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A few days ago I found this question here on math.stackexchange, which gave a sufficient criterion for a separable, algebraic extension $E/K$ to be an algebraic closure of $K$. However it was claimed by KCd, in a comment below the question I'm referring to, that we can drop the separability condition on the extension and still get the same result

My question is: how does the proof work in the non-separable case?

Put precisely: Let $E/K$ be an algebraic extension such that every non-constant polynomial in $K[X]$ has a root in $E$, then $E$ is the (up to isomorphism) algebraic closure of $K$.

It is pretty clear to me that Makotos proof in the separable case (which can be found on the page the link above is leading to) won't work for the case of a non-separable extension (e.g. because the primitive element theorem may fail). I had some ideas of working with the separable closure but didn't try much, because I didn't see a real perspective in my approach. In other words, I'm stuck.

Lastly an apology: I refrained from asking KCd this question directly because it might be of common interest.

Regards

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Whoops - I'm halfway through typing up an answer, but I have to take off for a bit. But I'll finish it up tonight. –  mixedmath May 11 '12 at 19:24
    
@mixedmath: bring it on!;) Or tell me at least roughly how the argument goes. –  Nils Matthes May 17 '12 at 11:09
    
@mixedmath I would be interested in seeing your answer if you are still interested in writing it down. –  M Turgeon Jun 16 '12 at 0:28
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@MTurgeon: Oh - I somehow completely forgot! I'll get back to it now. Thank you for that - –  mixedmath Jun 17 '12 at 1:45
    
@Nils , I've added an answer to your question. –  DonAntonio Jun 17 '12 at 11:30

1 Answer 1

up vote 3 down vote accepted

Let $\,E/K\,$ be not separable, so that $\,\operatorname{char} K=p>0\,$ . Then, $\,K^p\neq K\,$. Our basic assumption is that any non-constant polynomial in $\,K[x]\,$ has a root in $\,E\,$. Define $$F:=\{x\in E\;|\;x^{p^n}\in K\,,\,\text{for some natural}\,\,n\}$$

First exercise: Show that $\,F\,$ is a field and $\,K\leq F\leq E\,$.

Now, let us take $\,x\in F\,$, with say $\,\alpha=x^{p^n}\in K\Longrightarrow p(t):=t^{p^{n+1}}-\alpha\in K[t]\,$ . Our assumption is that $$\exists w\in E\,\,s.t.\,\,p(w)=0\Longrightarrow x^{p^n}=w^{p^{n+1}}=\alpha\in K\Longrightarrow w\in F $$

Second exercise: Prove that in fact $\,x=w^p\,$ and deduce that $\,F^p=F\,$ (i.e., $\,F\,$ is a perfect field)

Third exercise: Using our basic assumption, show that $\,E\,$ is perfect (hint: any irreducible polynomial over $\,E\,$ divides some irreducible pol. over $\,F\,$)

Let now $$h(t):=\sum_{i=0}^kc_it^i\in F[t]\Longrightarrow \,\exists N\in\mathbb{N}\,\,s.t.\,\,c_i^{p^N}\in K\Longrightarrow g(t)=\sum_{i=0}^kc_i^{p^N}t^i\in K[t]$$and our basic assumption says $\,\exists\,z\in E\,\,s.t.\,\,g(z)=0\,$ .

Finally, since $\,E^{p^n}=E\,\,\,\forall n\in\mathbb{N}\,$ (why?) , we have that $\,z=r^{p^n}\,,\,\text{for some}\,n\in\mathbb{N}\,,\,r\in E$

Fourth exercise: Prove that $\,h(t)\,$ has a root in $\,E\,$, ending thus the proof.

(Hint: Use Freshman's Dream Theorem: $$(a+b+c+...)^{p^k}=a^{p^k}+b^{p^k}+c^{p^k}+...$$ when working in fields with characteristic $\,p$. Of course, the sum above is finite)

The above is not original. I remember this (as it stunned me as it theoretically simplifies the proof I knew about the existence of an alg. closed overfield for any field) from some answer given by K. Conrad, who based it on some paper by Gilmer, if I remember correctly.

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Dear DonAntonio: Thanks for your answer. I will check it out in detail as soon as possible. I have some unrelated (but still mathematical! ;)) work to do before that, so don't be surprised if I don't accept your answer in the next few days. Your're not forgotten! Regards, Nils –  Nils Matthes Jun 18 '12 at 17:14
    
Thank you for your consideration, @Nils. Take the time you need to check the above. –  DonAntonio Jun 18 '12 at 17:22
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Dear @DonAntonio: Finally, I managed to go through what you wrote in detail, and I understand enough of it to accept your answer. Thanks very much for sharing this beautiful proof with me/us! –  Nils Matthes Oct 4 '12 at 18:14
    
@NilsMatthes can u please tell me a counter example which shows this statement is false if $E|K$ is not necessarily algebraic? –  user115608 Oct 24 at 13:05
    
@DonAntonio can u please tell me a counter example which shows this statement is false if $E|K$ is not necessarily algebraic? –  user115608 Oct 24 at 13:06

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