Mixing
$X^4-4X^2-1$ irreducible over $mathbb{Q}[X]$
$begingroup$
I want to show, that $X^4-4X^2-1$ is irreducible over $mathbb{Q}[X]$.
Since there are no roots in $mathbb{Q}$ it has to be:
$(X^4-4X^2-1)=(X^2+aX+b)(X^2+cX+d)$
Comparision of the coefficients shows that this can not hold over $mathbb{Q}$. But I am searching for an easier way to show this which uses less calculation.
I tried this:
$X^4-4X^2-1=X^4-4X^2+4-5=(X^2-2)^2-5=(X^2-2-sqrt{5})(X^2-2+sqrt{5})$
Which is obviously not in $mathbb{Q}[X]$ anymore.
But would this calculation be enough to show, that $X^4-4X^2-1$ is irreducible.
How do I know, that this is the only possible way to factor it?
Thanks in advance.
abstract-algebra polynomials field-theory irreducible-polynomials
edited Sep 26 '17 at 23:18
José Carlos Santos
172k22132239
asked Sep 26 '17 at 22:51
CornmanCornman
3,37121229
$endgroup$
|
show 2 more comments
$begingroup$
I want to show, that $X^4-4X^2-1$ is irreducible over $mathbb{Q}[X]$.
Since there are no roots in $mathbb{Q}$ it has to be:
$(X^4-4X^2-1)=(X^2+aX+b)(X^2+cX+d)$
Comparision of the coefficients shows that this can not hold over $mathbb{Q}$. But I am searching for an easier way to show this which uses less calculation.
I tried this:
$X^4-4X^2-1=X^4-4X^2+4-5=(X^2-2)^2-5=(X^2-2-sqrt{5})(X^2-2+sqrt{5})$
Which is obviously not in $mathbb{Q}[X]$ anymore.
But would this calculation be enough to show, that $X^4-4X^2-1$ is irreducible.
How do I know, that this is the only possible way to factor it?
Thanks in advance.
abstract-algebra polynomials field-theory irreducible-polynomials
edited Sep 26 '17 at 23:18
José Carlos Santos
172k22132239
asked Sep 26 '17 at 22:51
CornmanCornman
3,37121229
$endgroup$
1
$begingroup$
This doesn't answer the exact question being asked here, but as a tangential aside, this polynomial is irreducible $pmod{p}$ for a small $p$ and hence irreducible over $mathbb{Q}$.
$endgroup$
– Kaj Hansen
Sep 26 '17 at 22:59
1
$begingroup$
Look up the rational root theorem.
$endgroup$
– Doug M
Sep 26 '17 at 23:00
$begingroup$
@KajHansen: But using the reduction theorem (which I think you are refering too) not involve a similar calculation, which might be easier?
$endgroup$
– Cornman
Sep 26 '17 at 23:04
3
$begingroup$
@DougM: The rational root theroem should not work here, since the degree of the polynomial is 4. The non existence of a root does not mean the polynomial is irreducible.
$endgroup$
– Cornman
Sep 26 '17 at 23:05
$begingroup$
@Cornman, the nice thing about it is that the number of irreducible quadratic polynomials in this particular $mathbb{F}_p$ is quite small--small enough to just try them all. On the other hand, the infinitude of polynomials in $mathbb{Q}[x]$ makes the current endeavor tricky.
$endgroup$
– Kaj Hansen
Sep 26 '17 at 23:05
|
show 2 more comments
$begingroup$
I want to show, that $X^4-4X^2-1$ is irreducible over $mathbb{Q}[X]$.
Since there are no roots in $mathbb{Q}$ it has to be:
$(X^4-4X^2-1)=(X^2+aX+b)(X^2+cX+d)$
Comparision of the coefficients shows that this can not hold over $mathbb{Q}$. But I am searching for an easier way to show this which uses less calculation.
I tried this:
$X^4-4X^2-1=X^4-4X^2+4-5=(X^2-2)^2-5=(X^2-2-sqrt{5})(X^2-2+sqrt{5})$
Which is obviously not in $mathbb{Q}[X]$ anymore.
But would this calculation be enough to show, that $X^4-4X^2-1$ is irreducible.
How do I know, that this is the only possible way to factor it?
Thanks in advance.
abstract-algebra polynomials field-theory irreducible-polynomials
edited Sep 26 '17 at 23:18
José Carlos Santos
172k22132239
asked Sep 26 '17 at 22:51
CornmanCornman
3,37121229
$endgroup$
I want to show, that $X^4-4X^2-1$ is irreducible over $mathbb{Q}[X]$.
Since there are no roots in $mathbb{Q}$ it has to be:
$(X^4-4X^2-1)=(X^2+aX+b)(X^2+cX+d)$
Comparision of the coefficients shows that this can not hold over $mathbb{Q}$. But I am searching for an easier way to show this which uses less calculation.
I tried this:
$X^4-4X^2-1=X^4-4X^2+4-5=(X^2-2)^2-5=(X^2-2-sqrt{5})(X^2-2+sqrt{5})$
Which is obviously not in $mathbb{Q}[X]$ anymore.
But would this calculation be enough to show, that $X^4-4X^2-1$ is irreducible.
How do I know, that this is the only possible way to factor it?
Thanks in advance.
abstract-algebra polynomials field-theory irreducible-polynomials
abstract-algebra polynomials field-theory irreducible-polynomials
edited Sep 26 '17 at 23:18
José Carlos Santos
172k22132239
asked Sep 26 '17 at 22:51
CornmanCornman
3,37121229
edited Sep 26 '17 at 23:18
José Carlos Santos
172k22132239
asked Sep 26 '17 at 22:51
CornmanCornman
3,37121229
edited Sep 26 '17 at 23:18
José Carlos Santos
172k22132239
edited Sep 26 '17 at 23:18
José Carlos Santos
172k22132239
edited Sep 26 '17 at 23:18
José Carlos Santos
172k22132239
172k22132239
asked Sep 26 '17 at 22:51
CornmanCornman
3,37121229
asked Sep 26 '17 at 22:51
CornmanCornman
3,37121229
asked Sep 26 '17 at 22:51
CornmanCornman
3,37121229
3,37121229
1
$begingroup$
This doesn't answer the exact question being asked here, but as a tangential aside, this polynomial is irreducible $pmod{p}$ for a small $p$ and hence irreducible over $mathbb{Q}$.
$endgroup$
– Kaj Hansen
Sep 26 '17 at 22:59
1
$begingroup$
Look up the rational root theorem.
$endgroup$
– Doug M
Sep 26 '17 at 23:00
$begingroup$
@KajHansen: But using the reduction theorem (which I think you are refering too) not involve a similar calculation, which might be easier?
$endgroup$
– Cornman
Sep 26 '17 at 23:04
3
$begingroup$
@DougM: The rational root theroem should not work here, since the degree of the polynomial is 4. The non existence of a root does not mean the polynomial is irreducible.
$endgroup$
– Cornman
Sep 26 '17 at 23:05
$begingroup$
@Cornman, the nice thing about it is that the number of irreducible quadratic polynomials in this particular $mathbb{F}_p$ is quite small--small enough to just try them all. On the other hand, the infinitude of polynomials in $mathbb{Q}[x]$ makes the current endeavor tricky.
$endgroup$
– Kaj Hansen
Sep 26 '17 at 23:05
|
show 2 more comments
1
$begingroup$
This doesn't answer the exact question being asked here, but as a tangential aside, this polynomial is irreducible $pmod{p}$ for a small $p$ and hence irreducible over $mathbb{Q}$.
$endgroup$
– Kaj Hansen
Sep 26 '17 at 22:59
1
$begingroup$
Look up the rational root theorem.
$endgroup$
– Doug M
Sep 26 '17 at 23:00
$begingroup$
@KajHansen: But using the reduction theorem (which I think you are refering too) not involve a similar calculation, which might be easier?
$endgroup$
– Cornman
Sep 26 '17 at 23:04
3
$begingroup$
@DougM: The rational root theroem should not work here, since the degree of the polynomial is 4. The non existence of a root does not mean the polynomial is irreducible.
$endgroup$
– Cornman
Sep 26 '17 at 23:05
$begingroup$
@Cornman, the nice thing about it is that the number of irreducible quadratic polynomials in this particular $mathbb{F}_p$ is quite small--small enough to just try them all. On the other hand, the infinitude of polynomials in $mathbb{Q}[x]$ makes the current endeavor tricky.
$endgroup$
– Kaj Hansen
Sep 26 '17 at 23:05
1
1
$begingroup$
This doesn't answer the exact question being asked here, but as a tangential aside, this polynomial is irreducible $pmod{p}$ for a small $p$ and hence irreducible over $mathbb{Q}$.
$endgroup$
– Kaj Hansen
Sep 26 '17 at 22:59
$begingroup$
This doesn't answer the exact question being asked here, but as a tangential aside, this polynomial is irreducible $pmod{p}$ for a small $p$ and hence irreducible over $mathbb{Q}$.
$endgroup$
– Kaj Hansen
Sep 26 '17 at 22:59
1
1
$begingroup$
Look up the rational root theorem.
$endgroup$
– Doug M
Sep 26 '17 at 23:00
$begingroup$
Look up the rational root theorem.
$endgroup$
– Doug M
Sep 26 '17 at 23:00
$begingroup$
@KajHansen: But using the reduction theorem (which I think you are refering too) not involve a similar calculation, which might be easier?
$endgroup$
– Cornman
Sep 26 '17 at 23:04
$begingroup$
@KajHansen: But using the reduction theorem (which I think you are refering too) not involve a similar calculation, which might be easier?
$endgroup$
– Cornman
Sep 26 '17 at 23:04
3
3
$begingroup$
@DougM: The rational root theroem should not work here, since the degree of the polynomial is 4. The non existence of a root does not mean the polynomial is irreducible.
$endgroup$
– Cornman
Sep 26 '17 at 23:05
$begingroup$
@DougM: The rational root theroem should not work here, since the degree of the polynomial is 4. The non existence of a root does not mean the polynomial is irreducible.
$endgroup$
– Cornman
Sep 26 '17 at 23:05
$begingroup$
@Cornman, the nice thing about it is that the number of irreducible quadratic polynomials in this particular $mathbb{F}_p$ is quite small--small enough to just try them all. On the other hand, the infinitude of polynomials in $mathbb{Q}[x]$ makes the current endeavor tricky.
$endgroup$
– Kaj Hansen
Sep 26 '17 at 23:05
$begingroup$
@Cornman, the nice thing about it is that the number of irreducible quadratic polynomials in this particular $mathbb{F}_p$ is quite small--small enough to just try them all. On the other hand, the infinitude of polynomials in $mathbb{Q}[x]$ makes the current endeavor tricky.
$endgroup$
– Kaj Hansen
Sep 26 '17 at 23:05
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Your polynomial has no rational roots. Therefore, if it was reducible in $mathbb{Q}[x]$, then it would have to be the product of two quadratic polynomials $p_1(X),p_2(X)inmathbb{Q}[X]$. Each of them either has to real roots or two complex non-real roots.
The roots of your polynomial are $pmsqrt{sqrt5+2}$ and $pm isqrt{sqrt5-2}$. Of these, only the last two are complex non-real. But$$left(X-isqrt{sqrt5-2},right)left(X+isqrt{sqrt5-2},right)=X^2-2+sqrt5notinmathbb{Q}[X].$$Therefore, your polynomial is irreducible.
answered Sep 26 '17 at 23:09
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
$begingroup$
Therefore my factorisation would be enough to show that the polynomial is irreducible over $mathbb{Q}[X]$?
$endgroup$
– Cornman
Sep 26 '17 at 23:11
3
$begingroup$
@Cornman I don't see why. The fact that your decomposition does not work doesn't imply that no decomposition works. For instance, $$x^4-5x^2+6=left(x^2-left(sqrt2+sqrt3right)x+sqrt6right)left(x^2+left(sqrt2+sqrt3right)x+sqrt6right),$$but $x^4-5x^2+6$ is reducible, since it is equal to $(x^2-3)(x^2-2)$.
$endgroup$
– José Carlos Santos
Sep 26 '17 at 23:16
add a comment |
$begingroup$
Let's write $f(x)=x^4-4x^2-1$. Note that since $f(x)=f(-x)$, the roots of $f$ come in positive and negative pairs; say the roots are $a$, $-a$, $b$, and $-b$. Up to renaming the roots, then, there are two cases of factorizations with integer coefficients that $f$ could have. We could have $f(x)=g(x)h(x)$ where $g(x)=(x-a)(x+a)=x^2-a^2$ and $h(x)=(x-b)(x+b)=x^2-b^2$. But that would mean that $x^2-4x-1$ factors as $(x-a^2)(x-b^2)$, and we know $x^2-4x-1$ is irreducible.
The other possibility is that we have a factorization like $f(x)=g(x)h(x)$ with $g(x)=(x-a)(x-b)=x^2-(a+b)x+ab$ and $h(x)=(x+a)(x+b)=x^2+(a+b)x+ab$. In this case, though, $ab$ would be an integer and the constant term of $f(x)$ would be the square of $ab$. Since the constant term is $-1$, this is impossible.
More generally, a similar argument shows that if $e(x)inmathbb{Z}[x]$ is irreducible and monic of degree $n$, then $f(x)=e(x^2)$ is irreducible unless its constant term is $(-1)^n$ times a perfect square. See the answers to For which monic irreducible $f(x)in mathbb Z[x]$ , is $f(x^2)$ also irreducible in $mathbb Z[x]$? for more details.
answered Sep 26 '17 at 23:25
Eric WofseyEric Wofsey
192k14217351
$endgroup$
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%2f2446699%2fx4-4x2-1-irreducible-over-mathbbqx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your polynomial has no rational roots. Therefore, if it was reducible in $mathbb{Q}[x]$, then it would have to be the product of two quadratic polynomials $p_1(X),p_2(X)inmathbb{Q}[X]$. Each of them either has to real roots or two complex non-real roots.
The roots of your polynomial are $pmsqrt{sqrt5+2}$ and $pm isqrt{sqrt5-2}$. Of these, only the last two are complex non-real. But$$left(X-isqrt{sqrt5-2},right)left(X+isqrt{sqrt5-2},right)=X^2-2+sqrt5notinmathbb{Q}[X].$$Therefore, your polynomial is irreducible.
answered Sep 26 '17 at 23:09
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
$begingroup$
Therefore my factorisation would be enough to show that the polynomial is irreducible over $mathbb{Q}[X]$?
$endgroup$
– Cornman
Sep 26 '17 at 23:11
3
$begingroup$
@Cornman I don't see why. The fact that your decomposition does not work doesn't imply that no decomposition works. For instance, $$x^4-5x^2+6=left(x^2-left(sqrt2+sqrt3right)x+sqrt6right)left(x^2+left(sqrt2+sqrt3right)x+sqrt6right),$$but $x^4-5x^2+6$ is reducible, since it is equal to $(x^2-3)(x^2-2)$.
$endgroup$
– José Carlos Santos
Sep 26 '17 at 23:16
add a comment |
$begingroup$
Your polynomial has no rational roots. Therefore, if it was reducible in $mathbb{Q}[x]$, then it would have to be the product of two quadratic polynomials $p_1(X),p_2(X)inmathbb{Q}[X]$. Each of them either has to real roots or two complex non-real roots.
The roots of your polynomial are $pmsqrt{sqrt5+2}$ and $pm isqrt{sqrt5-2}$. Of these, only the last two are complex non-real. But$$left(X-isqrt{sqrt5-2},right)left(X+isqrt{sqrt5-2},right)=X^2-2+sqrt5notinmathbb{Q}[X].$$Therefore, your polynomial is irreducible.
answered Sep 26 '17 at 23:09
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
$begingroup$
Therefore my factorisation would be enough to show that the polynomial is irreducible over $mathbb{Q}[X]$?
$endgroup$
– Cornman
Sep 26 '17 at 23:11
3
$begingroup$
@Cornman I don't see why. The fact that your decomposition does not work doesn't imply that no decomposition works. For instance, $$x^4-5x^2+6=left(x^2-left(sqrt2+sqrt3right)x+sqrt6right)left(x^2+left(sqrt2+sqrt3right)x+sqrt6right),$$but $x^4-5x^2+6$ is reducible, since it is equal to $(x^2-3)(x^2-2)$.
$endgroup$
– José Carlos Santos
Sep 26 '17 at 23:16
add a comment |
$begingroup$
Your polynomial has no rational roots. Therefore, if it was reducible in $mathbb{Q}[x]$, then it would have to be the product of two quadratic polynomials $p_1(X),p_2(X)inmathbb{Q}[X]$. Each of them either has to real roots or two complex non-real roots.
The roots of your polynomial are $pmsqrt{sqrt5+2}$ and $pm isqrt{sqrt5-2}$. Of these, only the last two are complex non-real. But$$left(X-isqrt{sqrt5-2},right)left(X+isqrt{sqrt5-2},right)=X^2-2+sqrt5notinmathbb{Q}[X].$$Therefore, your polynomial is irreducible.
answered Sep 26 '17 at 23:09
José Carlos SantosJosé Carlos Santos
172k22132239
$endgroup$
Your polynomial has no rational roots. Therefore, if it was reducible in $mathbb{Q}[x]$, then it would have to be the product of two quadratic polynomials $p_1(X),p_2(X)inmathbb{Q}[X]$. Each of them either has to real roots or two complex non-real roots.
The roots of your polynomial are $pmsqrt{sqrt5+2}$ and $pm isqrt{sqrt5-2}$. Of these, only the last two are complex non-real. But$$left(X-isqrt{sqrt5-2},right)left(X+isqrt{sqrt5-2},right)=X^2-2+sqrt5notinmathbb{Q}[X].$$Therefore, your polynomial is irreducible.
answered Sep 26 '17 at 23:09
José Carlos SantosJosé Carlos Santos
172k22132239
edited Jan 30 at 21:52
answered Sep 26 '17 at 23:09
José Carlos SantosJosé Carlos Santos
172k22132239
answered Sep 26 '17 at 23:09
José Carlos SantosJosé Carlos Santos
172k22132239
answered Sep 26 '17 at 23:09
José Carlos SantosJosé Carlos Santos
172k22132239
172k22132239
$begingroup$
Therefore my factorisation would be enough to show that the polynomial is irreducible over $mathbb{Q}[X]$?
$endgroup$
– Cornman
Sep 26 '17 at 23:11
3
$begingroup$
@Cornman I don't see why. The fact that your decomposition does not work doesn't imply that no decomposition works. For instance, $$x^4-5x^2+6=left(x^2-left(sqrt2+sqrt3right)x+sqrt6right)left(x^2+left(sqrt2+sqrt3right)x+sqrt6right),$$but $x^4-5x^2+6$ is reducible, since it is equal to $(x^2-3)(x^2-2)$.
$endgroup$
– José Carlos Santos
Sep 26 '17 at 23:16
add a comment |
$begingroup$
Therefore my factorisation would be enough to show that the polynomial is irreducible over $mathbb{Q}[X]$?
$endgroup$
– Cornman
Sep 26 '17 at 23:11
3
$begingroup$
@Cornman I don't see why. The fact that your decomposition does not work doesn't imply that no decomposition works. For instance, $$x^4-5x^2+6=left(x^2-left(sqrt2+sqrt3right)x+sqrt6right)left(x^2+left(sqrt2+sqrt3right)x+sqrt6right),$$but $x^4-5x^2+6$ is reducible, since it is equal to $(x^2-3)(x^2-2)$.
$endgroup$
– José Carlos Santos
Sep 26 '17 at 23:16
$begingroup$
Therefore my factorisation would be enough to show that the polynomial is irreducible over $mathbb{Q}[X]$?
$endgroup$
– Cornman
Sep 26 '17 at 23:11
$begingroup$
Therefore my factorisation would be enough to show that the polynomial is irreducible over $mathbb{Q}[X]$?
$endgroup$
– Cornman
Sep 26 '17 at 23:11
3
3
$begingroup$
@Cornman I don't see why. The fact that your decomposition does not work doesn't imply that no decomposition works. For instance, $$x^4-5x^2+6=left(x^2-left(sqrt2+sqrt3right)x+sqrt6right)left(x^2+left(sqrt2+sqrt3right)x+sqrt6right),$$but $x^4-5x^2+6$ is reducible, since it is equal to $(x^2-3)(x^2-2)$.
$endgroup$
– José Carlos Santos
Sep 26 '17 at 23:16
$begingroup$
@Cornman I don't see why. The fact that your decomposition does not work doesn't imply that no decomposition works. For instance, $$x^4-5x^2+6=left(x^2-left(sqrt2+sqrt3right)x+sqrt6right)left(x^2+left(sqrt2+sqrt3right)x+sqrt6right),$$but $x^4-5x^2+6$ is reducible, since it is equal to $(x^2-3)(x^2-2)$.
$endgroup$
– José Carlos Santos
Sep 26 '17 at 23:16
add a comment |
$begingroup$
Let's write $f(x)=x^4-4x^2-1$. Note that since $f(x)=f(-x)$, the roots of $f$ come in positive and negative pairs; say the roots are $a$, $-a$, $b$, and $-b$. Up to renaming the roots, then, there are two cases of factorizations with integer coefficients that $f$ could have. We could have $f(x)=g(x)h(x)$ where $g(x)=(x-a)(x+a)=x^2-a^2$ and $h(x)=(x-b)(x+b)=x^2-b^2$. But that would mean that $x^2-4x-1$ factors as $(x-a^2)(x-b^2)$, and we know $x^2-4x-1$ is irreducible.
The other possibility is that we have a factorization like $f(x)=g(x)h(x)$ with $g(x)=(x-a)(x-b)=x^2-(a+b)x+ab$ and $h(x)=(x+a)(x+b)=x^2+(a+b)x+ab$. In this case, though, $ab$ would be an integer and the constant term of $f(x)$ would be the square of $ab$. Since the constant term is $-1$, this is impossible.
More generally, a similar argument shows that if $e(x)inmathbb{Z}[x]$ is irreducible and monic of degree $n$, then $f(x)=e(x^2)$ is irreducible unless its constant term is $(-1)^n$ times a perfect square. See the answers to For which monic irreducible $f(x)in mathbb Z[x]$ , is $f(x^2)$ also irreducible in $mathbb Z[x]$? for more details.
answered Sep 26 '17 at 23:25
Eric WofseyEric Wofsey
192k14217351
$endgroup$
add a comment |
$begingroup$
Let's write $f(x)=x^4-4x^2-1$. Note that since $f(x)=f(-x)$, the roots of $f$ come in positive and negative pairs; say the roots are $a$, $-a$, $b$, and $-b$. Up to renaming the roots, then, there are two cases of factorizations with integer coefficients that $f$ could have. We could have $f(x)=g(x)h(x)$ where $g(x)=(x-a)(x+a)=x^2-a^2$ and $h(x)=(x-b)(x+b)=x^2-b^2$. But that would mean that $x^2-4x-1$ factors as $(x-a^2)(x-b^2)$, and we know $x^2-4x-1$ is irreducible.
The other possibility is that we have a factorization like $f(x)=g(x)h(x)$ with $g(x)=(x-a)(x-b)=x^2-(a+b)x+ab$ and $h(x)=(x+a)(x+b)=x^2+(a+b)x+ab$. In this case, though, $ab$ would be an integer and the constant term of $f(x)$ would be the square of $ab$. Since the constant term is $-1$, this is impossible.
More generally, a similar argument shows that if $e(x)inmathbb{Z}[x]$ is irreducible and monic of degree $n$, then $f(x)=e(x^2)$ is irreducible unless its constant term is $(-1)^n$ times a perfect square. See the answers to For which monic irreducible $f(x)in mathbb Z[x]$ , is $f(x^2)$ also irreducible in $mathbb Z[x]$? for more details.
answered Sep 26 '17 at 23:25
Eric WofseyEric Wofsey
192k14217351
$endgroup$
add a comment |
$begingroup$
Let's write $f(x)=x^4-4x^2-1$. Note that since $f(x)=f(-x)$, the roots of $f$ come in positive and negative pairs; say the roots are $a$, $-a$, $b$, and $-b$. Up to renaming the roots, then, there are two cases of factorizations with integer coefficients that $f$ could have. We could have $f(x)=g(x)h(x)$ where $g(x)=(x-a)(x+a)=x^2-a^2$ and $h(x)=(x-b)(x+b)=x^2-b^2$. But that would mean that $x^2-4x-1$ factors as $(x-a^2)(x-b^2)$, and we know $x^2-4x-1$ is irreducible.
The other possibility is that we have a factorization like $f(x)=g(x)h(x)$ with $g(x)=(x-a)(x-b)=x^2-(a+b)x+ab$ and $h(x)=(x+a)(x+b)=x^2+(a+b)x+ab$. In this case, though, $ab$ would be an integer and the constant term of $f(x)$ would be the square of $ab$. Since the constant term is $-1$, this is impossible.
More generally, a similar argument shows that if $e(x)inmathbb{Z}[x]$ is irreducible and monic of degree $n$, then $f(x)=e(x^2)$ is irreducible unless its constant term is $(-1)^n$ times a perfect square. See the answers to For which monic irreducible $f(x)in mathbb Z[x]$ , is $f(x^2)$ also irreducible in $mathbb Z[x]$? for more details.
answered Sep 26 '17 at 23:25
Eric WofseyEric Wofsey
192k14217351
$endgroup$
Let's write $f(x)=x^4-4x^2-1$. Note that since $f(x)=f(-x)$, the roots of $f$ come in positive and negative pairs; say the roots are $a$, $-a$, $b$, and $-b$. Up to renaming the roots, then, there are two cases of factorizations with integer coefficients that $f$ could have. We could have $f(x)=g(x)h(x)$ where $g(x)=(x-a)(x+a)=x^2-a^2$ and $h(x)=(x-b)(x+b)=x^2-b^2$. But that would mean that $x^2-4x-1$ factors as $(x-a^2)(x-b^2)$, and we know $x^2-4x-1$ is irreducible.
The other possibility is that we have a factorization like $f(x)=g(x)h(x)$ with $g(x)=(x-a)(x-b)=x^2-(a+b)x+ab$ and $h(x)=(x+a)(x+b)=x^2+(a+b)x+ab$. In this case, though, $ab$ would be an integer and the constant term of $f(x)$ would be the square of $ab$. Since the constant term is $-1$, this is impossible.
More generally, a similar argument shows that if $e(x)inmathbb{Z}[x]$ is irreducible and monic of degree $n$, then $f(x)=e(x^2)$ is irreducible unless its constant term is $(-1)^n$ times a perfect square. See the answers to For which monic irreducible $f(x)in mathbb Z[x]$ , is $f(x^2)$ also irreducible in $mathbb Z[x]$? for more details.
answered Sep 26 '17 at 23:25
Eric WofseyEric Wofsey
192k14217351
edited Dec 29 '17 at 2:54
answered Sep 26 '17 at 23:25
Eric WofseyEric Wofsey
192k14217351
answered Sep 26 '17 at 23:25
Eric WofseyEric Wofsey
192k14217351
answered Sep 26 '17 at 23:25
Eric WofseyEric Wofsey
192k14217351
192k14217351
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%2f2446699%2fx4-4x2-1-irreducible-over-mathbbqx%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
$begingroup$
This doesn't answer the exact question being asked here, but as a tangential aside, this polynomial is irreducible $pmod{p}$ for a small $p$ and hence irreducible over $mathbb{Q}$.
$endgroup$
– Kaj Hansen
Sep 26 '17 at 22:59
1
$begingroup$
Look up the rational root theorem.
$endgroup$
– Doug M
Sep 26 '17 at 23:00
$begingroup$
@KajHansen: But using the reduction theorem (which I think you are refering too) not involve a similar calculation, which might be easier?
$endgroup$
– Cornman
Sep 26 '17 at 23:04
3
$begingroup$
@DougM: The rational root theroem should not work here, since the degree of the polynomial is 4. The non existence of a root does not mean the polynomial is irreducible.
$endgroup$
– Cornman
Sep 26 '17 at 23:05
$begingroup$
@Cornman, the nice thing about it is that the number of irreducible quadratic polynomials in this particular $mathbb{F}_p$ is quite small--small enough to just try them all. On the other hand, the infinitude of polynomials in $mathbb{Q}[x]$ makes the current endeavor tricky.
$endgroup$
– Kaj Hansen
Sep 26 '17 at 23:05