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3 added 119 characters in body

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.

2 added clarifying sentence

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.

1

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?
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