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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, $c^2=(-1)^2=1$.

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    @Noomi : I suggest that you "check" answers, it gives more vote points to the answerers. It is right under the downvote arrow for you to click on ; choose wisely for you can only check one answer! (You can always change the check to another answer later on though.) An upvote is 10 points, a downvote is -2 and a check is 25, so they're really worth something.2012-09-16
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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 $c^2=[2]^2=[1]$.

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    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.2012-09-16
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    You're right, I didn't take into account that we could a priori have two different ring structures on the same group. Thanks for the correction.2012-09-16