ac.commutative algebra - Is (relatively) algebraically closed stable under finite field extensions?

Counterexample:

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 $a^p,b^pin F$.

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 $Kneq F$ 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 $F^primeotimes_F L$ is not a field: the tensor product $F^primeotimes_K L$ equals $L(x)[Y]/(f)$. However $f$ is a $p$-th power in $L(x)[Y]$.

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