Finding a complete theory which contains the theory of sentences that hold for every finite field. I am trying to make some exercises for a introductory course to mathematical logic.
Let $T$ be the theory consisting of all sentences that are true in every finite field. It is not that hard to see that this is a consistent, but not a complete theory. I am being asked to give a concrete example of theory which contains $T$, but which is complete.
I know that I will need to add a sentence for every cardinality characteristic ,but I am unsure if this is enough to make the theory complete. If it is not I do not know how to give a concrete theory which is even larger.
I also thought about taking the set of all sentences that are true in a specific model of $T$, but again I do not know how to show completeness.
 A: You wrote:

I also thought about taking the set of all sentences that are true in a specific model of $T$, but again I do not know how to show completeness.

This is the easiest way to pick a completion of a consistent theory $T$. Let $M\models T$ and consider $$T' = \text{Th}(M) = \{\varphi\mid M\models \varphi\}.$$ The completeness of $T'$ is obvious - there's almost nothing to show. $T'$ is consistent, since $M\models T'$. And for any sentence $\psi$, either $M\models \psi$ or $M\models \lnot \psi$, so either $\psi$ or $\lnot\psi$ is in $T'$, so $T'$ is complete.
In fact, every completion of $T$ has this form. If $T'$ is a complete consistent theory extending $T$, then letting $M\models T'$, we have $T' = \text{Th}(M)$.
Ok, but your exercise asks you to give a concrete example of a completion of $T$. This is the problem with the strategy above. For a given model $M$, it may be hard (if not impossible) to give a concrete axiomatization of $\text{Th}(M)$.
Let's specialize now to the situation in your exercise, where $T$ is the theory consisting of all sentences which are true in every finite field. As pointed out in the comments, there's an easy way to axiomatize a complete consistent theory extending $T$: For every prime $p$ and natural number $n>0$, there is a unique finite field of cardinality $p^n$ up to isomorphism. So letting $T'$ be the field axioms together with a sentence asserting that a model has exactly $p^n$ elements, $T'$ has a unique model up to isomorphism. And it is clear that a theory with exactly one model up to isomorphism is consistent and complete.
Ok, can we solve your exercise in a less trivial way, by writing down a complete theory extending $T$ that has infinite models? It turns out the answer is yes, but it's pretty complicated and relies on some deep work by James Ax: The Elementary Theory of Finite fields (1968).
To begin with, Ax axiomatized the theory $T$: the models of $T$ are called pseudo-finite fields. These are characterized as fields which are perfect (i.e., have characteristic $0$, or have characteristic $p$ and every element has a $p^{\text{th}}$ root), have absolute Galois group $\hat{\mathbb{Z}}$ (equivalently, have a unique algebraic extension of degree $n$ for each $n$), and are pseudo algebraically closed. It is not immediately obvious that these conditions can be written down as first-order sentences in the language of fields, but with a little work they can be.
Now Ax also proved that two pseudo-finite fields $K$ and $L$ are elementarily equivalent if and only if $\text{Abs}(K) \cong \text{Abs}(L)$, where $\text{Abs}(F)$ is the relative algebraic closure of the prime field inside $F$, i.e., those elements of $F$ which are algebraic over $\mathbb{Q}$ in characteristic $0$ and over $\mathbb{F}_p$ in characteristic $p$. Now the isomorphism type of an algebraic extension $E$ of a perfect field $F$ is determined by which polynomials in $F[x]$ have roots in $E$ (see here). So we can axiomatize the complete theory of a pseudo-finite field $K$ by writing down:

*

*The axioms of pseudo-finite fields.

*Axioms fixing the characteristic ($0$ or a prime $p$).

*Letting $F$ be the prime subfield of $K$, axioms specifying which polynomials in $F[x]$ have roots in $K$.

