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












0














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







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  • 1




    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 $I$.
    – mouthetics
    Jan 3 at 22:52


















0














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




















  • 1




    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 $I$.
    – mouthetics
    Jan 3 at 22:52
















0












0








0


0





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















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

39.3k123983




39.3k123983










asked Jan 3 at 22:35









sarafi

884




884








  • 1




    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 $I$.
    – mouthetics
    Jan 3 at 22:52
















  • 1




    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 $I$.
    – mouthetics
    Jan 3 at 22:52










1




1




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 $I$.
– mouthetics
Jan 3 at 22:52






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 $I$.
– mouthetics
Jan 3 at 22:52












1 Answer
1






active

oldest

votes


















2














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





















  • 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
    – sarafi
    Jan 4 at 0:52











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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














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





















  • 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
    – sarafi
    Jan 4 at 0:52
















2














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





















  • 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
    – sarafi
    Jan 4 at 0:52














2












2








2






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












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









Bernard

118k639112




118k639112












  • 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
    – sarafi
    Jan 4 at 0:52


















  • 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
    – sarafi
    Jan 4 at 0:52
















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
– sarafi
Jan 4 at 0:52




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
– sarafi
Jan 4 at 0:52


















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