How to prove a polynomial in $mathbb{Z}[x,y]$ is irreducible












0












$begingroup$


I had the following question:




Let $f(x,y) = y^5+xy^2+x in mathbb{Z}[x,y]$ and let:



$I_2 = (f(x,y),x-1,2)$ and $I_3 = (f(x,y),x-1,3)$ be two ideals in $mathbb{Z}[x,y]$.



Prove that $f(x,y)$ is irreducible, and determine which of these ideals is maximal.




My ideas:




  1. I am guesssing for the first part apply Eisenstein with $x$ as $(x)$ is prime ideal in $mathbb{Z}[x,y]$? Is this correct?


  2. For part $2$, I know that an ideal $I$ of $R$ is maximal iff $R/I$ is a field. How would I apply this here?



Thanks for any help.






abstract-algebra ideals irreducible-polynomials







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    For the first one yes you can use the Eisenstein criteria with $x$. For the second note that if you mod out by $<I,p>$, where $I$ is an ideal and $p$ a prime, is the same as you mod out the coefficients ring by $p$ and then mod out by $I$.
    $endgroup$
    – mouthetics
    Jan 3 at 22:52


















0












$begingroup$


I had the following question:




Let $f(x,y) = y^5+xy^2+x in mathbb{Z}[x,y]$ and let:



$I_2 = (f(x,y),x-1,2)$ and $I_3 = (f(x,y),x-1,3)$ be two ideals in $mathbb{Z}[x,y]$.



Prove that $f(x,y)$ is irreducible, and determine which of these ideals is maximal.




My ideas:




  1. I am guesssing for the first part apply Eisenstein with $x$ as $(x)$ is prime ideal in $mathbb{Z}[x,y]$? Is this correct?


  2. For part $2$, I know that an ideal $I$ of $R$ is maximal iff $R/I$ is a field. How would I apply this here?



Thanks for any help.






abstract-algebra ideals irreducible-polynomials







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    For the first one yes you can use the Eisenstein criteria with $x$. For the second note that if you mod out by $<I,p>$, where $I$ is an ideal and $p$ a prime, is the same as you mod out the coefficients ring by $p$ and then mod out by $I$.
    $endgroup$
    – mouthetics
    Jan 3 at 22:52
















0












0








0


0



$begingroup$


I had the following question:




Let $f(x,y) = y^5+xy^2+x in mathbb{Z}[x,y]$ and let:



$I_2 = (f(x,y),x-1,2)$ and $I_3 = (f(x,y),x-1,3)$ be two ideals in $mathbb{Z}[x,y]$.



Prove that $f(x,y)$ is irreducible, and determine which of these ideals is maximal.




My ideas:




  1. I am guesssing for the first part apply Eisenstein with $x$ as $(x)$ is prime ideal in $mathbb{Z}[x,y]$? Is this correct?


  2. For part $2$, I know that an ideal $I$ of $R$ is maximal iff $R/I$ is a field. How would I apply this here?



Thanks for any help.






abstract-algebra ideals irreducible-polynomials







share|cite|improve this question











$endgroup$




I had the following question:




Let $f(x,y) = y^5+xy^2+x in mathbb{Z}[x,y]$ and let:



$I_2 = (f(x,y),x-1,2)$ and $I_3 = (f(x,y),x-1,3)$ be two ideals in $mathbb{Z}[x,y]$.



Prove that $f(x,y)$ is irreducible, and determine which of these ideals is maximal.




My ideas:




  1. I am guesssing for the first part apply Eisenstein with $x$ as $(x)$ is prime ideal in $mathbb{Z}[x,y]$? Is this correct?


  2. For part $2$, I know that an ideal $I$ of $R$ is maximal iff $R/I$ is a field. How would I apply this here?



Thanks for any help.






abstract-algebra ideals irreducible-polynomials




abstract-algebra ideals irreducible-polynomials






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 23:55









user26857

39.3k124183




39.3k124183










asked Jan 3 at 22:35









sarafisarafi

884




884








  • 1




    $begingroup$
    For the first one yes you can use the Eisenstein criteria with $x$. For the second note that if you mod out by $<I,p>$, where $I$ is an ideal and $p$ a prime, is the same as you mod out the coefficients ring by $p$ and then mod out by $I$.
    $endgroup$
    – mouthetics
    Jan 3 at 22:52
















  • 1




    $begingroup$
    For the first one yes you can use the Eisenstein criteria with $x$. For the second note that if you mod out by $<I,p>$, where $I$ is an ideal and $p$ a prime, is the same as you mod out the coefficients ring by $p$ and then mod out by $I$.
    $endgroup$
    – mouthetics
    Jan 3 at 22:52










1




1




$begingroup$
For the first one yes you can use the Eisenstein criteria with $x$. For the second note that if you mod out by $<I,p>$, where $I$ is an ideal and $p$ a prime, is the same as you mod out the coefficients ring by $p$ and then mod out by $I$.
$endgroup$
– mouthetics
Jan 3 at 22:52






$begingroup$
For the first one yes you can use the Eisenstein criteria with $x$. For the second note that if you mod out by $<I,p>$, where $I$ is an ideal and $p$ a prime, is the same as you mod out the coefficients ring by $p$ and then mod out by $I$.
$endgroup$
– mouthetics
Jan 3 at 22:52












1 Answer
1






active

oldest

votes


















2












$begingroup$

For part 1, it's perfectly correc.



Hint for part 2: $;mathbf Z[x,y]/I_2simeq (mathbf Z/2mathbf Z)[y]bigm/bigl(f(1,y)bigr)= (mathbf Z/2mathbf Z)[y]bigm/(y^5+y^2+1))$.



Obviously it has no root in $mathbf Z/2mathbf Z$, so you only have to determine whether it has a quadratic factor or not.



Similar method for $I_3$, but you'll have to compute in $mathbf Z/3mathbf Z$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much! So....awkward question, I don't exactly know how to factor over finite fields. I know Berlekamp's algorithm is a thing that exists but other than that, nothing. Is there an easy way to factor without resorting to the heavy machinery of Berlekamp? Thanks
    $endgroup$
    – sarafi
    Jan 4 at 0:52











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

For part 1, it's perfectly correc.



Hint for part 2: $;mathbf Z[x,y]/I_2simeq (mathbf Z/2mathbf Z)[y]bigm/bigl(f(1,y)bigr)= (mathbf Z/2mathbf Z)[y]bigm/(y^5+y^2+1))$.



Obviously it has no root in $mathbf Z/2mathbf Z$, so you only have to determine whether it has a quadratic factor or not.



Similar method for $I_3$, but you'll have to compute in $mathbf Z/3mathbf Z$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much! So....awkward question, I don't exactly know how to factor over finite fields. I know Berlekamp's algorithm is a thing that exists but other than that, nothing. Is there an easy way to factor without resorting to the heavy machinery of Berlekamp? Thanks
    $endgroup$
    – sarafi
    Jan 4 at 0:52
















2












$begingroup$

For part 1, it's perfectly correc.



Hint for part 2: $;mathbf Z[x,y]/I_2simeq (mathbf Z/2mathbf Z)[y]bigm/bigl(f(1,y)bigr)= (mathbf Z/2mathbf Z)[y]bigm/(y^5+y^2+1))$.



Obviously it has no root in $mathbf Z/2mathbf Z$, so you only have to determine whether it has a quadratic factor or not.



Similar method for $I_3$, but you'll have to compute in $mathbf Z/3mathbf Z$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you so much! So....awkward question, I don't exactly know how to factor over finite fields. I know Berlekamp's algorithm is a thing that exists but other than that, nothing. Is there an easy way to factor without resorting to the heavy machinery of Berlekamp? Thanks
    $endgroup$
    – sarafi
    Jan 4 at 0:52














2












2








2





$begingroup$

For part 1, it's perfectly correc.



Hint for part 2: $;mathbf Z[x,y]/I_2simeq (mathbf Z/2mathbf Z)[y]bigm/bigl(f(1,y)bigr)= (mathbf Z/2mathbf Z)[y]bigm/(y^5+y^2+1))$.



Obviously it has no root in $mathbf Z/2mathbf Z$, so you only have to determine whether it has a quadratic factor or not.



Similar method for $I_3$, but you'll have to compute in $mathbf Z/3mathbf Z$.






share|cite|improve this answer









$endgroup$



For part 1, it's perfectly correc.



Hint for part 2: $;mathbf Z[x,y]/I_2simeq (mathbf Z/2mathbf Z)[y]bigm/bigl(f(1,y)bigr)= (mathbf Z/2mathbf Z)[y]bigm/(y^5+y^2+1))$.



Obviously it has no root in $mathbf Z/2mathbf Z$, so you only have to determine whether it has a quadratic factor or not.



Similar method for $I_3$, but you'll have to compute in $mathbf Z/3mathbf Z$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 3 at 23:02









BernardBernard

122k740116




122k740116












  • $begingroup$
    Thank you so much! So....awkward question, I don't exactly know how to factor over finite fields. I know Berlekamp's algorithm is a thing that exists but other than that, nothing. Is there an easy way to factor without resorting to the heavy machinery of Berlekamp? Thanks
    $endgroup$
    – sarafi
    Jan 4 at 0:52


















  • $begingroup$
    Thank you so much! So....awkward question, I don't exactly know how to factor over finite fields. I know Berlekamp's algorithm is a thing that exists but other than that, nothing. Is there an easy way to factor without resorting to the heavy machinery of Berlekamp? Thanks
    $endgroup$
    – sarafi
    Jan 4 at 0:52
















$begingroup$
Thank you so much! So....awkward question, I don't exactly know how to factor over finite fields. I know Berlekamp's algorithm is a thing that exists but other than that, nothing. Is there an easy way to factor without resorting to the heavy machinery of Berlekamp? Thanks
$endgroup$
– sarafi
Jan 4 at 0:52




$begingroup$
Thank you so much! So....awkward question, I don't exactly know how to factor over finite fields. I know Berlekamp's algorithm is a thing that exists but other than that, nothing. Is there an easy way to factor without resorting to the heavy machinery of Berlekamp? Thanks
$endgroup$
– sarafi
Jan 4 at 0:52


















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