Is it possible to characterize the theory of Integral domains with first-order logic alone ?

Question

Is it possible to characterize general ring theory with first-order logic alone ? Is it possible to do so for the theory of Integral domains ?

Answer 1

The language of rings include $+,\cdot$ as two binary operators called "addition" and "multiplication" repsectively, and for simplicity we can also add two constant symbols $0$ and $1$ to denote the neutral elements (if you don't want to talk about unital rings, just omit $1$, of course).

Now the axioms state that both binary operators are associative, and $+$ is commutative and that $\cdot$ is distributive, and that $0$ is the additive neutral element, and that every element has an additive inverse; and finally $1$ is the multiplicative neutral element.

All those are first-order sentences in the language of rings, for example: $\forall x(x+0)=x$ is the statement that $0$ is the additive neutral element. And so on.

So the theory of rings is indeed a first-order theory.

If you want to add the statement that the ring is an integral domain, just recall that an integral domain is a ring that $\forall x\forall y(x\cdot y=0\rightarrow(x=0\lor y=0))$. Namely, whenever $x\cdot y=0$, one of them is $0$. Lo and behold, the statement we wrote is a first-order statement in the language of rings, and when adding it to the axioms mentioned above, we get exactly the theory of integral domains.

(As Zhen Lin indicates below, it is also standard to add that $0\neq 1$ for integral domain, but this is also a first-order statement in the language of rings, so it's not a problem.)