Equality field extensions
$begingroup$
How would you prove this:
If E is an extension of K and a and b are elements of EK, a^m belongs to K, b^n belongs to K, gcd (m, n) =1, then K(ab) =K(a,b)
??
I don't know which strategy to use to prove it.
abstract-algebra field-theory extension-field
asked Dec 30 '18 at 11:46
Luis Gimeno SoteloLuis Gimeno Sotelo
73
$endgroup$
add a comment |
$begingroup$
How would you prove this:
If E is an extension of K and a and b are elements of EK, a^m belongs to K, b^n belongs to K, gcd (m, n) =1, then K(ab) =K(a,b)
??
I don't know which strategy to use to prove it.
abstract-algebra field-theory extension-field
asked Dec 30 '18 at 11:46
Luis Gimeno SoteloLuis Gimeno Sotelo
73
$endgroup$
add a comment |
$begingroup$
How would you prove this:
If E is an extension of K and a and b are elements of EK, a^m belongs to K, b^n belongs to K, gcd (m, n) =1, then K(ab) =K(a,b)
??
I don't know which strategy to use to prove it.
abstract-algebra field-theory extension-field
asked Dec 30 '18 at 11:46
Luis Gimeno SoteloLuis Gimeno Sotelo
73
$endgroup$
How would you prove this:
If E is an extension of K and a and b are elements of EK, a^m belongs to K, b^n belongs to K, gcd (m, n) =1, then K(ab) =K(a,b)
??
I don't know which strategy to use to prove it.
abstract-algebra field-theory extension-field
abstract-algebra field-theory extension-field
asked Dec 30 '18 at 11:46
Luis Gimeno SoteloLuis Gimeno Sotelo
73
asked Dec 30 '18 at 11:46
Luis Gimeno SoteloLuis Gimeno Sotelo
73
asked Dec 30 '18 at 11:46
Luis Gimeno SoteloLuis Gimeno Sotelo
73
asked Dec 30 '18 at 11:46
Luis Gimeno SoteloLuis Gimeno Sotelo
73
asked Dec 30 '18 at 11:46
Luis Gimeno SoteloLuis Gimeno Sotelo
73
73
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We can assume that $m$ and $n$ are minimal such that $a^m,b^n in K$ (if not replace $m$ and $n$ by the minimal such. They will still be coprime). Since $m$ and $n$ are coprime, there exists a $k,l in mathbb{N}$ such that $mk = 1 + ln$. Then $$(ab)^{mk} = (b^n)^l(a^m)^kb in K(ab)$$
so since $a^m,b^n in K$, this shows that $b in K(ab)$. Similarly we can show that $a in K(ab)$ (just consider $(ab)^{nl}$) and hence $K(a,b) subseteq K(ab)$. Since $K(ab) subseteq K(a,b)$, they must be equal.
You could also use degrees to prove this : if you can show that both $[K(ab):K]=mn=[K(a,b):K]$, then since $K(ab) subseteq K(a,b)$, they must be equal.
answered Dec 30 '18 at 12:04
ODFODF
1,396510
$endgroup$
$begingroup$
Ok, thank you very much for your answer! My previous attempts were to show what you stated below, but it is harder
$endgroup$
– Luis Gimeno Sotelo
Dec 30 '18 at 12:33
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%2f3056751%2fequality-field-extensions%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
$begingroup$
We can assume that $m$ and $n$ are minimal such that $a^m,b^n in K$ (if not replace $m$ and $n$ by the minimal such. They will still be coprime). Since $m$ and $n$ are coprime, there exists a $k,l in mathbb{N}$ such that $mk = 1 + ln$. Then $$(ab)^{mk} = (b^n)^l(a^m)^kb in K(ab)$$
so since $a^m,b^n in K$, this shows that $b in K(ab)$. Similarly we can show that $a in K(ab)$ (just consider $(ab)^{nl}$) and hence $K(a,b) subseteq K(ab)$. Since $K(ab) subseteq K(a,b)$, they must be equal.
You could also use degrees to prove this : if you can show that both $[K(ab):K]=mn=[K(a,b):K]$, then since $K(ab) subseteq K(a,b)$, they must be equal.
answered Dec 30 '18 at 12:04
ODFODF
1,396510
$endgroup$
$begingroup$
Ok, thank you very much for your answer! My previous attempts were to show what you stated below, but it is harder
$endgroup$
– Luis Gimeno Sotelo
Dec 30 '18 at 12:33
add a comment |
$begingroup$
We can assume that $m$ and $n$ are minimal such that $a^m,b^n in K$ (if not replace $m$ and $n$ by the minimal such. They will still be coprime). Since $m$ and $n$ are coprime, there exists a $k,l in mathbb{N}$ such that $mk = 1 + ln$. Then $$(ab)^{mk} = (b^n)^l(a^m)^kb in K(ab)$$
so since $a^m,b^n in K$, this shows that $b in K(ab)$. Similarly we can show that $a in K(ab)$ (just consider $(ab)^{nl}$) and hence $K(a,b) subseteq K(ab)$. Since $K(ab) subseteq K(a,b)$, they must be equal.
You could also use degrees to prove this : if you can show that both $[K(ab):K]=mn=[K(a,b):K]$, then since $K(ab) subseteq K(a,b)$, they must be equal.
answered Dec 30 '18 at 12:04
ODFODF
1,396510
$endgroup$
$begingroup$
Ok, thank you very much for your answer! My previous attempts were to show what you stated below, but it is harder
$endgroup$
– Luis Gimeno Sotelo
Dec 30 '18 at 12:33
add a comment |
$begingroup$
We can assume that $m$ and $n$ are minimal such that $a^m,b^n in K$ (if not replace $m$ and $n$ by the minimal such. They will still be coprime). Since $m$ and $n$ are coprime, there exists a $k,l in mathbb{N}$ such that $mk = 1 + ln$. Then $$(ab)^{mk} = (b^n)^l(a^m)^kb in K(ab)$$
so since $a^m,b^n in K$, this shows that $b in K(ab)$. Similarly we can show that $a in K(ab)$ (just consider $(ab)^{nl}$) and hence $K(a,b) subseteq K(ab)$. Since $K(ab) subseteq K(a,b)$, they must be equal.
You could also use degrees to prove this : if you can show that both $[K(ab):K]=mn=[K(a,b):K]$, then since $K(ab) subseteq K(a,b)$, they must be equal.
answered Dec 30 '18 at 12:04
ODFODF
1,396510
$endgroup$
We can assume that $m$ and $n$ are minimal such that $a^m,b^n in K$ (if not replace $m$ and $n$ by the minimal such. They will still be coprime). Since $m$ and $n$ are coprime, there exists a $k,l in mathbb{N}$ such that $mk = 1 + ln$. Then $$(ab)^{mk} = (b^n)^l(a^m)^kb in K(ab)$$
so since $a^m,b^n in K$, this shows that $b in K(ab)$. Similarly we can show that $a in K(ab)$ (just consider $(ab)^{nl}$) and hence $K(a,b) subseteq K(ab)$. Since $K(ab) subseteq K(a,b)$, they must be equal.
You could also use degrees to prove this : if you can show that both $[K(ab):K]=mn=[K(a,b):K]$, then since $K(ab) subseteq K(a,b)$, they must be equal.
answered Dec 30 '18 at 12:04
ODFODF
1,396510
answered Dec 30 '18 at 12:04
ODFODF
1,396510
answered Dec 30 '18 at 12:04
ODFODF
1,396510
answered Dec 30 '18 at 12:04
ODFODF
1,396510
1,396510
$begingroup$
Ok, thank you very much for your answer! My previous attempts were to show what you stated below, but it is harder
$endgroup$
– Luis Gimeno Sotelo
Dec 30 '18 at 12:33
add a comment |
$begingroup$
Ok, thank you very much for your answer! My previous attempts were to show what you stated below, but it is harder
$endgroup$
– Luis Gimeno Sotelo
Dec 30 '18 at 12:33
$begingroup$
Ok, thank you very much for your answer! My previous attempts were to show what you stated below, but it is harder
$endgroup$
– Luis Gimeno Sotelo
Dec 30 '18 at 12:33
$begingroup$
Ok, thank you very much for your answer! My previous attempts were to show what you stated below, but it is harder
$endgroup$
– Luis Gimeno Sotelo
Dec 30 '18 at 12:33
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%2f3056751%2fequality-field-extensions%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
