# Is every algebraically closed field an algebraic extension of a nontrivial subfield?

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.

Let $$k$$ be the prime subfield of $$L$$ and let $$B$$ be a transcendence basis for $$L$$ over $$k$$ (that is, a maximal algebraically independent subset). Then $$L$$ is algebraic over the subfield $$K=k(B)$$ (if some $$x\in L$$ were transcendental over $$K$$ then $$B\cup\{x\}$$ would be algebraically independent, contradicting maximality of $$B$$). We can see that $$K$$ is not all of $$L$$ since $$K$$ cannot be algebraically closed. For instance, if $$x\in B$$, then $$x$$ cannot have a square root in $$K$$ (you can see this by using the explicit representation of elements of $$K$$ as formal rational functions in the elements of $$B$$ and unique factorization of polynomials, for instance). And if $$B$$ is empty, then $$K$$ would just be the prime field which is not algebraically closed.
(Incidentally, the question of whether $$L$$ is a nontrivial finite extension of a subfield is more interesting. By a theorem of Artin and Schreier, the answer turns out to be no if $$L$$ has positive characteristic, and yes if $$L$$ has characteristic $$0$$. Moreover, in the characteristic $$0$$ case, the only possibility is that the extension $$K\subset L$$ "looks like $$\mathbb{R}\subset\mathbb{C}$$": more precisely, $$K$$ will always be a real-closed field and $$L$$ is obtained from $$K$$ by adjoining a square root of $$-1$$.)