Mixing
an odd degree polynomial with cyclic Galois group has root all real
I am thinking about following question:
Let $finmathbb{Q}[x]$ be an odd degree polynomial with cyclic Galois group. Prove that all the roots of $f$ are real.
I tried to prove it by contraction.
Suppose $f$ has some complex roots. Let $K$ be its splitting field. Then $[K:mathbb{Q}]=odd$.
I tried to derive something from the cyclic Galois group, but I failed. Or maybe I am reviewing for the prelim so I am really terrified...
Any hints or explanations are really appreciated!!!
abstract-algebra field-theory galois-theory
asked May 2 '18 at 3:36
JacobsonRadical
402111
add a comment |
I am thinking about following question:
Let $finmathbb{Q}[x]$ be an odd degree polynomial with cyclic Galois group. Prove that all the roots of $f$ are real.
I tried to prove it by contraction.
Suppose $f$ has some complex roots. Let $K$ be its splitting field. Then $[K:mathbb{Q}]=odd$.
I tried to derive something from the cyclic Galois group, but I failed. Or maybe I am reviewing for the prelim so I am really terrified...
Any hints or explanations are really appreciated!!!
abstract-algebra field-theory galois-theory
asked May 2 '18 at 3:36
JacobsonRadical
402111
Is $ f $ supposed to be irreducible?
– Starfall
May 2 '18 at 3:55
@Starfall the question does not specify the irreducibility. I think we could assume that it is irreducible
– JacobsonRadical
May 2 '18 at 3:57
add a comment |
I am thinking about following question:
Let $finmathbb{Q}[x]$ be an odd degree polynomial with cyclic Galois group. Prove that all the roots of $f$ are real.
I tried to prove it by contraction.
Suppose $f$ has some complex roots. Let $K$ be its splitting field. Then $[K:mathbb{Q}]=odd$.
I tried to derive something from the cyclic Galois group, but I failed. Or maybe I am reviewing for the prelim so I am really terrified...
Any hints or explanations are really appreciated!!!
abstract-algebra field-theory galois-theory
asked May 2 '18 at 3:36
JacobsonRadical
402111
I am thinking about following question:
Let $finmathbb{Q}[x]$ be an odd degree polynomial with cyclic Galois group. Prove that all the roots of $f$ are real.
I tried to prove it by contraction.
Suppose $f$ has some complex roots. Let $K$ be its splitting field. Then $[K:mathbb{Q}]=odd$.
I tried to derive something from the cyclic Galois group, but I failed. Or maybe I am reviewing for the prelim so I am really terrified...
Any hints or explanations are really appreciated!!!
abstract-algebra field-theory galois-theory
abstract-algebra field-theory galois-theory
asked May 2 '18 at 3:36
JacobsonRadical
402111
asked May 2 '18 at 3:36
JacobsonRadical
402111
asked May 2 '18 at 3:36
JacobsonRadical
402111
asked May 2 '18 at 3:36
JacobsonRadical
402111
asked May 2 '18 at 3:36
JacobsonRadical
402111
402111
Is $ f $ supposed to be irreducible?
– Starfall
May 2 '18 at 3:55
@Starfall the question does not specify the irreducibility. I think we could assume that it is irreducible
– JacobsonRadical
May 2 '18 at 3:57
add a comment |
Is $ f $ supposed to be irreducible?
– Starfall
May 2 '18 at 3:55
@Starfall the question does not specify the irreducibility. I think we could assume that it is irreducible
– JacobsonRadical
May 2 '18 at 3:57
Is $ f $ supposed to be irreducible?
– Starfall
May 2 '18 at 3:55
Is $ f $ supposed to be irreducible?
– Starfall
May 2 '18 at 3:55
@Starfall the question does not specify the irreducibility. I think we could assume that it is irreducible
– JacobsonRadical
May 2 '18 at 3:57
@Starfall the question does not specify the irreducibility. I think we could assume that it is irreducible
– JacobsonRadical
May 2 '18 at 3:57
add a comment |
2 Answers
2
active
oldest
votes
Without the assumption that $ f $ is irreducible, the statement is false; for instance $ f(x) = x(x^2 + x + 1) $ is an odd degree polynomial with cyclic Galois group which has nonreal roots.
If we assume that $ f $ is irreducible then the statement is true: if $ n $ is the degree of $ f $ we know that the Galois group is a transitive subgroup of $ S_n $, and the only cyclic transitive subgroup of $ S_n $ is one generated by an $ n $-cycle, so has odd order $ n $. If $ f $ had nonreal roots, complex conjugation would give an automorphism of order $ 2 $ in the Galois group, which is impossible by Lagrange since the order of the Galois group is odd.
answered May 2 '18 at 3:59
Starfall
13.3k11139
add a comment |
Starfall's answer, of course, settles the question. I can't resist adding the following alternative argument. Assuming $f(x)$ is irreducible.
- As an odd degree polynomial $f(x)$ has a real zero $alpha$.
- The usual complex conjugation $tau$ is an automorphism, possibly trivial, of the splitting field of $f(x)$.
- If $beta$ is another zero of $f(x)$, there exists an automorphism $sigma$ of the splitting field such that $sigma(alpha)=beta$.
- Because the Galois group is abelian (don't need cyclicity!) $tausigma=sigmatau$, and hence $$tau(beta)=tau(sigma(alpha))=sigma(tau(alpha))=sigma(alpha)=beta.$$
answered May 2 '18 at 7:00
Jyrki Lahtonen
108k12166367
IOW, we also see that if an even degree irreducible polynomial has a real zero and an abelian Galois group then all its zeros must be real.
– Jyrki Lahtonen
May 2 '18 at 12:32
Oh! Thank you so much! It ends the whole question!
– JacobsonRadical
May 2 '18 at 22:45
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%2f2762812%2fan-odd-degree-polynomial-with-cyclic-galois-group-has-root-all-real%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
Without the assumption that $ f $ is irreducible, the statement is false; for instance $ f(x) = x(x^2 + x + 1) $ is an odd degree polynomial with cyclic Galois group which has nonreal roots.
If we assume that $ f $ is irreducible then the statement is true: if $ n $ is the degree of $ f $ we know that the Galois group is a transitive subgroup of $ S_n $, and the only cyclic transitive subgroup of $ S_n $ is one generated by an $ n $-cycle, so has odd order $ n $. If $ f $ had nonreal roots, complex conjugation would give an automorphism of order $ 2 $ in the Galois group, which is impossible by Lagrange since the order of the Galois group is odd.
answered May 2 '18 at 3:59
Starfall
13.3k11139
add a comment |
Without the assumption that $ f $ is irreducible, the statement is false; for instance $ f(x) = x(x^2 + x + 1) $ is an odd degree polynomial with cyclic Galois group which has nonreal roots.
If we assume that $ f $ is irreducible then the statement is true: if $ n $ is the degree of $ f $ we know that the Galois group is a transitive subgroup of $ S_n $, and the only cyclic transitive subgroup of $ S_n $ is one generated by an $ n $-cycle, so has odd order $ n $. If $ f $ had nonreal roots, complex conjugation would give an automorphism of order $ 2 $ in the Galois group, which is impossible by Lagrange since the order of the Galois group is odd.
answered May 2 '18 at 3:59
Starfall
13.3k11139
add a comment |
Without the assumption that $ f $ is irreducible, the statement is false; for instance $ f(x) = x(x^2 + x + 1) $ is an odd degree polynomial with cyclic Galois group which has nonreal roots.
If we assume that $ f $ is irreducible then the statement is true: if $ n $ is the degree of $ f $ we know that the Galois group is a transitive subgroup of $ S_n $, and the only cyclic transitive subgroup of $ S_n $ is one generated by an $ n $-cycle, so has odd order $ n $. If $ f $ had nonreal roots, complex conjugation would give an automorphism of order $ 2 $ in the Galois group, which is impossible by Lagrange since the order of the Galois group is odd.
answered May 2 '18 at 3:59
Starfall
13.3k11139
Without the assumption that $ f $ is irreducible, the statement is false; for instance $ f(x) = x(x^2 + x + 1) $ is an odd degree polynomial with cyclic Galois group which has nonreal roots.
If we assume that $ f $ is irreducible then the statement is true: if $ n $ is the degree of $ f $ we know that the Galois group is a transitive subgroup of $ S_n $, and the only cyclic transitive subgroup of $ S_n $ is one generated by an $ n $-cycle, so has odd order $ n $. If $ f $ had nonreal roots, complex conjugation would give an automorphism of order $ 2 $ in the Galois group, which is impossible by Lagrange since the order of the Galois group is odd.
answered May 2 '18 at 3:59
Starfall
13.3k11139
edited Nov 21 '18 at 2:13
answered May 2 '18 at 3:59
Starfall
13.3k11139
answered May 2 '18 at 3:59
Starfall
13.3k11139
answered May 2 '18 at 3:59
Starfall
13.3k11139
13.3k11139
add a comment |
add a comment |
Starfall's answer, of course, settles the question. I can't resist adding the following alternative argument. Assuming $f(x)$ is irreducible.
- As an odd degree polynomial $f(x)$ has a real zero $alpha$.
- The usual complex conjugation $tau$ is an automorphism, possibly trivial, of the splitting field of $f(x)$.
- If $beta$ is another zero of $f(x)$, there exists an automorphism $sigma$ of the splitting field such that $sigma(alpha)=beta$.
- Because the Galois group is abelian (don't need cyclicity!) $tausigma=sigmatau$, and hence $$tau(beta)=tau(sigma(alpha))=sigma(tau(alpha))=sigma(alpha)=beta.$$
answered May 2 '18 at 7:00
Jyrki Lahtonen
108k12166367
IOW, we also see that if an even degree irreducible polynomial has a real zero and an abelian Galois group then all its zeros must be real.
– Jyrki Lahtonen
May 2 '18 at 12:32
Oh! Thank you so much! It ends the whole question!
– JacobsonRadical
May 2 '18 at 22:45
add a comment |
Starfall's answer, of course, settles the question. I can't resist adding the following alternative argument. Assuming $f(x)$ is irreducible.
- As an odd degree polynomial $f(x)$ has a real zero $alpha$.
- The usual complex conjugation $tau$ is an automorphism, possibly trivial, of the splitting field of $f(x)$.
- If $beta$ is another zero of $f(x)$, there exists an automorphism $sigma$ of the splitting field such that $sigma(alpha)=beta$.
- Because the Galois group is abelian (don't need cyclicity!) $tausigma=sigmatau$, and hence $$tau(beta)=tau(sigma(alpha))=sigma(tau(alpha))=sigma(alpha)=beta.$$
answered May 2 '18 at 7:00
Jyrki Lahtonen
108k12166367
IOW, we also see that if an even degree irreducible polynomial has a real zero and an abelian Galois group then all its zeros must be real.
– Jyrki Lahtonen
May 2 '18 at 12:32
Oh! Thank you so much! It ends the whole question!
– JacobsonRadical
May 2 '18 at 22:45
add a comment |
Starfall's answer, of course, settles the question. I can't resist adding the following alternative argument. Assuming $f(x)$ is irreducible.
- As an odd degree polynomial $f(x)$ has a real zero $alpha$.
- The usual complex conjugation $tau$ is an automorphism, possibly trivial, of the splitting field of $f(x)$.
- If $beta$ is another zero of $f(x)$, there exists an automorphism $sigma$ of the splitting field such that $sigma(alpha)=beta$.
- Because the Galois group is abelian (don't need cyclicity!) $tausigma=sigmatau$, and hence $$tau(beta)=tau(sigma(alpha))=sigma(tau(alpha))=sigma(alpha)=beta.$$
answered May 2 '18 at 7:00
Jyrki Lahtonen
108k12166367
Starfall's answer, of course, settles the question. I can't resist adding the following alternative argument. Assuming $f(x)$ is irreducible.
- As an odd degree polynomial $f(x)$ has a real zero $alpha$.
- The usual complex conjugation $tau$ is an automorphism, possibly trivial, of the splitting field of $f(x)$.
- If $beta$ is another zero of $f(x)$, there exists an automorphism $sigma$ of the splitting field such that $sigma(alpha)=beta$.
- Because the Galois group is abelian (don't need cyclicity!) $tausigma=sigmatau$, and hence $$tau(beta)=tau(sigma(alpha))=sigma(tau(alpha))=sigma(alpha)=beta.$$
answered May 2 '18 at 7:00
Jyrki Lahtonen
108k12166367
answered May 2 '18 at 7:00
Jyrki Lahtonen
108k12166367
answered May 2 '18 at 7:00
Jyrki Lahtonen
108k12166367
answered May 2 '18 at 7:00
Jyrki Lahtonen
108k12166367
108k12166367
IOW, we also see that if an even degree irreducible polynomial has a real zero and an abelian Galois group then all its zeros must be real.
– Jyrki Lahtonen
May 2 '18 at 12:32
Oh! Thank you so much! It ends the whole question!
– JacobsonRadical
May 2 '18 at 22:45
add a comment |
IOW, we also see that if an even degree irreducible polynomial has a real zero and an abelian Galois group then all its zeros must be real.
– Jyrki Lahtonen
May 2 '18 at 12:32
Oh! Thank you so much! It ends the whole question!
– JacobsonRadical
May 2 '18 at 22:45
IOW, we also see that if an even degree irreducible polynomial has a real zero and an abelian Galois group then all its zeros must be real.
– Jyrki Lahtonen
May 2 '18 at 12:32
IOW, we also see that if an even degree irreducible polynomial has a real zero and an abelian Galois group then all its zeros must be real.
– Jyrki Lahtonen
May 2 '18 at 12:32
Oh! Thank you so much! It ends the whole question!
– JacobsonRadical
May 2 '18 at 22:45
Oh! Thank you so much! It ends the whole question!
– JacobsonRadical
May 2 '18 at 22:45
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%2f2762812%2fan-odd-degree-polynomial-with-cyclic-galois-group-has-root-all-real%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
Is $ f $ supposed to be irreducible?
– Starfall
May 2 '18 at 3:55
@Starfall the question does not specify the irreducibility. I think we could assume that it is irreducible
– JacobsonRadical
May 2 '18 at 3:57