Fdtxjr

How is the claim "I am in New York only if I am in America" the same as "If I am in New York, then I am in America?

LaTeX closing $ signs makes cursor jump

Why Is Death Allowed In the Matrix?

I’m planning on buying a laser printer but concerned about the life cycle of toner in the machine

Prove that NP is closed under karp reduction?

Theorems that impeded progress

Risk of getting Chronic Wasting Disease (CWD) in the United States?

How old can references or sources in a thesis be?

Why was the small council so happy for Tyrion to become the Master of Coin?

If I cast Expeditious Retreat, can I Dash as a bonus action on the same turn?

Python: next in for loop

Smoothness of finite-dimensional functional calculus

Approximately how much travel time was saved by the opening of the Suez Canal in 1869?

How to test if a transaction is standard without spending real money?

can i play a electric guitar through a bass amp?

the place where lots of roads meet

Why are electrically insulating heatsinks so rare? Is it just cost?

What typically incentivizes a professor to change jobs to a lower ranking university?

Why did the Germans forbid the possession of pet pigeons in Rostov-on-Don in 1941?

What are these boxed doors outside store fronts in New York?

Have astronauts in space suits ever taken selfies? If so, how?

To string or not to string

Do I have a twin with permutated remainders?

Arthur Somervell: 1000 Exercises - Meaning of this notation

Separability over intersection of intermediate fields

“Intersection” of separable subfieldsProve this field extension is separableExtension over the intersection of intermediate fields is also separableGalois extension of intersection of fieldsSplitting fields of polynomials over finite fieldsIntermediate fields of splitting fieldNumber of intermediate field of a finite separable extensionExistence of an intermediate field $Ksubseteq M subseteq L$ such that $[L:M]=p$Intermediate field of a Galois extensionA finite extension is simple iff the purely inseparable closure is simple?Classification of fields with only separable finite degree fields extensionsLet $L:K$ be a Galois extention, show that $L:M$ is a normal.Comparative size of lattice of subgroups and lattice of the intermediate fieldsExtension over the intersection of intermediate fields is also separable

2

$begingroup$

Let $Emathop/F$ be a finite-degree normal extension of fields. Let $K$ and $L$ be intermediate fields (that is, $Fsubseteq Ksubseteq E$ and $Fsubseteq Lsubseteq E$) such that $E$ is separable over both $K$ and $L$. Show that $E$ is separable over $Kcap L$.

So I was thinking that if $f=min_Kcap L(alpha)$. Then $min_K(alpha)$ divides $f$. But if $f$ has repeated roots then I don't see what goes wrong.

Thanks!

galois-theory

share|cite|improve this question

edited Mar 22 at 2:45

Saad

20.4k92352

asked Jan 29 '15 at 8:58

ZeusZeus

936

$endgroup$

add a comment | 

2

$begingroup$

Let $Emathop/F$ be a finite-degree normal extension of fields. Let $K$ and $L$ be intermediate fields (that is, $Fsubseteq Ksubseteq E$ and $Fsubseteq Lsubseteq E$) such that $E$ is separable over both $K$ and $L$. Show that $E$ is separable over $Kcap L$.

So I was thinking that if $f=min_Kcap L(alpha)$. Then $min_K(alpha)$ divides $f$. But if $f$ has repeated roots then I don't see what goes wrong.

Thanks!

galois-theory

share|cite|improve this question

edited Mar 22 at 2:45

Saad

20.4k92352

asked Jan 29 '15 at 8:58

ZeusZeus

936

$endgroup$

add a comment | 

$begingroup$

Let $Emathop/F$ be a finite-degree normal extension of fields. Let $K$ and $L$ be intermediate fields (that is, $Fsubseteq Ksubseteq E$ and $Fsubseteq Lsubseteq E$) such that $E$ is separable over both $K$ and $L$. Show that $E$ is separable over $Kcap L$.

So I was thinking that if $f=min_Kcap L(alpha)$. Then $min_K(alpha)$ divides $f$. But if $f$ has repeated roots then I don't see what goes wrong.

Thanks!

galois-theory

share|cite|improve this question

edited Mar 22 at 2:45

Saad

20.4k92352

asked Jan 29 '15 at 8:58

ZeusZeus

936

$endgroup$

share|cite|improve this question

edited Mar 22 at 2:45

Saad

20.4k92352

asked Jan 29 '15 at 8:58

ZeusZeus

936

share|cite|improve this question

edited Mar 22 at 2:45

Saad

20.4k92352

asked Jan 29 '15 at 8:58

ZeusZeus

936

edited Mar 22 at 2:45

Saad

20.4k92352

edited Mar 22 at 2:45

Saad

20.4k92352

asked Jan 29 '15 at 8:58

ZeusZeus

936

asked Jan 29 '15 at 8:58

ZeusZeus

936

1 Answer
1

active

oldest

votes

1

$begingroup$

$E/F$ is normal implies $E/K$ and $E/L$ are both normal. Since they are also separable, we get that $E/K$ and $E/L$ are Galois. Set $G_1=Gal(E/K)$ and $G_2= Gal(E/L)$. Note that $E^G_1=K$ and $E^G_2=L$.

Now consider $G=Aut(E/Kcap L)$ and $E^G$. Obviously, $Kcap Lleq E^G$. Since obviously $G_1leq G$ and $G_2leq G$, we get that $E^Gleq E^G_1=K$ and $E^Gleq E^G_2=L$, hence $E^Gleq Kcap L$.

So, $E^G=Kcap L$, i.e. $E^Aut(E/Kcap L)=Kcap L$, hence $E/Kcap L$ is Galois, and therefore separable.

share|cite|improve this answer

edited Jan 30 '15 at 22:03

answered Jan 30 '15 at 21:55

SMMSMM

3,058512

$endgroup$

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
);



);

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

Name

Email

Required, but never shown

StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1124716%2fseparability-over-intersection-of-intermediate-fields%23new-answer', 'question_page');

);

Post as a guest

Name

Email

Required, but never shown

1

$begingroup$

$E/F$ is normal implies $E/K$ and $E/L$ are both normal. Since they are also separable, we get that $E/K$ and $E/L$ are Galois. Set $G_1=Gal(E/K)$ and $G_2= Gal(E/L)$. Note that $E^G_1=K$ and $E^G_2=L$.

Now consider $G=Aut(E/Kcap L)$ and $E^G$. Obviously, $Kcap Lleq E^G$. Since obviously $G_1leq G$ and $G_2leq G$, we get that $E^Gleq E^G_1=K$ and $E^Gleq E^G_2=L$, hence $E^Gleq Kcap L$.

So, $E^G=Kcap L$, i.e. $E^Aut(E/Kcap L)=Kcap L$, hence $E/Kcap L$ is Galois, and therefore separable.

share|cite|improve this answer

edited Jan 30 '15 at 22:03

answered Jan 30 '15 at 21:55

SMMSMM

3,058512

$endgroup$

add a comment | 

1

$begingroup$

$E/F$ is normal implies $E/K$ and $E/L$ are both normal. Since they are also separable, we get that $E/K$ and $E/L$ are Galois. Set $G_1=Gal(E/K)$ and $G_2= Gal(E/L)$. Note that $E^G_1=K$ and $E^G_2=L$.

Now consider $G=Aut(E/Kcap L)$ and $E^G$. Obviously, $Kcap Lleq E^G$. Since obviously $G_1leq G$ and $G_2leq G$, we get that $E^Gleq E^G_1=K$ and $E^Gleq E^G_2=L$, hence $E^Gleq Kcap L$.

So, $E^G=Kcap L$, i.e. $E^Aut(E/Kcap L)=Kcap L$, hence $E/Kcap L$ is Galois, and therefore separable.

share|cite|improve this answer

edited Jan 30 '15 at 22:03

answered Jan 30 '15 at 21:55

SMMSMM

3,058512

$endgroup$

add a comment | 

$begingroup$

$E/F$ is normal implies $E/K$ and $E/L$ are both normal. Since they are also separable, we get that $E/K$ and $E/L$ are Galois. Set $G_1=Gal(E/K)$ and $G_2= Gal(E/L)$. Note that $E^G_1=K$ and $E^G_2=L$.

Now consider $G=Aut(E/Kcap L)$ and $E^G$. Obviously, $Kcap Lleq E^G$. Since obviously $G_1leq G$ and $G_2leq G$, we get that $E^Gleq E^G_1=K$ and $E^Gleq E^G_2=L$, hence $E^Gleq Kcap L$.

So, $E^G=Kcap L$, i.e. $E^Aut(E/Kcap L)=Kcap L$, hence $E/Kcap L$ is Galois, and therefore separable.

share|cite|improve this answer

edited Jan 30 '15 at 22:03

answered Jan 30 '15 at 21:55

SMMSMM

3,058512

$endgroup$

share|cite|improve this answer

edited Jan 30 '15 at 22:03

answered Jan 30 '15 at 21:55

SMMSMM

3,058512

answered Jan 30 '15 at 21:55

SMMSMM

3,058512

answered Jan 30 '15 at 21:55

SMMSMM

3,058512