2

Suppose that $R$ is a ring. Prove that if $a\in R$, $a^2 = a$, then R is a commutative ring.

So, I know that this means that the ring multiplication is commutative. So... is this saying that for ANY $a\in R$, $a^2 = a$? Which means that every element of R is its own multiplicative inverse... But inverses, if they exist, are unique, so then the only element of R is unity. And if all you have is unity, then of course multiplication is commutative?

I'm probably way off here.

EDIT:

So, following the advice in the comments, I produced something that looks like this:

Since R is a ring, if $a,b\in R$, then $ab\in R$. Thus

$(ab)^2 = ab$

$\to (ab)(ab)=ab $

$\to (ab)(ab)a=aba $

$\to a(ba)(ba)=a(ba)$

$\to a(ba)^2=a(ba)$

So, AM I allowed to just say "by cancellation" $(ba)^2=ba$ therefore R is a commutative ring?

Indigo
  • 398

2 Answers2

5

You need to prove that $ab=ba$ for all $a, b \in R$. I'm not sure I follow the approach that people are attempting to outline in the comments, and certainly the phrase "by cancellation" is very fishy. Here's the approach I know works:

  1. Prove that $R$ has characteristic $2$ (or $1$, but that's pretty boring). This follows from $2^2=2$. Edit: Or, as pointed out in the comments, if you don't assume $R$ has unity, you can still show that $a+a=(a+a)^2 = a^2+a^2+a^2+a^2 = a+a+a+a$, witnessing $a+a=0$ for every $a \in R$.
  2. Compare $(a+b)$ with $(a+b)^2$, which must be equal. Since you don't yet know that multiplication is commutative, you have $(a+b)^2 = a^2+ab+ba+b^2 = a+ab+ba+b$. Now subtract $a$ and $b$ to get $ab+ba=0$. In characteristic $2$, this is equivalent to $ab=ba$.
Dustan Levenstein
  • 13,219
  • Why is this true for characteristic 2? And how can the ring be of characteristic 2 and also be characteristic 1? Am I misunderstanding? – Indigo Dec 05 '15 at 06:49
  • I'm saying that knowing $2=0$ in $R$ gives two possibilities for the characteristic of $R$, namely $2$ or $1$. There's only one ring of characteristic $1$, so that's not really a very interesting case. And to see why $ab+ba=0$ is equivalent to $ab=ba$, simply subtract to get $ab=-ba$, and observe that $1=-1$ in characteristic $2$. – Dustan Levenstein Dec 05 '15 at 06:52
  • Okay. Sorry, not trying to be dense. I understand that the characteristic of a ring is the least positive integer $n$ such that $na=0$ for all $a\in R$.

    Ah, I get it. Well, I think I get it? Aren't we assuming that $2\in R$?

     – Indigo Dec 05 '15 at 06:59
  • 1
    @Indigo: $2$ is, by definition, $1 + 1$. It's canonically an element of every ring. – Qiaochu Yuan Dec 05 '15 at 07:07
  • 2
    It is not necessary to assume that $R$ has unity. Just use that $a+a=0$ for all $a\in R$. – user26857 Dec 05 '15 at 10:52
  • 2
    That is, $2a$ is defined in any ring, even if 2 isn't – rschwieb Dec 05 '15 at 14:49
-4

@Indigo i hope this is usefull to you, for $a,b\in R$ , $a^{2}=a$ and $b^{2}=b$ , since $R$ is ring, $ab\in R$. $ab=(a^{2})(b^{2})$ and $ab=(ab)^{2}$ We get $(ab)^{2}=(a^{2})(b^{2})$ $abab=aabb$ By cancelation $ba=ab$

AUA
  • 59
  • 7
  • But, is it allowed to just cancel things? Doesn't cancelation assume unity and inverses? Or am I thinking way too hard and canceling things is perfectly allowed? Because if it IS, then I am pretty sure that I have it. – Indigo Dec 05 '15 at 06:34
  • 1
    I think this ring should has add thing which is no zero divisor (to make sure the cancelation still hold on, it can be proven by using the "+" operation) – AUA Dec 05 '15 at 06:47
  • 1
    If you ask for no zero divisors, the ring becomes the field of two elements or the zero ring, utterly ruining the problem. Cancelation and Boolean rings are extremely incompatible. – rschwieb Dec 05 '15 at 16:31