If $L/F$ and $F/E$ are separable, then $L/E$ is separable

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.

• Can you do this assuming all extensions are finite? This is an easy reduction. Jun 7 '16 at 5:12

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.

• 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.
– Xam
Jun 7 '16 at 10:16
• 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. Jun 7 '16 at 11:28
• (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.) Jun 7 '16 at 14:18
• Are the $F$s and $E$s swapped when compared with the question? The question asked to show $L/E$ was separable Oct 11 '20 at 4:14

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.