Finding an example of field extensions satisfying certain properties

Question

This is problem 3.18 from Patrick Morandi's Field and Galois Theory.

Give an example of fields $k \subseteq K \subseteq L$ and $l \subseteq L$ for which $l/k$ and $L/K$ are algebraic, $k$ is algebraically closed in $K$, and $lK = L$, but $l$ is not algebraically closed in $L$.

This question has been asked twice, namely here and here, though neither had any answers, and only some inconclusive comments. Normally I wouldn't ask the question again, but after struggling with it for weeks at this point, I emailed Morandi, and he gave me some hints.

He told me that he remembers finding an example, but that he isn't sure. He told me that it would need to be over characteristic $p$ and that the extensions (at least some of them I suppose) could not be separable. Moreover, he told me that $K = k(x)(a)$, where $a$ is a root of an irreducible polynomial $f(x,y) \in k(x,y)$ (I'm not 100% what the root of a two variable polynomial is). Finally, he said that $l$ is an algebraic extension of $k$, where $f$ becomes reducible over $l$. Still however, I can't come up with anything. I was hoping that with these extra hints, someone would be able to figure it out.

He did point out that there is a chance that there is an error in the question, but I haven't managed to prove that such an example cannot exist.

Answer 1

I think he might be wrong, because his claim fails in characteristic $0$ and if his claims holds in characteristic $p$ then the minimal polynomials can be lifted to a setting where his claim holds in characteristic $0$.

• Here is why his claim fails in characteristic $0$

$K/k$ is a non-algebraic extension, $l/k$ is a finite extension and $K\cap \overline{k}=k$.

With the primitive element theorem $l=k(a)$.

Then $k(a)=k[x]/(f)$ and $K(a)=K[x]/(g)$ where $f,g$ are the monic minimal polynomials of $a$. Since $g$ divides $f$ then $g\in \overline{k}[x]$ thus $g\in k[x]$ and $g=f$.

His claim is that it may happen that $b\in K(a)\cap \overline{k}$, $b\not \in k(a)$.

If so then by the primitive element theorem $k(a,b)=k(c)$. The same argument as before gives $$[k(a):k]=[K(a):K]=[K(c):K] = [k(c):k]$$ whence $k(a)=k(c)$ and $b\in k(a)$ which is a contradiction.

Thus $lK\cap \overline{k}=l$ and since it holds whenever $l/k$ is a finite extension it holds whenever $l/k$ is algebraic.