A unital ring$R$ is an integral domain (or simply domain) if it is nontrivial and has no non-zero zero divisors (i.e., $1 \ne 0$ and $a b = 0$ implies $a = 0$ or $b = 0$). For example, the ring of integers, any skewfield, the ring of global sections of the structure sheaf of any integral scheme, an Ore extension of any other integral domain.

In constructive mathematics, one wants to phrase the condition as $a b \neq 0$ whenever $a \neq 0$ and $b \neq 0$, where $\neq$ is a tight apartness relation relative to which the ring operations are strongly extensional. (Of course, if the underlying set of the ring has decidable equality —as is true of $\mathbf{Z}$, $\mathbf{Q}$, $\mathbf{Z}/n$, finite fields, etc— then this is trivial.)

The trivial ring is too simple to be an integral domain. You can see this by phrasing the definition without bias as: any product of (finitely many) nonzero elements of $R$ (which includes the empty product $1$) must be nonzero.

Some authors require an integral domain to be commutative, even when they do not require this of rings in general. Commutative integral domains are precisely subrings of fields.

In principle, one could just as easily consider a rig or semiring$R$. In that case, however, a zero divisor is not what the name literally implies: the definition is that multiplication by a nonzero element (on either side) is injective. Furthermore, we should add the additional requirement that addition in $R$ is cancellable (that is, addition by any element is injective), to make the analogue of the previous paragraph correct. Since ‘integral domain’ is too specific and ‘integral ring’ is not standard (and means something else in the phrase ‘integral ring extension’), it's not clear exactly what these should be called; perhaps integral cancellable rig/semiring is sufficiently unambiguous.

An integral domain $R$ is an Ore domain if the set of all nonzero elements is an Ore set in $R$. In that case the Ore localized ring is called the Ore quotient ring? of $R$.