therefore $m(X)$ is irreducible.



Would this be correct?






abstract-algebra polynomials proof-verification finite-fields irreducible-polynomials







share|cite|improve this question











$endgroup$




Question: $mathbb{F}_5={0,1,2,3,4,5}$ be the field with $5$ elements, let $mathbb{F}_5[X]$ be the polynomial ring over $mathbb{F}_5$. Let $m(X) = X^2+X+1$. Prove that $m(X)$ is irreducible over $mathbb{F}_5$.



Proof.



Suppose $m(X)$ is reducible then there exist $r(X),q(X)$ where $deg r(X)=deg q(X)= 1$



So by the quadratic formula we have



$$m(X)= left(X-left(frac{-1+3i}{2}right)right)left(X-left(frac{-1-3i}{2}right)right)$$



However $$frac{-1+3i}{2},frac{-1-3i}{2}notin mathbb{F}_5.$$



Therefore $m(X)$ is irreducible.



Would this be correct?






abstract-algebra polynomials proof-verification finite-fields irreducible-polynomials




abstract-algebra polynomials proof-verification finite-fields irreducible-polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Oct 7 '16 at 13:53









user26857

39.3k124183




39.3k124183










asked Oct 7 '16 at 10:43









aoshdosiaoshdosi

111




111








  • 2




    $begingroup$
    Notice that $Bbb F_5={0,1,2,3,4}$
    $endgroup$
    – Masacroso
    Oct 7 '16 at 10:52






  • 1




    $begingroup$
    You can show (by simply plugging in all values) that $m(X)$ has no roots over $mathbb{F}_5$.
    $endgroup$
    – Janik
    Oct 7 '16 at 10:59






  • 2




    $begingroup$
    Not sure how you applied the quadratic formula. $i$, as in $sqrt {-1}$ is an element of $mathbb F_{,5}$ as $2^2=-1pmod 5$. But you should have had $sqrt {-3}$ and $-3$ is not a square $pmod 5$.
    $endgroup$
    – lulu
    Oct 7 '16 at 11:03












  • $begingroup$
    I'm sorry, I don't quite understand how modulus is involved? Isn't this a polynomial ring?
    $endgroup$
    – aoshdosi
    Oct 7 '16 at 11:12










  • $begingroup$
    Working in $mathbb F_{,5}$ is the same as working in the integers $pmod 5$.
    $endgroup$
    – lulu
    Oct 7 '16 at 11:20














  • 2




    $begingroup$
    Notice that $Bbb F_5={0,1,2,3,4}$
    $endgroup$
    – Masacroso
    Oct 7 '16 at 10:52






  • 1




    $begingroup$
    You can show (by simply plugging in all values) that $m(X)$ has no roots over $mathbb{F}_5$.
    $endgroup$
    – Janik
    Oct 7 '16 at 10:59






  • 2




    $begingroup$
    Not sure how you applied the quadratic formula. $i$, as in $sqrt {-1}$ is an element of $mathbb F_{,5}$ as $2^2=-1pmod 5$. But you should have had $sqrt {-3}$ and $-3$ is not a square $pmod 5$.
    $endgroup$
    – lulu
    Oct 7 '16 at 11:03












  • $begingroup$
    I'm sorry, I don't quite understand how modulus is involved? Isn't this a polynomial ring?
    $endgroup$
    – aoshdosi
    Oct 7 '16 at 11:12










  • $begingroup$
    Working in $mathbb F_{,5}$ is the same as working in the integers $pmod 5$.
    $endgroup$
    – lulu
    Oct 7 '16 at 11:20








2




2




$begingroup$
Notice that $Bbb F_5={0,1,2,3,4}$
$endgroup$
– Masacroso
Oct 7 '16 at 10:52




$begingroup$
Notice that $Bbb F_5={0,1,2,3,4}$
$endgroup$
– Masacroso
Oct 7 '16 at 10:52




1




1




$begingroup$
You can show (by simply plugging in all values) that $m(X)$ has no roots over $mathbb{F}_5$.
$endgroup$
– Janik
Oct 7 '16 at 10:59




$begingroup$
You can show (by simply plugging in all values) that $m(X)$ has no roots over $mathbb{F}_5$.
$endgroup$
– Janik
Oct 7 '16 at 10:59




2




2




$begingroup$
Not sure how you applied the quadratic formula. $i$, as in $sqrt {-1}$ is an element of $mathbb F_{,5}$ as $2^2=-1pmod 5$. But you should have had $sqrt {-3}$ and $-3$ is not a square $pmod 5$.
$endgroup$
– lulu
Oct 7 '16 at 11:03






$begingroup$
Not sure how you applied the quadratic formula. $i$, as in $sqrt {-1}$ is an element of $mathbb F_{,5}$ as $2^2=-1pmod 5$. But you should have had $sqrt {-3}$ and $-3$ is not a square $pmod 5$.
$endgroup$
– lulu
Oct 7 '16 at 11:03














$begingroup$
I'm sorry, I don't quite understand how modulus is involved? Isn't this a polynomial ring?
$endgroup$
– aoshdosi
Oct 7 '16 at 11:12




$begingroup$
I'm sorry, I don't quite understand how modulus is involved? Isn't this a polynomial ring?
$endgroup$
– aoshdosi
Oct 7 '16 at 11:12












$begingroup$
Working in $mathbb F_{,5}$ is the same as working in the integers $pmod 5$.
$endgroup$
– lulu
Oct 7 '16 at 11:20




$begingroup$
Working in $mathbb F_{,5}$ is the same as working in the integers $pmod 5$.
$endgroup$
– lulu
Oct 7 '16 at 11:20










1 Answer
1






active

oldest

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1












$begingroup$

A quadratic polynomial with integer coefficients is irreducible over a finite field $mathbb{F}_p$ iff it has no root in $mathbb{F}_p$, i.e. iff its discriminant is not a square in $mathbb{F}_p$. Since the discriminant of $x^2+x+1$ is $-3$ and $-3equiv 2$ is not a square $!!pmod{5}$ (the set of squares is made by the residue classes $0,1,4$ only), it follows that $x^2+x+1$ is irreducible over $mathbb{F}_5$.



Notice that the complex roots of $x^2+x+1=Phi_3(x)$ are $frac{1pmsqrt{-3}}{2}$ and not what you wrote.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your help! I don't quite understand what you've done, where does the modulus get involved?
    $endgroup$
    – aoshdosi
    Oct 7 '16 at 11:13






  • 1




    $begingroup$
    @aoshdosi: $mathbb{F}_5$ is the finite field with $5$ elements. The prime $p$ defining $mathbb{F}_p$ has to play some role! In our case $p=5$. The squares $pmod{5}$ are $0,1,4$; the discriminant of our polynomial is not $0,1$ or $4pmod{5}$.
    $endgroup$
    – Jack D'Aurizio
    Oct 7 '16 at 11:16













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1 Answer
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$begingroup$

A quadratic polynomial with integer coefficients is irreducible over a finite field $mathbb{F}_p$ iff it has no root in $mathbb{F}_p$, i.e. iff its discriminant is not a square in $mathbb{F}_p$. Since the discriminant of $x^2+x+1$ is $-3$ and $-3equiv 2$ is not a square $!!pmod{5}$ (the set of squares is made by the residue classes $0,1,4$ only), it follows that $x^2+x+1$ is irreducible over $mathbb{F}_5$.



Notice that the complex roots of $x^2+x+1=Phi_3(x)$ are $frac{1pmsqrt{-3}}{2}$ and not what you wrote.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your help! I don't quite understand what you've done, where does the modulus get involved?
    $endgroup$
    – aoshdosi
    Oct 7 '16 at 11:13






  • 1




    $begingroup$
    @aoshdosi: $mathbb{F}_5$ is the finite field with $5$ elements. The prime $p$ defining $mathbb{F}_p$ has to play some role! In our case $p=5$. The squares $pmod{5}$ are $0,1,4$; the discriminant of our polynomial is not $0,1$ or $4pmod{5}$.
    $endgroup$
    – Jack D'Aurizio
    Oct 7 '16 at 11:16


















1












$begingroup$

A quadratic polynomial with integer coefficients is irreducible over a finite field $mathbb{F}_p$ iff it has no root in $mathbb{F}_p$, i.e. iff its discriminant is not a square in $mathbb{F}_p$. Since the discriminant of $x^2+x+1$ is $-3$ and $-3equiv 2$ is not a square $!!pmod{5}$ (the set of squares is made by the residue classes $0,1,4$ only), it follows that $x^2+x+1$ is irreducible over $mathbb{F}_5$.



Notice that the complex roots of $x^2+x+1=Phi_3(x)$ are $frac{1pmsqrt{-3}}{2}$ and not what you wrote.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you for your help! I don't quite understand what you've done, where does the modulus get involved?
    $endgroup$
    – aoshdosi
    Oct 7 '16 at 11:13






  • 1




    $begingroup$
    @aoshdosi: $mathbb{F}_5$ is the finite field with $5$ elements. The prime $p$ defining $mathbb{F}_p$ has to play some role! In our case $p=5$. The squares $pmod{5}$ are $0,1,4$; the discriminant of our polynomial is not $0,1$ or $4pmod{5}$.
    $endgroup$
    – Jack D'Aurizio
    Oct 7 '16 at 11:16
















1












1








1





$begingroup$

A quadratic polynomial with integer coefficients is irreducible over a finite field $mathbb{F}_p$ iff it has no root in $mathbb{F}_p$, i.e. iff its discriminant is not a square in $mathbb{F}_p$. Since the discriminant of $x^2+x+1$ is $-3$ and $-3equiv 2$ is not a square $!!pmod{5}$ (the set of squares is made by the residue classes $0,1,4$ only), it follows that $x^2+x+1$ is irreducible over $mathbb{F}_5$.



Notice that the complex roots of $x^2+x+1=Phi_3(x)$ are $frac{1pmsqrt{-3}}{2}$ and not what you wrote.






share|cite|improve this answer









$endgroup$



A quadratic polynomial with integer coefficients is irreducible over a finite field $mathbb{F}_p$ iff it has no root in $mathbb{F}_p$, i.e. iff its discriminant is not a square in $mathbb{F}_p$. Since the discriminant of $x^2+x+1$ is $-3$ and $-3equiv 2$ is not a square $!!pmod{5}$ (the set of squares is made by the residue classes $0,1,4$ only), it follows that $x^2+x+1$ is irreducible over $mathbb{F}_5$.



Notice that the complex roots of $x^2+x+1=Phi_3(x)$ are $frac{1pmsqrt{-3}}{2}$ and not what you wrote.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Oct 7 '16 at 11:09









Jack D'AurizioJack D'Aurizio

291k33284666




291k33284666












  • $begingroup$
    Thank you for your help! I don't quite understand what you've done, where does the modulus get involved?
    $endgroup$
    – aoshdosi
    Oct 7 '16 at 11:13






  • 1




    $begingroup$
    @aoshdosi: $mathbb{F}_5$ is the finite field with $5$ elements. The prime $p$ defining $mathbb{F}_p$ has to play some role! In our case $p=5$. The squares $pmod{5}$ are $0,1,4$; the discriminant of our polynomial is not $0,1$ or $4pmod{5}$.
    $endgroup$
    – Jack D'Aurizio
    Oct 7 '16 at 11:16




















  • $begingroup$
    Thank you for your help! I don't quite understand what you've done, where does the modulus get involved?
    $endgroup$
    – aoshdosi
    Oct 7 '16 at 11:13






  • 1




    $begingroup$
    @aoshdosi: $mathbb{F}_5$ is the finite field with $5$ elements. The prime $p$ defining $mathbb{F}_p$ has to play some role! In our case $p=5$. The squares $pmod{5}$ are $0,1,4$; the discriminant of our polynomial is not $0,1$ or $4pmod{5}$.
    $endgroup$
    – Jack D'Aurizio
    Oct 7 '16 at 11:16


















$begingroup$
Thank you for your help! I don't quite understand what you've done, where does the modulus get involved?
$endgroup$
– aoshdosi
Oct 7 '16 at 11:13




$begingroup$
Thank you for your help! I don't quite understand what you've done, where does the modulus get involved?
$endgroup$
– aoshdosi
Oct 7 '16 at 11:13




1




1




$begingroup$
@aoshdosi: $mathbb{F}_5$ is the finite field with $5$ elements. The prime $p$ defining $mathbb{F}_p$ has to play some role! In our case $p=5$. The squares $pmod{5}$ are $0,1,4$; the discriminant of our polynomial is not $0,1$ or $4pmod{5}$.
$endgroup$
– Jack D'Aurizio
Oct 7 '16 at 11:16






$begingroup$
@aoshdosi: $mathbb{F}_5$ is the finite field with $5$ elements. The prime $p$ defining $mathbb{F}_p$ has to play some role! In our case $p=5$. The squares $pmod{5}$ are $0,1,4$; the discriminant of our polynomial is not $0,1$ or $4pmod{5}$.
$endgroup$
– Jack D'Aurizio
Oct 7 '16 at 11:16




















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