Mixing
Splitting field of an irreductible polynomial $f(X) in F_{q}[X]$
0
$begingroup$
Let $F_q$ be a finite field ($q$ is a power a prime) and irreductible polynomial $f(X)in F_q[X]$ with degree $ngeq 2$.
I have to see that $F_{q^n}$ is the splitting field of $f$ over $F_q$, and that all the roots of $f$ have the same order under the multiplicative group $F^*_{q^n}$.
What I know so far:
$F_{q^n}$ is the finite field of $q^n$ elements and its elements are the roots of the polynomial $g(X) = X^{q^n}-X in F_q[X]$. I read here that $g$ is the product of all the monic polynomials of $F_q[X]$ which divide $g$. So $f$ must divide $g$.
Because of this, if $f | g$ we have that, as $g$ generates the elements of $F_{q^n}$, thus $f$ splits in $F_{q^n}$.- About the order of the roots, I know that $|F^*_{q^n}| = phi(q^n)$. But this group is the Galois group of the extension over $F_p$, where $q = p^m$, for some $mgeq 1$. So if you consider the extension $F_{q^n}$ over $F_q$, you have that its degree is $n$, but... here I made a mess in my head and I really don't know how to finish this.
I would appreciate if someone points out something about the second point or anything I could have done wrong on the first one.
Many thanks in advance!
galois-theory finite-fields irreducible-polynomials splitting-field galois-extensions
asked Jan 29 at 16:06
2pac2pac
665
$endgroup$
add a comment |
0
$begingroup$
Let $F_q$ be a finite field ($q$ is a power a prime) and irreductible polynomial $f(X)in F_q[X]$ with degree $ngeq 2$.
I have to see that $F_{q^n}$ is the splitting field of $f$ over $F_q$, and that all the roots of $f$ have the same order under the multiplicative group $F^*_{q^n}$.
What I know so far:
$F_{q^n}$ is the finite field of $q^n$ elements and its elements are the roots of the polynomial $g(X) = X^{q^n}-X in F_q[X]$. I read here that $g$ is the product of all the monic polynomials of $F_q[X]$ which divide $g$. So $f$ must divide $g$.
Because of this, if $f | g$ we have that, as $g$ generates the elements of $F_{q^n}$, thus $f$ splits in $F_{q^n}$.- About the order of the roots, I know that $|F^*_{q^n}| = phi(q^n)$. But this group is the Galois group of the extension over $F_p$, where $q = p^m$, for some $mgeq 1$. So if you consider the extension $F_{q^n}$ over $F_q$, you have that its degree is $n$, but... here I made a mess in my head and I really don't know how to finish this.
I would appreciate if someone points out something about the second point or anything I could have done wrong on the first one.
Many thanks in advance!
galois-theory finite-fields irreducible-polynomials splitting-field galois-extensions
asked Jan 29 at 16:06
2pac2pac
665
$endgroup$
2
$begingroup$
The things you listed in item 2 reveal some confusion. For one $|F_{q^n}^*|=q^n-1$ because in a field zero is the only non-invertible element. The Galois group of $F_{q^n}$ over $F_q$, is cyclic of order $n$. It is generated by the Frobenius automorphism $zmapsto z^q$. The key to success is that if $alpha$ is a zero of $f(X)$, then $alpha^q$ must be another. Rinse. Repeat.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 17:41
$begingroup$
But this question has been answered on our site.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 17:41
$begingroup$
For example here or here. Not voting to close as a dupe because A) I think I've seen a better version also, B) in such a case I should not pick a target I answered myself.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 21:31
add a comment |
0
0
0
$begingroup$
Let $F_q$ be a finite field ($q$ is a power a prime) and irreductible polynomial $f(X)in F_q[X]$ with degree $ngeq 2$.
I have to see that $F_{q^n}$ is the splitting field of $f$ over $F_q$, and that all the roots of $f$ have the same order under the multiplicative group $F^*_{q^n}$.
What I know so far:
$F_{q^n}$ is the finite field of $q^n$ elements and its elements are the roots of the polynomial $g(X) = X^{q^n}-X in F_q[X]$. I read here that $g$ is the product of all the monic polynomials of $F_q[X]$ which divide $g$. So $f$ must divide $g$.
Because of this, if $f | g$ we have that, as $g$ generates the elements of $F_{q^n}$, thus $f$ splits in $F_{q^n}$.- About the order of the roots, I know that $|F^*_{q^n}| = phi(q^n)$. But this group is the Galois group of the extension over $F_p$, where $q = p^m$, for some $mgeq 1$. So if you consider the extension $F_{q^n}$ over $F_q$, you have that its degree is $n$, but... here I made a mess in my head and I really don't know how to finish this.
I would appreciate if someone points out something about the second point or anything I could have done wrong on the first one.
Many thanks in advance!
galois-theory finite-fields irreducible-polynomials splitting-field galois-extensions
asked Jan 29 at 16:06
2pac2pac
665
$endgroup$
Let $F_q$ be a finite field ($q$ is a power a prime) and irreductible polynomial $f(X)in F_q[X]$ with degree $ngeq 2$.
I have to see that $F_{q^n}$ is the splitting field of $f$ over $F_q$, and that all the roots of $f$ have the same order under the multiplicative group $F^*_{q^n}$.
What I know so far:
$F_{q^n}$ is the finite field of $q^n$ elements and its elements are the roots of the polynomial $g(X) = X^{q^n}-X in F_q[X]$. I read here that $g$ is the product of all the monic polynomials of $F_q[X]$ which divide $g$. So $f$ must divide $g$.
Because of this, if $f | g$ we have that, as $g$ generates the elements of $F_{q^n}$, thus $f$ splits in $F_{q^n}$.- About the order of the roots, I know that $|F^*_{q^n}| = phi(q^n)$. But this group is the Galois group of the extension over $F_p$, where $q = p^m$, for some $mgeq 1$. So if you consider the extension $F_{q^n}$ over $F_q$, you have that its degree is $n$, but... here I made a mess in my head and I really don't know how to finish this.
I would appreciate if someone points out something about the second point or anything I could have done wrong on the first one.
Many thanks in advance!
galois-theory finite-fields irreducible-polynomials splitting-field galois-extensions
galois-theory finite-fields irreducible-polynomials splitting-field galois-extensions
asked Jan 29 at 16:06
2pac2pac
665
asked Jan 29 at 16:06
2pac2pac
665
asked Jan 29 at 16:06
2pac2pac
665
asked Jan 29 at 16:06
2pac2pac
665
asked Jan 29 at 16:06
2pac2pac
665
665
2
$begingroup$
The things you listed in item 2 reveal some confusion. For one $|F_{q^n}^*|=q^n-1$ because in a field zero is the only non-invertible element. The Galois group of $F_{q^n}$ over $F_q$, is cyclic of order $n$. It is generated by the Frobenius automorphism $zmapsto z^q$. The key to success is that if $alpha$ is a zero of $f(X)$, then $alpha^q$ must be another. Rinse. Repeat.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 17:41
$begingroup$
But this question has been answered on our site.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 17:41
$begingroup$
For example here or here. Not voting to close as a dupe because A) I think I've seen a better version also, B) in such a case I should not pick a target I answered myself.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 21:31
add a comment |
2
$begingroup$
The things you listed in item 2 reveal some confusion. For one $|F_{q^n}^*|=q^n-1$ because in a field zero is the only non-invertible element. The Galois group of $F_{q^n}$ over $F_q$, is cyclic of order $n$. It is generated by the Frobenius automorphism $zmapsto z^q$. The key to success is that if $alpha$ is a zero of $f(X)$, then $alpha^q$ must be another. Rinse. Repeat.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 17:41
$begingroup$
But this question has been answered on our site.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 17:41
$begingroup$
For example here or here. Not voting to close as a dupe because A) I think I've seen a better version also, B) in such a case I should not pick a target I answered myself.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 21:31
2
2
$begingroup$
The things you listed in item 2 reveal some confusion. For one $|F_{q^n}^*|=q^n-1$ because in a field zero is the only non-invertible element. The Galois group of $F_{q^n}$ over $F_q$, is cyclic of order $n$. It is generated by the Frobenius automorphism $zmapsto z^q$. The key to success is that if $alpha$ is a zero of $f(X)$, then $alpha^q$ must be another. Rinse. Repeat.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 17:41
$begingroup$
The things you listed in item 2 reveal some confusion. For one $|F_{q^n}^*|=q^n-1$ because in a field zero is the only non-invertible element. The Galois group of $F_{q^n}$ over $F_q$, is cyclic of order $n$. It is generated by the Frobenius automorphism $zmapsto z^q$. The key to success is that if $alpha$ is a zero of $f(X)$, then $alpha^q$ must be another. Rinse. Repeat.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 17:41
$begingroup$
But this question has been answered on our site.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 17:41
$begingroup$
But this question has been answered on our site.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 17:41
$begingroup$
For example here or here. Not voting to close as a dupe because A) I think I've seen a better version also, B) in such a case I should not pick a target I answered myself.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 21:31
$begingroup$
For example here or here. Not voting to close as a dupe because A) I think I've seen a better version also, B) in such a case I should not pick a target I answered myself.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 21:31
add a comment |
0
active
oldest
votes
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
});
}
});
draft saved
draft discarded
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%2f3092355%2fsplitting-field-of-an-irreductible-polynomial-fx-in-f-qx%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
draft saved
draft discarded
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.
draft saved
draft discarded
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%2f3092355%2fsplitting-field-of-an-irreductible-polynomial-fx-in-f-qx%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
2
$begingroup$
The things you listed in item 2 reveal some confusion. For one $|F_{q^n}^*|=q^n-1$ because in a field zero is the only non-invertible element. The Galois group of $F_{q^n}$ over $F_q$, is cyclic of order $n$. It is generated by the Frobenius automorphism $zmapsto z^q$. The key to success is that if $alpha$ is a zero of $f(X)$, then $alpha^q$ must be another. Rinse. Repeat.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 17:41
$begingroup$
But this question has been answered on our site.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 17:41
$begingroup$
For example here or here. Not voting to close as a dupe because A) I think I've seen a better version also, B) in such a case I should not pick a target I answered myself.
$endgroup$
– Jyrki Lahtonen
Jan 29 at 21:31