Angus Macintyre proved that if is a totally transcendental field (or, a field definable in a totally transcendental theory), then is algebraically closed. Cherlin and Shelah generalized this to superstable fields.

## Proof sketch

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

Using this, it follows that the maps are surjective for every , essentially because if is a generic of , then is interalgebraic with . Similarly, in positive characteristic, the Artin-Schreier map 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 is a totally transcendental field. By induction on , we can see that all the th roots of unity are present in . (If not, then the extension obtained by adding the th roots is cyclic of degree dividing . 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 has no cyclic extensions.

Finally, suppose is a totally transcendental field that is not algebraically closed. Then is perfect. Let be a nontrivial finite Galois extension. Let , and let be a cyclic subgroup of . By Galois theory, for some between and . But is interpretable in Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K} , as it is finite dimensional over , so Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle K'} is totally transcendental. And is a cyclic extension of Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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. ::: ::: :::