Mixing
$x^2+y^2-1$ is irreducible in $mathbb{Q}[x,y]$.
$begingroup$
To solve the Dummit-Foote's exercise I've stuck here with this problem : (P-312)
Qn11. Show that $p(x)=x^2+y^2-1$ is irreducible in $mathbb{Q}[x,y]$.
This is a polynomial of degree 2. But I can not think of zeros of this polynomial.
My another approach is : $p(x)=x^2-1in mathbb{Q}[x,y]/langle yrangle$ and also $p(x)=y^2-1in mathbb{Q}[x,y]/langle xrangle$ and $langle xrangle$, $langle yrangle$ are proper ideal of $mathbb{Q}[x,y]$. Also, $p(x)=x^2-1$ can be factored with $(x+1)(x-1)$ in $mathbb{Q}[x,y]/langle yrangle$. Then how I go ahead?
I think my attempt is stupid. Can any one suggest any references to understand/solve this?
Thanks in advance.
abstract-algebra ring-theory field-theory irreducible-polynomials
edited Feb 1 at 12:10
user3482749
4,3291119
asked Jan 31 at 16:41
BijanDattaBijanDatta
309113
$endgroup$
add a comment |
$begingroup$
To solve the Dummit-Foote's exercise I've stuck here with this problem : (P-312)
Qn11. Show that $p(x)=x^2+y^2-1$ is irreducible in $mathbb{Q}[x,y]$.
This is a polynomial of degree 2. But I can not think of zeros of this polynomial.
My another approach is : $p(x)=x^2-1in mathbb{Q}[x,y]/langle yrangle$ and also $p(x)=y^2-1in mathbb{Q}[x,y]/langle xrangle$ and $langle xrangle$, $langle yrangle$ are proper ideal of $mathbb{Q}[x,y]$. Also, $p(x)=x^2-1$ can be factored with $(x+1)(x-1)$ in $mathbb{Q}[x,y]/langle yrangle$. Then how I go ahead?
I think my attempt is stupid. Can any one suggest any references to understand/solve this?
Thanks in advance.
abstract-algebra ring-theory field-theory irreducible-polynomials
edited Feb 1 at 12:10
user3482749
4,3291119
asked Jan 31 at 16:41
BijanDattaBijanDatta
309113
$endgroup$
3
$begingroup$
What is your trying?
$endgroup$
– Michael Rozenberg
Jan 31 at 16:47
$begingroup$
Do you know a powerful criterion for irreducibility for Q ?
$endgroup$
– Maman
Jan 31 at 16:53
$begingroup$
> But I can not think of zeros of this polynomial. Really? There are four very obvious ones.
$endgroup$
– user3482749
Feb 1 at 12:11
add a comment |
$begingroup$
To solve the Dummit-Foote's exercise I've stuck here with this problem : (P-312)
Qn11. Show that $p(x)=x^2+y^2-1$ is irreducible in $mathbb{Q}[x,y]$.
This is a polynomial of degree 2. But I can not think of zeros of this polynomial.
My another approach is : $p(x)=x^2-1in mathbb{Q}[x,y]/langle yrangle$ and also $p(x)=y^2-1in mathbb{Q}[x,y]/langle xrangle$ and $langle xrangle$, $langle yrangle$ are proper ideal of $mathbb{Q}[x,y]$. Also, $p(x)=x^2-1$ can be factored with $(x+1)(x-1)$ in $mathbb{Q}[x,y]/langle yrangle$. Then how I go ahead?
I think my attempt is stupid. Can any one suggest any references to understand/solve this?
Thanks in advance.
abstract-algebra ring-theory field-theory irreducible-polynomials
edited Feb 1 at 12:10
user3482749
4,3291119
asked Jan 31 at 16:41
BijanDattaBijanDatta
309113
$endgroup$
To solve the Dummit-Foote's exercise I've stuck here with this problem : (P-312)
Qn11. Show that $p(x)=x^2+y^2-1$ is irreducible in $mathbb{Q}[x,y]$.
This is a polynomial of degree 2. But I can not think of zeros of this polynomial.
My another approach is : $p(x)=x^2-1in mathbb{Q}[x,y]/langle yrangle$ and also $p(x)=y^2-1in mathbb{Q}[x,y]/langle xrangle$ and $langle xrangle$, $langle yrangle$ are proper ideal of $mathbb{Q}[x,y]$. Also, $p(x)=x^2-1$ can be factored with $(x+1)(x-1)$ in $mathbb{Q}[x,y]/langle yrangle$. Then how I go ahead?
I think my attempt is stupid. Can any one suggest any references to understand/solve this?
Thanks in advance.
abstract-algebra ring-theory field-theory irreducible-polynomials
abstract-algebra ring-theory field-theory irreducible-polynomials
edited Feb 1 at 12:10
user3482749
4,3291119
asked Jan 31 at 16:41
BijanDattaBijanDatta
309113
edited Feb 1 at 12:10
user3482749
4,3291119
asked Jan 31 at 16:41
BijanDattaBijanDatta
309113
edited Feb 1 at 12:10
user3482749
4,3291119
edited Feb 1 at 12:10
user3482749
4,3291119
edited Feb 1 at 12:10
user3482749
4,3291119
4,3291119
asked Jan 31 at 16:41
BijanDattaBijanDatta
309113
asked Jan 31 at 16:41
BijanDattaBijanDatta
309113
asked Jan 31 at 16:41
BijanDattaBijanDatta
309113
309113
3
$begingroup$
What is your trying?
$endgroup$
– Michael Rozenberg
Jan 31 at 16:47
$begingroup$
Do you know a powerful criterion for irreducibility for Q ?
$endgroup$
– Maman
Jan 31 at 16:53
$begingroup$
> But I can not think of zeros of this polynomial. Really? There are four very obvious ones.
$endgroup$
– user3482749
Feb 1 at 12:11
add a comment |
3
$begingroup$
What is your trying?
$endgroup$
– Michael Rozenberg
Jan 31 at 16:47
$begingroup$
Do you know a powerful criterion for irreducibility for Q ?
$endgroup$
– Maman
Jan 31 at 16:53
$begingroup$
> But I can not think of zeros of this polynomial. Really? There are four very obvious ones.
$endgroup$
– user3482749
Feb 1 at 12:11
3
3
$begingroup$
What is your trying?
$endgroup$
– Michael Rozenberg
Jan 31 at 16:47
$begingroup$
What is your trying?
$endgroup$
– Michael Rozenberg
Jan 31 at 16:47
$begingroup$
Do you know a powerful criterion for irreducibility for Q ?
$endgroup$
– Maman
Jan 31 at 16:53
$begingroup$
Do you know a powerful criterion for irreducibility for Q ?
$endgroup$
– Maman
Jan 31 at 16:53
$begingroup$
> But I can not think of zeros of this polynomial. Really? There are four very obvious ones.
$endgroup$
– user3482749
Feb 1 at 12:11
$begingroup$
> But I can not think of zeros of this polynomial. Really? There are four very obvious ones.
$endgroup$
– user3482749
Feb 1 at 12:11
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Hint: consider $x^2+y^2-1inmathbb{Q}[y][x]$ and use a well-known irreducibility criterion holding over every PID.
answered Feb 1 at 12:29
egregegreg
185k1486208
$endgroup$
$begingroup$
And what is that powerful irreducibility criterion that works over any PID? I'm asking just for my own learning. And does it work over any $R[x_1,cdots,x_n]$ where $R$ is a PID?
$endgroup$
– stressed out
Feb 14 at 15:43
1
$begingroup$
@stressedout Eisenstein's criterion. Only on a PID.
$endgroup$
– egreg
Feb 14 at 16:30
$begingroup$
Thanks. The Wikipedia link doesn't discuss its generalization to PIDs, but it says something about generalizing it to $D[x]$ where $D$ is an integral domain. I hope I'm not bothering you by asking too many questions, but could you elaborate further? And the statement is for single variable polynomials. How does it apply to multivariable polynomials?
$endgroup$
– stressed out
Feb 14 at 16:43
add a comment |
$begingroup$
Just use the definition of irreducible element. Suppose
$$
x^2+y^2-1 = p(x,y)cdot q(x,y),
$$
with $p,qin mathbb{Q}[x,y]$, so that $2 = deg_x p + deg_x q$ and $2 = deg_y p + deg_y q$. The only case to consider is when both $p$ and $q$ are linear in $x$ and $y$. We can write them explicitely and reach a contradiction when imposing that their product must be equal to $x^2+y^2-1$.
An approach which is not brute force is to look at $x^2+y^2-1$ as a polynomial in $mathbb{Q}[y][x]$ and use one irreducibility criterion.
answered Feb 1 at 12:32
AlessioDVAlessioDV
1,208114
$endgroup$
add a comment |
$begingroup$
Hint: if a polynomial $p(x,y)$ over $K$ (any field, or in fact any domain) is reducible, then either for all $kin K$, the polynomials $p(x,k)$ are reducible or for all $kin K$, the polynomials $p(k,y)$ are reducible. This is clearly not true for $K=mathbf Q$ and your $p$.
answered Feb 1 at 12:40
tomasztomasz
24k23482
$endgroup$
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%2f3095115%2fx2y2-1-is-irreducible-in-mathbbqx-y%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint: consider $x^2+y^2-1inmathbb{Q}[y][x]$ and use a well-known irreducibility criterion holding over every PID.
answered Feb 1 at 12:29
egregegreg
185k1486208
$endgroup$
$begingroup$
And what is that powerful irreducibility criterion that works over any PID? I'm asking just for my own learning. And does it work over any $R[x_1,cdots,x_n]$ where $R$ is a PID?
$endgroup$
– stressed out
Feb 14 at 15:43
1
$begingroup$
@stressedout Eisenstein's criterion. Only on a PID.
$endgroup$
– egreg
Feb 14 at 16:30
$begingroup$
Thanks. The Wikipedia link doesn't discuss its generalization to PIDs, but it says something about generalizing it to $D[x]$ where $D$ is an integral domain. I hope I'm not bothering you by asking too many questions, but could you elaborate further? And the statement is for single variable polynomials. How does it apply to multivariable polynomials?
$endgroup$
– stressed out
Feb 14 at 16:43
add a comment |
$begingroup$
Hint: consider $x^2+y^2-1inmathbb{Q}[y][x]$ and use a well-known irreducibility criterion holding over every PID.
answered Feb 1 at 12:29
egregegreg
185k1486208
$endgroup$
$begingroup$
And what is that powerful irreducibility criterion that works over any PID? I'm asking just for my own learning. And does it work over any $R[x_1,cdots,x_n]$ where $R$ is a PID?
$endgroup$
– stressed out
Feb 14 at 15:43
1
$begingroup$
@stressedout Eisenstein's criterion. Only on a PID.
$endgroup$
– egreg
Feb 14 at 16:30
$begingroup$
Thanks. The Wikipedia link doesn't discuss its generalization to PIDs, but it says something about generalizing it to $D[x]$ where $D$ is an integral domain. I hope I'm not bothering you by asking too many questions, but could you elaborate further? And the statement is for single variable polynomials. How does it apply to multivariable polynomials?
$endgroup$
– stressed out
Feb 14 at 16:43
add a comment |
$begingroup$
Hint: consider $x^2+y^2-1inmathbb{Q}[y][x]$ and use a well-known irreducibility criterion holding over every PID.
answered Feb 1 at 12:29
egregegreg
185k1486208
$endgroup$
Hint: consider $x^2+y^2-1inmathbb{Q}[y][x]$ and use a well-known irreducibility criterion holding over every PID.
answered Feb 1 at 12:29
egregegreg
185k1486208
answered Feb 1 at 12:29
egregegreg
185k1486208
answered Feb 1 at 12:29
egregegreg
185k1486208
answered Feb 1 at 12:29
egregegreg
185k1486208
185k1486208
$begingroup$
And what is that powerful irreducibility criterion that works over any PID? I'm asking just for my own learning. And does it work over any $R[x_1,cdots,x_n]$ where $R$ is a PID?
$endgroup$
– stressed out
Feb 14 at 15:43
1
$begingroup$
@stressedout Eisenstein's criterion. Only on a PID.
$endgroup$
– egreg
Feb 14 at 16:30
$begingroup$
Thanks. The Wikipedia link doesn't discuss its generalization to PIDs, but it says something about generalizing it to $D[x]$ where $D$ is an integral domain. I hope I'm not bothering you by asking too many questions, but could you elaborate further? And the statement is for single variable polynomials. How does it apply to multivariable polynomials?
$endgroup$
– stressed out
Feb 14 at 16:43
add a comment |
$begingroup$
And what is that powerful irreducibility criterion that works over any PID? I'm asking just for my own learning. And does it work over any $R[x_1,cdots,x_n]$ where $R$ is a PID?
$endgroup$
– stressed out
Feb 14 at 15:43
1
$begingroup$
@stressedout Eisenstein's criterion. Only on a PID.
$endgroup$
– egreg
Feb 14 at 16:30
$begingroup$
Thanks. The Wikipedia link doesn't discuss its generalization to PIDs, but it says something about generalizing it to $D[x]$ where $D$ is an integral domain. I hope I'm not bothering you by asking too many questions, but could you elaborate further? And the statement is for single variable polynomials. How does it apply to multivariable polynomials?
$endgroup$
– stressed out
Feb 14 at 16:43
$begingroup$
And what is that powerful irreducibility criterion that works over any PID? I'm asking just for my own learning. And does it work over any $R[x_1,cdots,x_n]$ where $R$ is a PID?
$endgroup$
– stressed out
Feb 14 at 15:43
$begingroup$
And what is that powerful irreducibility criterion that works over any PID? I'm asking just for my own learning. And does it work over any $R[x_1,cdots,x_n]$ where $R$ is a PID?
$endgroup$
– stressed out
Feb 14 at 15:43
1
1
$begingroup$
@stressedout Eisenstein's criterion. Only on a PID.
$endgroup$
– egreg
Feb 14 at 16:30
$begingroup$
@stressedout Eisenstein's criterion. Only on a PID.
$endgroup$
– egreg
Feb 14 at 16:30
$begingroup$
Thanks. The Wikipedia link doesn't discuss its generalization to PIDs, but it says something about generalizing it to $D[x]$ where $D$ is an integral domain. I hope I'm not bothering you by asking too many questions, but could you elaborate further? And the statement is for single variable polynomials. How does it apply to multivariable polynomials?
$endgroup$
– stressed out
Feb 14 at 16:43
$begingroup$
Thanks. The Wikipedia link doesn't discuss its generalization to PIDs, but it says something about generalizing it to $D[x]$ where $D$ is an integral domain. I hope I'm not bothering you by asking too many questions, but could you elaborate further? And the statement is for single variable polynomials. How does it apply to multivariable polynomials?
$endgroup$
– stressed out
Feb 14 at 16:43
add a comment |
$begingroup$
Just use the definition of irreducible element. Suppose
$$
x^2+y^2-1 = p(x,y)cdot q(x,y),
$$
with $p,qin mathbb{Q}[x,y]$, so that $2 = deg_x p + deg_x q$ and $2 = deg_y p + deg_y q$. The only case to consider is when both $p$ and $q$ are linear in $x$ and $y$. We can write them explicitely and reach a contradiction when imposing that their product must be equal to $x^2+y^2-1$.
An approach which is not brute force is to look at $x^2+y^2-1$ as a polynomial in $mathbb{Q}[y][x]$ and use one irreducibility criterion.
answered Feb 1 at 12:32
AlessioDVAlessioDV
1,208114
$endgroup$
add a comment |
$begingroup$
Just use the definition of irreducible element. Suppose
$$
x^2+y^2-1 = p(x,y)cdot q(x,y),
$$
with $p,qin mathbb{Q}[x,y]$, so that $2 = deg_x p + deg_x q$ and $2 = deg_y p + deg_y q$. The only case to consider is when both $p$ and $q$ are linear in $x$ and $y$. We can write them explicitely and reach a contradiction when imposing that their product must be equal to $x^2+y^2-1$.
An approach which is not brute force is to look at $x^2+y^2-1$ as a polynomial in $mathbb{Q}[y][x]$ and use one irreducibility criterion.
answered Feb 1 at 12:32
AlessioDVAlessioDV
1,208114
$endgroup$
add a comment |
$begingroup$
Just use the definition of irreducible element. Suppose
$$
x^2+y^2-1 = p(x,y)cdot q(x,y),
$$
with $p,qin mathbb{Q}[x,y]$, so that $2 = deg_x p + deg_x q$ and $2 = deg_y p + deg_y q$. The only case to consider is when both $p$ and $q$ are linear in $x$ and $y$. We can write them explicitely and reach a contradiction when imposing that their product must be equal to $x^2+y^2-1$.
An approach which is not brute force is to look at $x^2+y^2-1$ as a polynomial in $mathbb{Q}[y][x]$ and use one irreducibility criterion.
answered Feb 1 at 12:32
AlessioDVAlessioDV
1,208114
$endgroup$
Just use the definition of irreducible element. Suppose
$$
x^2+y^2-1 = p(x,y)cdot q(x,y),
$$
with $p,qin mathbb{Q}[x,y]$, so that $2 = deg_x p + deg_x q$ and $2 = deg_y p + deg_y q$. The only case to consider is when both $p$ and $q$ are linear in $x$ and $y$. We can write them explicitely and reach a contradiction when imposing that their product must be equal to $x^2+y^2-1$.
An approach which is not brute force is to look at $x^2+y^2-1$ as a polynomial in $mathbb{Q}[y][x]$ and use one irreducibility criterion.
answered Feb 1 at 12:32
AlessioDVAlessioDV
1,208114
answered Feb 1 at 12:32
AlessioDVAlessioDV
1,208114
answered Feb 1 at 12:32
AlessioDVAlessioDV
1,208114
answered Feb 1 at 12:32
AlessioDVAlessioDV
1,208114
1,208114
add a comment |
add a comment |
$begingroup$
Hint: if a polynomial $p(x,y)$ over $K$ (any field, or in fact any domain) is reducible, then either for all $kin K$, the polynomials $p(x,k)$ are reducible or for all $kin K$, the polynomials $p(k,y)$ are reducible. This is clearly not true for $K=mathbf Q$ and your $p$.
answered Feb 1 at 12:40
tomasztomasz
24k23482
$endgroup$
add a comment |
$begingroup$
Hint: if a polynomial $p(x,y)$ over $K$ (any field, or in fact any domain) is reducible, then either for all $kin K$, the polynomials $p(x,k)$ are reducible or for all $kin K$, the polynomials $p(k,y)$ are reducible. This is clearly not true for $K=mathbf Q$ and your $p$.
answered Feb 1 at 12:40
tomasztomasz
24k23482
$endgroup$
add a comment |
$begingroup$
Hint: if a polynomial $p(x,y)$ over $K$ (any field, or in fact any domain) is reducible, then either for all $kin K$, the polynomials $p(x,k)$ are reducible or for all $kin K$, the polynomials $p(k,y)$ are reducible. This is clearly not true for $K=mathbf Q$ and your $p$.
answered Feb 1 at 12:40
tomasztomasz
24k23482
$endgroup$
Hint: if a polynomial $p(x,y)$ over $K$ (any field, or in fact any domain) is reducible, then either for all $kin K$, the polynomials $p(x,k)$ are reducible or for all $kin K$, the polynomials $p(k,y)$ are reducible. This is clearly not true for $K=mathbf Q$ and your $p$.
answered Feb 1 at 12:40
tomasztomasz
24k23482
answered Feb 1 at 12:40
tomasztomasz
24k23482
answered Feb 1 at 12:40
tomasztomasz
24k23482
answered Feb 1 at 12:40
tomasztomasz
24k23482
24k23482
add a comment |
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.
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%2f3095115%2fx2y2-1-is-irreducible-in-mathbbqx-y%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
3
$begingroup$
What is your trying?
$endgroup$
– Michael Rozenberg
Jan 31 at 16:47
$begingroup$
Do you know a powerful criterion for irreducibility for Q ?
$endgroup$
– Maman
Jan 31 at 16:53
$begingroup$
> But I can not think of zeros of this polynomial. Really? There are four very obvious ones.
$endgroup$
– user3482749
Feb 1 at 12:11