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I am trying to prove the following statement:

Let $l \subseteq E \subseteq L$ and $l \subseteq F \subseteq L$ be a towers of field extensions, and suppose $E/l$ is a separable extension. $EF/F$ is necessarily separable.

For the sake of completion, I am posting a slightly expanded version of Georeges' solution suggestion:

Note that $EF$ is generated over $F$ by $\alpha \in E$. By hypothesis, $\alpha \in E$ is separable over $k$, and therefore $\alpha$ is separable over $F$ since $m_{\alpha,F}(X)$ divides $m_{\alpha,k}(X)$ and $m_{\alpha,k}$ has no repeated roots. Since $EF/F$ is generated by separable elements, it must be a separable extension.

EDIT: Thanks to a comment by Chris Eagle, the attempt below has no reason to be a workable strategy, since if it were $F/l$ would be separable, which it does not have to be.

Note that we have the following tower of extensions:

$$ l \subseteq E \subseteq EF $$

Thus, if we can show $EF/l$ is separable, $EF/F$ is necessarily a separable extension. Assuming $\alpha \in EF$ is not separable over $l$, we can deduce that the minimal polynomial of $\alpha$ over $l$, $m(x)$, is not separable, hence has repeated roots. We can then conclude that $m(x)=g(x^p)$ for some separable, irreducible polynomial $g(x) \in l(x)$. This leads to the conclusion that $\alpha^p$ is separable over $l$, despite $\alpha$ being inseparable. From here, I am unsure how to derive a contradiction. Maybe it just hasn't hit me yet. Can anyone see a way to continue, or is this attempt doomed to failure?

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Why do you think $EF/l$ will be separable? Were this true, then $F/l$ would be separable, and it's pretty obvious that the condition that $E/l$ is separable isn't going to guarantee that. –  Chris Eagle May 21 '12 at 11:49
    
@ChrisEagle Thank you for pointing that out, that saved me a lot of wasted time. I guess, then, I have made no progress on the problem. –  Holdsworth88 May 21 '12 at 11:58
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2 Answers 2

up vote 6 down vote accepted

The field $EF$ is generated over $F$ by the elements $a\in E $ i.e. $ EF=F(E)$

Since each $a\in E$ is separable over $l$ it is a fortiori separable over $F$:
Indeed the minimal polynomial $Irr(a,F)$ of $a$ over $F$ divides $Irr(a, l)$, the minimal polynomial of $a$ over $l$ , and so has simple roots too.

Since an extension generated by separable elements is separable (Lang, Algebra, Ch.V, Thm. 4.4), we conclude that $EF/F$ is separable .

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Thank you for your help. Sometimes these things are just staring you in the face. –  Holdsworth88 May 21 '12 at 12:29
    
Don't be too harsh on yourself, Holdersworth88: things are easy when taken in the proper order, but I too often find myself puzzled by results which, when explained later by someone else, look painfully evident. –  Georges Elencwajg May 21 '12 at 12:34
    
Dear Georges, I was looking at Thm 4.4 in Lang, doesn't it require that the extension be algebraic as well before you can apply the theorem? –  Hana Bailey Mar 3 at 20:08
    
Dear @Hana, in the present context, where elements have minimal polynomials, $E$ is algebraic over $l$ and thus $F(E)$ is separable over $F$. –  Georges Elencwajg Mar 3 at 21:10
    
Sorry for my confusion, but do we know $EF/F$ is algebraic beforehand? It is certainly generated over $F$ by separable elements, but don't you need to have that $EF/F$ is algebraic, not just $E/l$ is algebraic? –  Hana Bailey Mar 3 at 21:26
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It seems like something's missing here, or the claim is plainly false: taking $\,E=l\,$, we get that $\,l\subset F\Longrightarrow EF=lF=F\,$, it would follow that any extension's separable...!

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Taking $E=l$, the claim is that $F/F$ is separable, which it plainly is. –  Chris Eagle May 21 '12 at 12:08
    
Of course. For some reason I was sure it was written in the OP that "... $\,EF/l\,$ is separable", which is exactly what you mention in your first comment above...Either the OP was edited or I misread. –  DonAntonio May 21 '12 at 12:15
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