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.
Best Answer
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 $-1$.)