If $A$ and $B$ are integral domains, how to make $A\times B$ an integral domain?

Question

I've been lately reading a bit of ring theory and when I reached the 'Integral Domains' section a question suddenly arised and it seems pretty natural to me. It is clear that if $A$ and $B$ are integral domains, then $A\times B$ is not neccesarily an integral domain (if the product on $A\times B$ is defined by $(a,b)(a',b')=(aa',bb')$). In fact, if $A$, and $B$ are rings with unit, $A\times B$ is NOT an integral domain as $(1,0')(0,1')=(0,0')$. The question is

Is it possible to define a 'canonical' product on $A\times B$ in order to make it into an integral domain?

With 'canonical' I mean 'definable from the operations on $A$ and $B$, or from some fixed relation between them' (for example, when the semi-direct product on groups is defined it only depends on a certain homomorphism from a group $G$ to the automorphism group of another group $H$).

I would appreciate any ideas. Thanks in advance.

Answer 1

This is not possible in general. Two well-known facts are that a finite integral domain is a field, and that a finite field must have a prime power number of elements. Thus, for instance, there is no way to give $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/3\mathbb{Z}$ the structure of an integral domain.

Answer 2

If $B=A$ and there exists a polynomial $x^2+ax+b\in A[x]$ which has no zeros in the field of fractions of $A$, then we can define a multiplication on $A\times A$ by mimicking the multiplication on the integral domain $A[x]/\left(x^2+ax+b\right)$. That is, we can define a multiplication on $A\times A$ as $$(p,q)\cdot (r,s):=(pr-bqs,ps+qr-aqs)\,,$$ for all $p,q,r,s\in A$. This is very much how we get $\mathbb{C}$ as $\mathbb{R}\times\mathbb{R}$, $\mathbb{F}_4$ as $\mathbb{F}_2\times\mathbb{F}_2$, $\mathbb{Z}\left[\frac{1+\sqrt5}{2}\right]$ as $\mathbb{Z}\times \mathbb{Z}$, etc.