Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $c^2=c$?

share|improve this question

2 Answers 2

up vote 12 down vote accepted

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

share|improve this answer
    
Thanks tarnation. –  Noomi Holloway Sep 16 '12 at 5:07
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. –  Patrick Da Silva Sep 16 '12 at 6:10

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 $c^2=[2]^2=[1]$.

share|improve this answer
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
1  
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. –  Niccolò Sep 16 '12 at 21:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.