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A field is called separably closed if the only separable algebraic extension is the trivial one. A separable closure of a field $K$ is a separable algebraic extension $K ⊆ K^{\text{sep}}$ with $K^{\text{sep}}$ separably closed.

I want to show that a separable closure $K^{\text{sep}}$ of $K$ is algebraically closed iff $K$ is perfect (i.e. every algebraic extension of $K$ is separable).

I have already proven

I) Every field has a separable closure.

II) Every pair of separable closures of $K$ is $K$-isomorphic.

III) For every tower $K ⊆ L ⊆ M$ of algebraic extensions we have $$ K⊆ L \text{ and } L ⊆ M \text{ are both separable } \iff K ⊆ M \text{ is separable.}$$

So here we go. Suppose a separable closure $K^{\text{sep}}$ of $K$ is algebraically closed. Let $L/K$ be an arbitrary algebraic extension. By (I) $L$ has a separable closure $L^{\text{sep}}$. Now if $L^{\text{sep}}$ were a separable extension of $K$ as well, then by (II) $K^{\text{sep}}$ and $L^{\text{sep}}$ would be $K$-isomorphic, so we would obtain a tower

$$K ⊆ L ⊆ L^{\text{sep}} \cong_K K^{\text{sep}}. $$

of algebraic extensions. As $K ⊆ K^{\text{sep}}$ is separable by definition, (III) implies that $K ⊆ L$ must be too, so $K$ is perfect.

However

All we really know is that $L ⊆ L^{\text{sep}}$ is separable. i.e. for all $x ∈ L^{\text{sep}}$ the minimal polynomial $f^x_L ∈ L[X]$ has no multiple roots in $\bar{L}$. This unfortunately does not imply that $f^x_K ∈ K[X]$ has no multiple roots in $\bar{K}$, as $f^x_L$ divides $f^x_K$, not the other way around. Is the reasoning so far correct? If so, how do I show the final ingredient?

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  • $\begingroup$ It may be helpful to instead of working with isomorphisms of algebraic closures, work with extending your field embeddings. Eg, if $K$ is perfect, then any algebraic extension $L/K$ lifts to a map $L$ to $K^{sep}$, and vice versa. $\endgroup$ – Chris H Apr 19 '20 at 23:39
  • $\begingroup$ Alright, but that seems to be the other implication, which I have not even addressed yet. $\endgroup$ – Jos van Nieuwman Apr 19 '20 at 23:50
  • $\begingroup$ It seems to me that you never use the fact that $K^{sep}$ is algebraically closed. It implies that $L^{sep}$ embeds in $K^{sep}$ since $L^{sep}/K$ is at least algebraic. $\endgroup$ – Captain Lama Apr 23 '20 at 16:23
  • $\begingroup$ If L is algebraic it lives in (up to a choice of embedding) the algebraic closure, which you’ve assumed is separable, so L is certain separable. $\endgroup$ – user208649 Apr 23 '20 at 16:37
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Let's do the direction you attempted. I think your difficulty is that you are confused about the following point: generally one fixes a particular algebraic closure of the ground field, and then an extension of that ground field is taken to mean an extension in that particular choice of algebraic closure. This just ensures that things like the compositum always "make sense". One can do this because every algebraic extension (living anywhere) admits an inclusion into any particular algebraic closure; in particular all algebraic closures are (non-canonically) isomorphic.

Let $L/K$ algebraic. So either we take $L$ to already be inside of $\bar K = K^{sep}$ (under your assumption) or there is an inclusion of $L$ into $\bar K = K^{sep}$. Either way, $L$ is separable (in the second case it is isomorphic to a separable extension, which is the same thing). So all algebraic extensions are separable, which means $K$ is perfect.

Conversely, if $K$ is perfect then all of its algebraic extensions are separable, so in particular the algebraic closure is separable!

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    $\begingroup$ Yes, of course it can be shown to exist. This is usually one of the first things proven in a course on field theory. For instance, see Lang Chapter V section 2 (which precedes the discussion of separability). The general idea is very simple - reducing to the case of single elements, $K(\alpha)$ embeds into $\bar K$ by choosing a root of the minimal polynomial of $\alpha$ that lives in $\bar K$ and sending $\alpha$ to it. The referenced section has all the details $\endgroup$ – user208649 Apr 23 '20 at 17:37
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    $\begingroup$ Zorn is necessary because the embedding is really a choice of infinitely many compatible embeddings that are defined on each primitive sub extension. A compositum is defined for any number of subfields; the compositum of a collection $\{E_\alpha\}_{\alpha\in A}$ all contained in a bigger field $M$ is the smallest subfield of $M$ which contains all of the $E_\alpha$'s. Although I'm not sure what primitive element you are talking about since I didn't use one. $\endgroup$ – user208649 Apr 24 '20 at 18:54
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    $\begingroup$ Oh I thought you used that a (finite) extension is separable iff there exists a primitive element iff there are finitely many intermediate fields. But if you don't need to have finitely many subfields to take the compositum, none of that is required. $\endgroup$ – Jos van Nieuwman Apr 24 '20 at 19:00
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    $\begingroup$ My definition is that either char$(K) = 0$ or char$(K)$ is prime and the Frobenius endomorphism is surjective, immediately followed by the theorem that this is equivalent to every algebraic extension being separable.. I often run into problems with the 'nonimplication' ($L/K$ algebraic $\nRightarrow L/K$ finite). $\endgroup$ – Jos van Nieuwman Apr 24 '20 at 19:33
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    $\begingroup$ No it isn’t; I was thinking of taking the composition of all the finite extensions since I had in mind the definition that a perfect field has all its finite extensions separable, but since you have the result about all algebraic extensions being separable already it isn’t necessary (I sort of mixed the two definitions). I will edit that out. $\endgroup$ – user208649 Apr 25 '20 at 18:23

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