# Algebraic closure is in some sense minimal

Suppose $T \leq S$ is a field extension. If $S$ is algebraically closed and algebraic over $T$, then we call $S$ an algebraic closure of $T$. Is this the same condition as: $S$ is minimal amongst field extensions of $T$ that are algebraically closed?

The latter would seem like a more natural definition of "algebraic closure", but after a quick check I didn't see this used anywhere.

An attempt at showing that they are in some sense equivalent:

Suppose $S$ is algebraically closed and algebraic over $T$. Let $L$ satisfying $T \leq L \leq S$ be an algebraically closed extension. Any $s \in S \setminus L$ is algebraic over $T$, and all polynomials in $T[x]$ are contained in $L[x]$, so they split in $L$. Thus, $s \in L$, which shows that $S$ is minimal amongst field extensions that are algebraically closed.

Suppose $S$ is minimal amongst extensions of $T$ that are algebraically closed. Put $U = \{ s \in S : s \text{ is algebraic over } T \}$. Then $U$ is an algebraic closure (by the first definition). From minimality of $S$ it follows that $U=S$.

• This is true and your proofs look good. Commented May 14, 2018 at 2:18

You're right and you can do one better : if $S$ is an algebraic closure of $T$ and $L$ is any field with roots to all polynomials of $T$ (so $T\leq S, T\leq L$), then there is a field morphism (hence an embedding) $S\to L$ fixing $T$ pointwise.
So you don't even need the comparison between $S,L$ : it's a minimum