Pseudo-finite field vs Nonstandard finite field

Question

Let $\mathbb{N}^*$ be a countable non-standard model of Peano arithmetic (PA) and let $\mathbb{Z}^*$ be the integers extended from $\mathbb{N}^*$. A non-standard finite field would be a ring $\mathbb{Z}^* /n^* \mathbb{Z}^*$ where $n^*$ is a non-standard prime number larger than any standard natural number.

How is a nonstandard finite field different from a pseudo-finite field? Can there be a mapping from one to the other?

Answer 1

The answer is negative. Not every pseudo-finite field, i.e a model of the theory of finite fields, is "nonstandard integers modulo a nonstandard prime". Every field of this form has characteristic zero. By contrast, there are pseudo-finite fields of positive characteristic:

http://www.logique.jussieu.fr/~zoe/papiers/Helsinki.pdf

(see page 17, example 5.1)

Answer 2

Quadratic residuosity gives an easy way to prove different pseudofinite fields non-isomorphic. e.g. $-1$ has a square root in the field if and only if $p \equiv 1 \bmod 4$. More generally, you have a primitive $n$-th root of unity if and only if $p \equiv 1 \bmod n$.

My first instinct that you could prove that two such fields are (externally) isomorphic if and only if the two primes lie in the same equivalence class modulo every standard prime. The only if part is more likely than the if part, in my opinion.