Given ring $(S,+,*)$ and $S=\{(x,y)\in \mathbb{Z} \times \mathbb{Z} \mid x\equiv y \pmod 2\}$, where $\mathbb{Z}$ is the set of integers. Find all idempotent elements of $S$.
How to approach this? I'm not even sure how to multiply ordered pairs. Basically I have to find x for which $(x,x)*(x,x)=(x,x)$, correct? My attempt was $(x^2, x^2) = (x,x)$ thus $x^2 = x$ and $x=0$ or $x=1$. And elements are $(1,1)$ and $(0,0)$.
An idempotent in a subring is also such in the bigger ring.
The idempotents in the ring $\mathbb{Z}\times\mathbb{Z}$ are the elements $(x,y)$ such that $$ (x,y)=(x,y)^2=(x^2,y^2) $$ so $x=0$ or $x=1$ and $y=0$ or $y=1$. These four elements are indeed idempotents; but only two belong to the subring $S$ which consists of the elements $(x,y)$ where $x$ and $y$ are either both even or both odd.