Past exam of the mathematics course of the University of Cambridge / 2023 / ib / Paper 2 / 1E / Solution

Forthe parity condition is preserved by componentwise addition, negation, and multiplication, and $(1,1)$ is the identity. Thus $R$ is a ring.

It is not an integral domain, since $(2,0)$ and $(0,2)$ are nonzero elements of $R$ whose product is zero. It is also not a product of two nontrivial rings. Indeed, an idempotent in ${\mathbb{Z}}^2$ has each coordinate in $\{0,1\}$, and the parity condition leaves only $(0,0)$ and $(1,1)$. The proved equivalence then excludes a nontrivial product decomposition.