Show existence of $sigma$
$begingroup$
Let $K$ be a normal extension of $F$ and $fin F[x]$ be irreducible over $F$.
Let $g_1, g_2$ be irreducible factors of $f$ in the ring $K[x]$.
How could we show that there exists $sigma in G(K/F)$ such that $g_2=sigma (g_1)$ ?
Could you give me a hint for that?
abstract-algebra galois-theory
asked Oct 29 '18 at 19:42
Mary StarMary Star
3,02582268
$endgroup$
add a comment |
$begingroup$
Let $K$ be a normal extension of $F$ and $fin F[x]$ be irreducible over $F$.
Let $g_1, g_2$ be irreducible factors of $f$ in the ring $K[x]$.
How could we show that there exists $sigma in G(K/F)$ such that $g_2=sigma (g_1)$ ?
Could you give me a hint for that?
abstract-algebra galois-theory
asked Oct 29 '18 at 19:42
Mary StarMary Star
3,02582268
$endgroup$
$begingroup$
IMHO the same question has appeared here. No answers, though. I'll keep looking...
$endgroup$
– Jyrki Lahtonen
Oct 30 '18 at 6:39
$begingroup$
A more promising match. Further checking required to decide whether this is a dupe. May be this?
$endgroup$
– Jyrki Lahtonen
Oct 30 '18 at 6:44
add a comment |
$begingroup$
Let $K$ be a normal extension of $F$ and $fin F[x]$ be irreducible over $F$.
Let $g_1, g_2$ be irreducible factors of $f$ in the ring $K[x]$.
How could we show that there exists $sigma in G(K/F)$ such that $g_2=sigma (g_1)$ ?
Could you give me a hint for that?
abstract-algebra galois-theory
asked Oct 29 '18 at 19:42
Mary StarMary Star
3,02582268
$endgroup$
Let $K$ be a normal extension of $F$ and $fin F[x]$ be irreducible over $F$.
Let $g_1, g_2$ be irreducible factors of $f$ in the ring $K[x]$.
How could we show that there exists $sigma in G(K/F)$ such that $g_2=sigma (g_1)$ ?
Could you give me a hint for that?
abstract-algebra galois-theory
abstract-algebra galois-theory
asked Oct 29 '18 at 19:42
Mary StarMary Star
3,02582268
asked Oct 29 '18 at 19:42
Mary StarMary Star
3,02582268
asked Oct 29 '18 at 19:42
Mary StarMary Star
3,02582268
asked Oct 29 '18 at 19:42
Mary StarMary Star
3,02582268
asked Oct 29 '18 at 19:42
Mary StarMary Star
3,02582268
3,02582268
$begingroup$
IMHO the same question has appeared here. No answers, though. I'll keep looking...
$endgroup$
– Jyrki Lahtonen
Oct 30 '18 at 6:39
$begingroup$
A more promising match. Further checking required to decide whether this is a dupe. May be this?
$endgroup$
– Jyrki Lahtonen
Oct 30 '18 at 6:44
add a comment |
$begingroup$
IMHO the same question has appeared here. No answers, though. I'll keep looking...
$endgroup$
– Jyrki Lahtonen
Oct 30 '18 at 6:39
$begingroup$
A more promising match. Further checking required to decide whether this is a dupe. May be this?
$endgroup$
– Jyrki Lahtonen
Oct 30 '18 at 6:44
$begingroup$
IMHO the same question has appeared here. No answers, though. I'll keep looking...
$endgroup$
– Jyrki Lahtonen
Oct 30 '18 at 6:39
$begingroup$
IMHO the same question has appeared here. No answers, though. I'll keep looking...
$endgroup$
– Jyrki Lahtonen
Oct 30 '18 at 6:39
$begingroup$
A more promising match. Further checking required to decide whether this is a dupe. May be this?
$endgroup$
– Jyrki Lahtonen
Oct 30 '18 at 6:44
$begingroup$
A more promising match. Further checking required to decide whether this is a dupe. May be this?
$endgroup$
– Jyrki Lahtonen
Oct 30 '18 at 6:44
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $L/K$ be the splitting field of $f$. Let $alpha_1$ be a root of $g_1$ and $alpha_2$ be a root of $g_2$. From field theory, since $alpha_1$ and $alpha_2$ have the same minimal polynomial over $F$ and $L/F$ is normal, we get $sigma in operatorname{Aut}(L/F)$ such that $sigma(alpha_1) = alpha_2$. Since $K/F$ is normal, $sigma|_K in operatorname{Aut}(K/F)$, still with $sigma|_K(alpha_1) = alpha_2$.
Now $g_2$ is irreducible in $K$ and $g_2(alpha_2) = 0 in L$, so $g_2$ is the minimal poolynomial of $alpha_2$ over $K$. However, $sigma|_K(g_1)$ is also irreducible in $K$, and $(sigma|_K(g_1))(alpha_2) = (sigma|_K(g_1))(sigma(alpha_1)) = sigma|_K(g_1(alpha_1)) = 0$, so $sigma|_K(g_1)$ is also the minimal polynomial of $alpha_2$ over $K$.
Since minimal polynomial is unique, we conclude $sigma|_K(g_1) = g_2$.
A counter-example when $K/F$ is not normal is $F=Bbb Q$ and $K=Bbb Q(sqrt[3]2)$ and $f = X^3-2 in F[X]$, with $f = (X-sqrt[3]2)(X^2+sqrt[3]2X+sqrt[3]4) in K[X]$.
answered Jan 1 at 23:53
Kenny LauKenny Lau
19.9k2159
$endgroup$
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%2f2976583%2fshow-existence-of-sigma%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
$begingroup$
Let $L/K$ be the splitting field of $f$. Let $alpha_1$ be a root of $g_1$ and $alpha_2$ be a root of $g_2$. From field theory, since $alpha_1$ and $alpha_2$ have the same minimal polynomial over $F$ and $L/F$ is normal, we get $sigma in operatorname{Aut}(L/F)$ such that $sigma(alpha_1) = alpha_2$. Since $K/F$ is normal, $sigma|_K in operatorname{Aut}(K/F)$, still with $sigma|_K(alpha_1) = alpha_2$.
Now $g_2$ is irreducible in $K$ and $g_2(alpha_2) = 0 in L$, so $g_2$ is the minimal poolynomial of $alpha_2$ over $K$. However, $sigma|_K(g_1)$ is also irreducible in $K$, and $(sigma|_K(g_1))(alpha_2) = (sigma|_K(g_1))(sigma(alpha_1)) = sigma|_K(g_1(alpha_1)) = 0$, so $sigma|_K(g_1)$ is also the minimal polynomial of $alpha_2$ over $K$.
Since minimal polynomial is unique, we conclude $sigma|_K(g_1) = g_2$.
A counter-example when $K/F$ is not normal is $F=Bbb Q$ and $K=Bbb Q(sqrt[3]2)$ and $f = X^3-2 in F[X]$, with $f = (X-sqrt[3]2)(X^2+sqrt[3]2X+sqrt[3]4) in K[X]$.
answered Jan 1 at 23:53
Kenny LauKenny Lau
19.9k2159
$endgroup$
add a comment |
$begingroup$
Let $L/K$ be the splitting field of $f$. Let $alpha_1$ be a root of $g_1$ and $alpha_2$ be a root of $g_2$. From field theory, since $alpha_1$ and $alpha_2$ have the same minimal polynomial over $F$ and $L/F$ is normal, we get $sigma in operatorname{Aut}(L/F)$ such that $sigma(alpha_1) = alpha_2$. Since $K/F$ is normal, $sigma|_K in operatorname{Aut}(K/F)$, still with $sigma|_K(alpha_1) = alpha_2$.
Now $g_2$ is irreducible in $K$ and $g_2(alpha_2) = 0 in L$, so $g_2$ is the minimal poolynomial of $alpha_2$ over $K$. However, $sigma|_K(g_1)$ is also irreducible in $K$, and $(sigma|_K(g_1))(alpha_2) = (sigma|_K(g_1))(sigma(alpha_1)) = sigma|_K(g_1(alpha_1)) = 0$, so $sigma|_K(g_1)$ is also the minimal polynomial of $alpha_2$ over $K$.
Since minimal polynomial is unique, we conclude $sigma|_K(g_1) = g_2$.
A counter-example when $K/F$ is not normal is $F=Bbb Q$ and $K=Bbb Q(sqrt[3]2)$ and $f = X^3-2 in F[X]$, with $f = (X-sqrt[3]2)(X^2+sqrt[3]2X+sqrt[3]4) in K[X]$.
answered Jan 1 at 23:53
Kenny LauKenny Lau
19.9k2159
$endgroup$
add a comment |
$begingroup$
Let $L/K$ be the splitting field of $f$. Let $alpha_1$ be a root of $g_1$ and $alpha_2$ be a root of $g_2$. From field theory, since $alpha_1$ and $alpha_2$ have the same minimal polynomial over $F$ and $L/F$ is normal, we get $sigma in operatorname{Aut}(L/F)$ such that $sigma(alpha_1) = alpha_2$. Since $K/F$ is normal, $sigma|_K in operatorname{Aut}(K/F)$, still with $sigma|_K(alpha_1) = alpha_2$.
Now $g_2$ is irreducible in $K$ and $g_2(alpha_2) = 0 in L$, so $g_2$ is the minimal poolynomial of $alpha_2$ over $K$. However, $sigma|_K(g_1)$ is also irreducible in $K$, and $(sigma|_K(g_1))(alpha_2) = (sigma|_K(g_1))(sigma(alpha_1)) = sigma|_K(g_1(alpha_1)) = 0$, so $sigma|_K(g_1)$ is also the minimal polynomial of $alpha_2$ over $K$.
Since minimal polynomial is unique, we conclude $sigma|_K(g_1) = g_2$.
A counter-example when $K/F$ is not normal is $F=Bbb Q$ and $K=Bbb Q(sqrt[3]2)$ and $f = X^3-2 in F[X]$, with $f = (X-sqrt[3]2)(X^2+sqrt[3]2X+sqrt[3]4) in K[X]$.
answered Jan 1 at 23:53
Kenny LauKenny Lau
19.9k2159
$endgroup$
Let $L/K$ be the splitting field of $f$. Let $alpha_1$ be a root of $g_1$ and $alpha_2$ be a root of $g_2$. From field theory, since $alpha_1$ and $alpha_2$ have the same minimal polynomial over $F$ and $L/F$ is normal, we get $sigma in operatorname{Aut}(L/F)$ such that $sigma(alpha_1) = alpha_2$. Since $K/F$ is normal, $sigma|_K in operatorname{Aut}(K/F)$, still with $sigma|_K(alpha_1) = alpha_2$.
Now $g_2$ is irreducible in $K$ and $g_2(alpha_2) = 0 in L$, so $g_2$ is the minimal poolynomial of $alpha_2$ over $K$. However, $sigma|_K(g_1)$ is also irreducible in $K$, and $(sigma|_K(g_1))(alpha_2) = (sigma|_K(g_1))(sigma(alpha_1)) = sigma|_K(g_1(alpha_1)) = 0$, so $sigma|_K(g_1)$ is also the minimal polynomial of $alpha_2$ over $K$.
Since minimal polynomial is unique, we conclude $sigma|_K(g_1) = g_2$.
A counter-example when $K/F$ is not normal is $F=Bbb Q$ and $K=Bbb Q(sqrt[3]2)$ and $f = X^3-2 in F[X]$, with $f = (X-sqrt[3]2)(X^2+sqrt[3]2X+sqrt[3]4) in K[X]$.
answered Jan 1 at 23:53
Kenny LauKenny Lau
19.9k2159
answered Jan 1 at 23:53
Kenny LauKenny Lau
19.9k2159
answered Jan 1 at 23:53
Kenny LauKenny Lau
19.9k2159
answered Jan 1 at 23:53
Kenny LauKenny Lau
19.9k2159
19.9k2159
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%2f2976583%2fshow-existence-of-sigma%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
$begingroup$
IMHO the same question has appeared here. No answers, though. I'll keep looking...
$endgroup$
– Jyrki Lahtonen
Oct 30 '18 at 6:39
$begingroup$
A more promising match. Further checking required to decide whether this is a dupe. May be this?
$endgroup$
– Jyrki Lahtonen
Oct 30 '18 at 6:44