# For a ring $\{0,1,c\}$, does $c^2=1$?

Say you have an arbitrary ring with three elements, $\{0,1,c\}$. Why does it have to be that $c^2=1$? If we don't assume that $c$ is invertible, what goes wrong if $c^2=0$ or $c^2=c$?

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Notice that we must have $c+1=0$, since $c+1=1$ implies $c=0$ and $c+1=c$ implies $1=0$, both contradictions. Thus we have $c=-1$. Therefore, $c^2=(-1)^2=1$.
As a different way to see it, a ring with three elements is in particular a group with three elements. Now the only group of order $3$ is $\mathbb Z / 3 \mathbb Z$, so clearly your $c$ must be $[2]$ and thus $c^2=[2]^2=[1]$.
The only group of order $3$ is $\mathbb Z \backslash 3\mathbb Z$, but is that the only ring with three elements? That is what Tarnation proved, but you haven't. What you could do though to complete your argument really neatly is that since in the group you have $2+2 = 1$, you can also say that $2(1+1) = 2 \cdot 2 = 1$ and since you've shown $c=2$ using your argument you're done. Nonetheless good idea, I +1'ed it. –  Patrick Da Silva Sep 16 '12 at 6:09