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Let $F\subset F'$ be a field extension such that $F$ is algebraically closed inside $F'$, i.e. if $x\in F'$ is algebraic over $F$ then $x$ belongs to $F$ itself. Let now $F\subset L$ be a finite field extension generated by one element, i.e. $L=F(\alpha)$ for some $\alpha\in L$. Is then $L$ also algebraically closed inside $F'\otimes_{F}L$ ?

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let $F$ be a non-perfect field of characteristic $p$.

Let $L$ be an extension of $F$ of degree $p^2$ such that $L=F(a,b)$ with $a^p,b^p\in F$.

The polynomial $f(Y):=Y^p-(a^px^p+b^p)\in F(x)[Y]$ is irreducible, where $F(x)$ is the rational function field in the variable $x$.

Consider $F^\prime := F(x,y)$, where $y$ is a root of $f$.

Then $F$ is algebraically closed in $F^\prime$: let $K$ be the algebraic closure of $F$ in $F^\prime$. Then $[K:F]=[K(x):F(x)]\leq [F^\prime :F(x)]=p$. Hence $K\neq F$ implies $F^\prime =K(x)$ and thus $y=g(x)\in K[x]$ with $[K:F]=p$ -- in contradiction to the choice of $y$.

The tensor product $F^\prime\otimes_F L$ is not a field: the tensor product $F^\prime\otimes_K L$ equals $L(x)[Y]/(f)$. However $f$ is a $p$-th power in $L(x)[Y]$.


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Very nice. Now I see why my proof idea didn't pan out! – Kevin Buzzard Dec 2 '09 at 19:53

I might was a bit too quick. Assuming your first statement, we then apply it to the case $L\subset F'\otimes_{F}L$, right!? So don't we first need to know that $F'\otimes_{F} L$ is a field to apply your criterion? Which is actually also the original statement I was interested in and which brought me to the above question. So maybe I should state my original problem:

Given $F$ algebraically closed inside $F'$ and $F\subset L$ a finite field extension, is then $F'\otimes_{F}L$ again a field?

If $L$ is simple, then this is easy seen to be true. Then I wanted to do an induction argument on the number of generators and for this I needed the above statement (given in my first question) to be true.

I am not quite sure it is actually true. Normally one needs something like $F\subset F'$ to be separable for general $L$. But one might can drop this condition for finite extensions $F\subset L$. So any suggestions regarding this problem/question?

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If F' tensor F-bar is a field, then F' tensor L is a subring, so it's an integral domain, and it's finite over a field, so it's a field by the standard evil trick. – Kevin Buzzard Dec 1 '09 at 20:50
[hence if my assertion "$F$ alg closed in $F'$ iff $\overline{F}\otimes_FF'$ is a field" is true then it solves all your problems.] – Kevin Buzzard Dec 1 '09 at 21:45

Let me have a punt at this. $F$ alg closed inside $F'$ iff $\overline{F}\otimes_FF'$ is a field, right? So now it's easy because $\overline{L}$ is an algebraic closure of $F$, and I don't think I even assumed $L$ was simple over $F$. Did I miss something?

Edit: my first assertion needs justification and I can't justify it so I could be mistaken. It's clear that $\overline{F}\otimes_FF'$ is a field iff $L\otimes_FF'$ is a field for all finite extensions $L$ of $F$ (union of fields is a field; integral domain finite over a field is a field). Moreover, if $F$ is not algebraically closed in $F'$ then choose some $\alpha\in F'$ algebraic over $F$ but not in $F$, and $L=F(\alpha)$ is finite but $L\otimes_FF'$ is not a field (it contains $L\otimes L$), so one way is OK. The problem is the other way. First say $K=F(\beta)$ is finite and simple over $F$. Then $F$ alg closed in $F'$ implies $K\otimes_FF'$ is a field because if the min poly of $\beta$ factors in $F'$ then the factors are algebraic over $F$, so in $F$. But as the OP quite rightly points out, that isn't enough for me. So this is not yet an answer to the question.

Second edit: I couldn't justify my first assertion because it is false. The counterexample posted looks good to me.

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And why is the first statement true? The rest is clear. – Jason Dec 1 '09 at 19:54
Lemme check: is this homework?? – Kevin Buzzard Dec 1 '09 at 19:56
No, don't worry ;). Just something I came across. – Jason Dec 1 '09 at 20:05
I keep trying to write down the trivial proof and getting bogged down, and now I have to put the kids to bed :-/ Maybe I made a mistake. I'll be back in about 45 mins, you can let me know if I have :-) – Kevin Buzzard Dec 1 '09 at 20:41
OK back. I still don't see an argument though, so maybe I am simply mistaken and there's some weird inseparability issue. I'll edit. – Kevin Buzzard Dec 1 '09 at 21:31

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