## FANDOM

78 Pages

Angus Macintyre proved that if $K$ is a totally transcendental field (or, a field definable in a totally transcendental theory), then $K$ is algebraically closed. Poizat or someone else generalized this to superstable fields.

## Proof sketch Edit

First one shows that $K$ has Morley degree 1. Indeed, if $I$ denotes the connected component of the additive group $(K,+)$, then for each $a \in K$, multiplication by $a$ is an automorphism of the group $(K,+)$, hence sends $I$ to $I$. Thus $I$ is an ideal, so $I$ must be all of $K$, and $(K,+)$ is connected, therefore having Morley degree 1.

Using this, it follows that the maps $x \mapsto x^n$ are surjective for every $n$, essentially because if $a$ is a generic of $K^\times$, then $a^n$ is interalgebraic with $a$. Similarly, in positive characteristic, the Artin-Schreier map $x \mapsto x^p - x$ is onto. From this, one sees that totally transcendental fields are always perfect, and that they never admit Kummer extensions or Artin-Schreier extensions.

Suppose $K$ is a totally transcendental field. By induction on $n$, we can see that all the $n$th roots of unity are present in $K$. (If not, then the extension obtained by adding the $n$th roots is cyclic of degree dividing $n - 1$. By the inductive hypothesis and Kummer theory, this is a non-trivial Kummer extension, a contradiction.) Now, by Kummer theory and Artin-Schreier theory, it follows that $K$ has no cyclic extensions.

Finally, suppose $K$ is a totally transcendental field that is not algebraically closed. Then $K$ is perfect. Let $L/K$ be a nontrivial finite Galois extension. Let $G = \operatorname{Gal}(L/K)$, and let $H$ be a cyclic subgroup of $G$. By Galois theory, $H = \operatorname{Gal}(L/K')$ for some $K'$ between $K$ and $L$. But $K'$ is interpretable in $K$, as it is finite dimensional over $K$, so $K'$ is totally transcendental. And $L$ is a cyclic extension of $K'$, contradicting the fact proven above that no totally transcendental field has any non-trivial cyclic extension.

By being more careful, the above proof can be made to work just as well for superstable fields.

On the other hand, strictly stable fields can fail to be algebraically closed. For example, separably closed fields, as pure fields, are stable.