How to prove a polynomial in $mathbb{Z}[x,y]$ is irreducible
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:
I am guesssing for the first part apply Eisenstein with $x$ as $(x)$ is prime ideal in $mathbb{Z}[x,y]$? Is this correct?
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
edited Jan 3 at 23:55
user26857
39.3k123983
asked Jan 3 at 22:35
sarafi
884
add a comment |
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:
I am guesssing for the first part apply Eisenstein with $x$ as $(x)$ is prime ideal in $mathbb{Z}[x,y]$? Is this correct?
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
edited Jan 3 at 23:55
user26857
39.3k123983
asked Jan 3 at 22:35
sarafi
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
add a comment |
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:
I am guesssing for the first part apply Eisenstein with $x$ as $(x)$ is prime ideal in $mathbb{Z}[x,y]$? Is this correct?
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
edited Jan 3 at 23:55
user26857
39.3k123983
asked Jan 3 at 22:35
sarafi
884
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:
I am guesssing for the first part apply Eisenstein with $x$ as $(x)$ is prime ideal in $mathbb{Z}[x,y]$? Is this correct?
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
edited Jan 3 at 23:55
user26857
39.3k123983
asked Jan 3 at 22:35
sarafi
884
edited Jan 3 at 23:55
user26857
39.3k123983
asked Jan 3 at 22:35
sarafi
884
edited Jan 3 at 23:55
user26857
39.3k123983
edited Jan 3 at 23:55
user26857
39.3k123983
edited Jan 3 at 23:55
user26857
39.3k123983
39.3k123983
asked Jan 3 at 22:35
sarafi
884
asked Jan 3 at 22:35
sarafi
884
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
add a comment |
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
add a comment |
1 Answer
1
active
oldest
votes
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 $mathbf Z/3mathbf Z$.
answered Jan 3 at 23:02
Bernard
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
add a comment |
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061100%2fhow-to-prove-a-polynomial-in-mathbbzx-y-is-irreducible%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $mathbf Z/3mathbf Z$.
answered Jan 3 at 23:02
Bernard
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
add a comment |
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 $mathbf Z/3mathbf Z$.
answered Jan 3 at 23:02
Bernard
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
add a comment |
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 $mathbf Z/3mathbf Z$.
answered Jan 3 at 23:02
Bernard
118k639112
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 $mathbf Z/3mathbf Z$.
answered Jan 3 at 23:02
Bernard
118k639112
answered Jan 3 at 23:02
Bernard
118k639112
answered Jan 3 at 23:02
Bernard
118k639112
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
add a comment |
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
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3061100%2fhow-to-prove-a-polynomial-in-mathbbzx-y-is-irreducible%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
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