Model for a theory of commutative rings having quantifier elimination Let $T$ be a theory of commutative rings , in the first order language of rings, having quantifier elimination . Let $R$ be a model for $T$ , so that $R$ is a commutative ring. Then is it necessarily true that all non-zero polynomials in $R[X]$ has only finitely many roots in $R$ ?
 A: No, this is not true. A commutative ring $R$ is Boolean if $x^2 = x$ for all $x\in R$ (from which it also follows that $x + x = 0$ for all $x\in R$). A Boolean ring $R$ is atomless if for all $x\neq 0$, there exists $y$ such that $y\neq 0$, $y\neq x$, and $xy = y$. Now there is a unique countable atomless Boolean ring $\mathcal{R}$ up to isomorphism (so the theory of atomless Boolean rings is $\aleph_0$-categorical and complete), and the theory of this ring has quantifier elimination. But the polynomial $x^2 - x$ has infinitely many roots in $\mathcal{R}$ (namely every element of $\mathcal{R}$!).
The motivation for this example comes from the well-known fact that Boolean rings and Boolean algebras are essentially the same objects. On any Boolean ring, we can define a partial ordering by $x\leq y\iff xy = x$. Under this ordering, $R$ is a Boolean algebra, with meet, join, and complement operations defined by $x\wedge y = xy$, $x\vee y = x+y+xy$, and $\lnot x = 1+x$. And conversely, given a Boolean algebra $(B,0,1,\wedge,\vee,\lnot)$, we can give $B$ the structure of a Boolean ring, by defining $xy = x\wedge y$, $x+y = (x\wedge \lnot y)\vee (\lnot x\wedge y)$, and $-x = x$. So quantifier-free formulas in the language of rings are equivalent (over Boolean rings/algebras) to quantifier-free formulas in the language of Boolean algebras, and vice versa.
Now Boolean algebras are a well-known example in Fraïssé theory: the class of finite Boolean algebras is a Fraïssé class, whose Fraïssé limit can be axiomatized by the theory of atomless Boolean algebras. So we get (from Fraïssé theory) that the theory of atomless Boolean algebras is complete, $\aleph_0$-categorical, and admits quantifier elimination. Then using our translation, it's easy to see that the theory of atomless Boolean rings satisfies the same properties.
You can learn a lot about rings with quantifier elimination in the paper Rings which admit quantifier elimination by Bruce I. Rose (doi: 10.2307/2271952, jstor). In this paper, Rose shows that for several classes of rings, the only infinite rings in these classes with quantifier elimination are algebraically closed fields. On the other hand, he generalizes the above observation about atomless Boolean rings to the class of atomless $p$-rings, showing that these also admit quantifier elimination.
