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Let $R$ be a ring with identity such that $r^2=r$ for all $r\in R$. Show that the characteristic of $R$ is $2$ and that $R$ is commutative.

My argument was the following. But, the thing I am not sure about is, whether we can say that $f$ has at most two roots in any ring.

So, let $f(x)=x^2-x\in R[x]$. Then $f$ has at most $2$ roots. $0,1\in R$ and they are roots of $f$. So, $R=\{0,1\}$. Hence Char $R=2$ and it is commutative.

Appreciate if point out any errors.

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  • $\begingroup$ $R$ can have more than two elements. Consider $(\mathbb{Z}/2\mathbb{Z})^n$, which has $2^n$ elements. $\endgroup$
    – Eric Thoma
    Jan 3, 2016 at 2:27

1 Answer 1

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It is not clear that $f$ has at most two roots. For this you need that $R$ is a domain. For example, $p(x) = 2x \in \mathbb{Z}/4[x]$ has two roots.

Instead notice that for all $a \in R$ $$ 1+a = (1+a)^2 = 1 + 2a + a^2 = 1 + 2a + a, $$ so $2a = 0$. Therefore $\mathrm{char}(R) = 2$. For all $a,b \in R$ we now have $$ ab-ba = ab+ba = (a+b)^2 - a^2 - b^2 = a+b - a - b = 0, $$ so $R$ is also commutative.

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