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Well, I'm stuck trying to prove the following about separable extensions.

If $L/E$ is a extension (not necessarily finite) such that $L/F$ and $F/E$ are both separable, then $L/E$ is also separable.

My problem is I need to prove this without using the primitive element theorem, facts about degree of separability-inseparability or the fundamental theorem of Galois Theory because this appears as an exercise before those results.

My idea was to prove that $L/F$ and $F/E$ are Galois extensions and then show $E=L^{G(L/E)}$. So, for a well-known theorem $L/E$ is a Galois extension and in particular, $L/E$ is separable, but I can't prove that $L/F$ is normal. Is it correct if I take $L$ to be the normal closure of $F$? Thanks in advance.

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  • $\begingroup$ Can you do this assuming all extensions are finite? This is an easy reduction. $\endgroup$
    – xyzzyz
    Jun 7 '16 at 5:12
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Careful! Separable extensions need not be Galois. Pick some $z\in L$ and let $f=m_{z,F}$ the minimal polynomial of $z$ over $F$, which is, by hypothesis, separable. The coefficients of $f$ lie in $F$, and are all separable over $E$. This means that we can assume $z$ is algebraic over an extension of the form $E'=F(a_1,\ldots,a_r)$ where $E'/F$ is now finite. Then $E'(z)/E'/F$ is a tower of finite separable extensions, and it suffices we prove this in this case. But then the multiplicative formula for the separable degree over finite towers shows that $E'(z)$ is separable over $F$. In particular, $z$ is separable over $F$. Because $z$ was an arbitrary element of $L$, this proves the claim.

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  • $\begingroup$ Well, honestly I knew your proof, but as I said I need one that doesn't use the multiplicative formula for the separable degree. Anyways, I appreciate your time for writing the proof. Thanks. $\endgroup$
    – Xam
    Jun 7 '16 at 10:16
  • $\begingroup$ Look at Lang's book. The separable degree is just encoding another situation you can make explicit. This is, I think, quite an elementary proof. $\endgroup$
    – Pedro Tamaroff
    Jun 7 '16 at 11:28
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    $\begingroup$ (My point here is that many "things with a name" are just a convenient way of packing up a perhaps long but simple procedure, and people tend to forget this.) $\endgroup$
    – Pedro Tamaroff
    Jun 7 '16 at 14:18
  • $\begingroup$ Are the $F$s and $E$s swapped when compared with the question? The question asked to show $L/E$ was separable $\endgroup$
    – user428487
    Oct 11 '20 at 4:14
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Lemma 1. Let $E/k$ be separable, and let $\sigma: k\rightarrow k^a$. Then $\sigma$ extends to an embedding of $E$ in $k^a$, and there are $[E:k]$ distinct extensions of $\sigma$. The number of such distinct monomorphism is $\lt [E:k]$ if $E/k$ is not separable.

Proof. We will prove the lemma by induction. Let $\alpha\in E$, then the minimal polynomial $\text{Irr}(\alpha,k)$ has no multiple roots. Any embedding of $k(\alpha)$ sends $\alpha$ to a root of $\text{Irr}(\alpha,k)$, so there are $\text{deg}(\text{Irr}(\alpha,k))=[k(\alpha):k]$ distinct extensions. For any $\beta\in k(\alpha)$, since $\text{Irr}(\beta,k(\alpha))\mid \text{Irr}(\beta,k)$, it follows from our assumption that $\text{Irr}(\beta,k(\alpha))$ is separable. By induction hypothesis, the number of distinct extensions of a given embedding of $k(\alpha)$ on $E$ is $[E:k(\alpha)]$. Hence the number of distinct extensions of $\sigma$ on $E$ is $[E:k(\alpha)]$. The second assertion is proven by essentially the same proof.

Lemma 2. Let $k\subset k(a_1)\subset k(a_1,a_2)\subset\cdots\subset k(a_1,\cdots,a_n)$ where $a_i$ is separable over $k(a_1,\cdots, a_{i-1})=E_{i-1}$, and $E=k(a_1,\cdots,a_n)$. Then the identity on $k$ extends to $[E:k]$ distinct monomorphism $\sigma: E/k\hookrightarrow k^a/k$.

Proof. By induction.

Proposition. Let $k\subset k(a_1)\subset k(a_1,a_2)\subset\cdots\subset k(a_1,\cdots,a_n)$ where $a_i$ is separable over $k(a_1,\cdots, a_{i-1})=E_{i-1}$, and $E=k(a_1,\cdots,a_n)$. Then $E/k$ is separable.

Proof. By Lemma $2$, the identity on $k$ extends to $[E:k]$ distinct monomorphism $\sigma: E/k\hookrightarrow k^a/k$. By Lemma $1$, this implies that $E/k$ is separable.

Theorem. Let $F/k$ and $E/F$ be finite separable. Then $E/k$ is separable.

Proof. By our previous proposition, $F(a)$ is separable over $k$ for each $a\in E$. This shows that $E/k$ is separable.

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