Let's talk about axiomatizability in first-order logic, with examples in ring theory. Let's call a class $\Ccl$ of rings **axiomatizable** if there exists a set $T$ of first order sentences such that $C\in\Ccl$ if and only if $C$ is a model of $T$ (that is, satisfies every sentence in $T$.)

What are some examples? The class of *all* rings is axiomatizable, because a rings are defined by a set of first-order axioms! The class of commutative rings is also axiomatizable because it is the class of rings that satisfies the additional sentence

$$\forall x\forall y(xy = yx).$$The class of fields is also axiomatizable, since it is the class of commutative rings that satisfies the additional sentence

$$\forall x(\lnot(x=0)\rightarrow \exists y(xy = 1)).$$ In fact, many classes of rings are axiomatizable. Try and think of a few more. On the other hand, there are some examples that are just not axiomatizable. For example, if you take any infinite ring $R$, then the class $\Ccl$ of all rings *isomorphic* to $R$ is not axiomatizable. Why is this? It is because if you have an infinite model of a set of sentences, then there exists models of arbitrary infinite cardinality, but every ring in $\Ccl$ has the same cardinality as $R$.

When is a class of rings is axiomatizable? Here is one set of criteria:

**Theorem.**A class of structures $\Ccl$ is axiomatizable if and only if it closed under elementary equivalence and the formation of ultraproducts.

*Proof.*$\Rightarrow$ Suppose $A\in\Ccl$ and $B$ is elementarily equivalent to $A$. Then $B$ satisfies all the axioms of $A$ and in particular satisfies all the axioms defining $\Ccl$. Therefore $B\in \Ccl$. Similarly, any ultraproduct of structures in $\Ccl$ satisfies all the axioms defining $\Ccl$, and hence also is in $\Ccl$.

$\Leftarrow$ Suppose $\Ccl$ is a class of structures closed under elementary equivalence and taking ultraproducts. Let $A$ be a model of ${\rm Th}(\Ccl)$. That is, $A$ is a model of the set of sentences that are true in every model in $\Ccl$. To finish the proof, we have to show that $A\in \Ccl$, which would show that $\Ccl$ is the set of all models satisifying ${\rm Th}(\Ccl)$.

To do this, for each $\alpha\in{\rm Th}(A)$, fix a model $C_\alpha\in\Ccl$ satisfying $\alpha$ and define

$$I_\alpha = \{ \beta\in {\rm Th}(A) : \alpha~\text{is true in}~C_\beta \}.$$ Then $\alpha\in I_\alpha$ so each $I_\alpha$ is nonempty, and satisfies $I_\alpha\cap I_{\alpha'} = I_{\alpha\wedge\alpha'}$. Hence, there exists an ultrafilter $\Ucl$ that contains all the $I_\alpha$ for $\alpha\in{\rm Th}(A)$. Since $\Ccl$ is closed under the formation of ultraproducts, the ultraproduct $C = \prod_\alpha C_\alpha/\Ucl$ is a model in $\Ccl$.

By construction of the ultrafilter $\Ucl$ and Łos's theorem, we see that $C$ is a model of ${\rm Th}(A)$, and conversely, every sentence that holds in $C$ also holds in $A$. Therefore $A$ and $C$ are elementarily equivalent, and so $A\in\Ccl$.

This criterion gives us another way to look at axiomatizability. For example, the class of fields of positive characteristic cannot be axiomatizable, since any ultraproduct of the form

$$\F_2\times\F_3\times \F_5\times\cdots/\Ucl$$ where $\Ucl$ is a nonprincipal ultrafilter is a field of characteristic zero. The class of fields of a fixed characteristic $p$ a prime are axiomatizable, which shows that an arbitrary union of axiomatizable classes is not necessarily axiomatizable.

**Exercise:** if $R$ is a finite ring, and $\Ccl$ is the class of all rings isomorphic to $R$, is the class $\Ccl$ axiomatizable?