Example of a bounded imperfect PAC field that is not separably closed How do you construct a bounded (meaning there are only finitely many separable finite extensions of any given degree) imperfect pseudo-algebraically closed field that is not separably closed? I assume they must exist, since for instance in this paper by Zoé Chatzidakis and Anand Pillay they are shown to be simple - in the model theory sense, see Forking and Dividing - and this would not be an interesting assertion about separably closed fields since they were known to be stable for a long time. In Fried-Jarden there is a chapter on PAC fields of positive characteristics, but it does not seem to bother with constructions.
 A: To me it was slightly surprising to find out that in a sense this is a very typical example. The $\mathrm{PAC}$ Nullstellensatz, for instance treated in Fried-Jarden [1] implies:

Let $K$ be a countable Hilbertian field and $e$ a positive integer. Then the fixed field of an automorphism $\sigma \in \mathrm{Gal}(K)$ is $\mathrm{PAC}$ for almost all $\sigma \in \mathrm{Gal}(K)$.

Also note Theorem 18.5.6. of the same book:

Let $K$ be a Hilbertian field and $e$ a positive integer. Then $\langle \sigma_1, \ldots, \sigma_n \rangle \simeq \hat{F}_e$ for almost all $(\sigma_1, \ldots, \sigma_n) \in \mathrm{Gal}(K)^e$

Note that in both of these theorem "for almost all" means "the set of elements not fulfilling this requirement has Haar-measure 0".
Together this gives that for almost all $\sigma \in \mathrm{Gal}(K)$ the fixed field of $\sigma$ is $\mathrm{PAC}$ and has absolute Galois group $\hat{\mathbb{Z}}$, so it is bounded. If we start with an imperfect countable Hilbertian field, for instance every function field over an infinite field of characteristic $p$, then this will give many imperfect bounded $\mathrm{PAC}$ fields, although in a very non-constructive manner.
[1]: Field Arithmetic, Michael D. Fried, Moshe Jarden, DOI: https://doi.org/10.1007/978-3-540-77270-5 
