I was wondering if every algebraically closed field is an algebraic extension of a nontrivial subfield. Suppose $L$ is an algebraically closed field. By nontrivial subfield, I mean a subfield $K \subset L, K \neq L$.
I ask this because I was wondering if every algebraically closed field was the algebraic closure of some subfield. This does seem like the case sometimes. Suppose we take some $\alpha \in L$ such that $\alpha$ has nontrivial degree over the base field of $L$. Take the maximal subfield of $L$ that does not contain $\alpha$, $K'$. Then $K$' has finite degree under $L$ and we're done. The obvious problem with this is when every element of $L$ is transcendental over the base field, in which case the claim seems less straightforward. But honestly I doubt that the claim is true because otherwise we would probably equate the two terms.
Any guidance would be greatly appreciated.