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I'm familiar that fields and integral domains of finite characteristic have prime characteristic. What about the case when $R$ is a simple ring, so that the only ideals are $0$ and $R$?

I've been trying to use the usual tricks of assuming the characteristic is composite, but even though $(a)=R$ for $a\neq 0$, I'm having trouble since I can't assume that $a$ is left or right invertible since $R$ is not necessarily commutative. Is it still possible to conclude that simple rings of finite characteristic have necessarily prime characteristic?

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up vote 5 down vote accepted

Yes. The result holds.

Note that for every $n\gt 0$, the set $nR=\{na\mid a\in R\}$ is an ideal; indeed, it is a subgroup, and if $r\in R$ and $na\in nR$, then $r(na) = n(ra)\in nR$ and $(na)r = n(ar)\in nR$.

Note that $RR$ is an ideal. Therefore, since we are assuming that $R$ is simple, either $RR=R$ or $RR=0$. Let's consider the former case first (which will hold, e.g., if $R$ has a unity).

Suppose that the characteristic is $n$ and $n=pq$; if $1\lt p$, then $qR\neq \{0\}$, and so by the simplicity of $R$ it follows that $qR=R$. Therefore, $pR = p(RR) = (pR)R = (pR)(qR) \subseteq (pq)R = nR = (0)$, which shows that $pR=0$, so $n|p$, showing that $p=n$. Thus, $n$ is prime: in any factorization $n=pq$, either $p=1$ or $p=n$.

If $R^2\neq R$, then by simplicity of $R$ we must have $R^2=0$. In particular, $ab=0$ for all $a,b\in R$, so $R$ is a ring with zero multiplication. Therefore, every subgroup of $R$ is an ideal, and so the underlying additive group of $R$ must be simple. Since the only simple abelian groups have prime order, it follows that the characteristic of $R$ is prime. (I do not consider the trivial ring to be "simple", since it does not have exactly two ideals).

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Thank you! ${}$ –  Lena Richman Jun 17 '12 at 6:17
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