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 $c^2=c$?
For a ring $\{0,1,c\}$, does $c^2=1$?
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ring-theory
2 Answers
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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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2@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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1You'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