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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 $(0,0)$.

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  • $\begingroup$ You are almost correct: the idempotent elements in general are the $a$ such that $a \star a = a$. In this case of $\mathbb{Z} \times \mathbb{Z}$, we would have to find $(x,y)$ such that $(x,y)^2 = (x,y)$. You can probably alter your previous attempt quite easily. $\endgroup$ – ToucanNapoleon Oct 3 '16 at 8:39
  • $\begingroup$ Ah yes, I noticed the error. Silly mistake, thank you for pointing it out. Would I have then, four elements (1,1), (0,0), (1,0), (0,1)? $\endgroup$ – user354021 Oct 3 '16 at 8:50
  • $\begingroup$ Seems good to me. $\endgroup$ – ToucanNapoleon Oct 3 '16 at 8:53
  • $\begingroup$ @DarkoDekan Does $(1,0)$ belong to $S$? I don't think so $\endgroup$ – egreg Oct 3 '16 at 9:07
  • $\begingroup$ @egreg I completely forgot about the S properties. So only (0,0) and (1,1) should be. $\endgroup$ – user354021 Oct 3 '16 at 9:57
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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.

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