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I'm aware that every commutative simple ring with unity is a field. Is this true even for rings for which we don't know if they have a unity?

Jose Brox
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ante.ceperic
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    @user35603 Yeah, I worded it clumsily. I thought about the case where we don't know if rings has a unity a priori... But there is a chance that being simple forces a ring to have unity, that's what I'm interested in. – ante.ceperic May 17 '14 at 10:44
  • Just curious; how do you define a field if there no unity (in particular, no $1$ in the ring)? – Zuriel Mar 12 '18 at 20:30
  • I have just found this similar question:

    https://math.stackexchange.com/questions/2500075/the-center-of-a-simple-ring-is-either-0-or-a-field

     – Jose Brox Nov 17 '20 at 07:26

2 Answers2

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Note that the in the setting of nonunital rings, a ring $R$ is called simple if its multiplication is not trivial (i.e., $R^2\neq0$) and it has no ideals other than $0$ and $R$. Hence we have two possible answers here:

  1. We allow for trivial multiplication. If $R$ has trivial multiplication then the ideals are exactly the subgroups of $(R,+)$, so $R$ is simple iff $(R,+)$ has no nontrivial proper subgroups, hence it is $0$ or a cyclic group of prime order. Therefore $R$ is $0$ or $(\mathbb Z_p,+)$ with trivial multiplication for some prime $p$.

  2. We restrict to the usual definition of simplicity given above. Then (as I show below) $R$ must have identity element, so it is a simple commutative unital ring, hence a field by the usual proof.

Proposition. A simple commutative ring is unital.

Proof: Since $R^2\neq0$ there is $0\neq a\in R$ such that $Ra=R$, as $Ra$ is a nonzero ideal. That is, $R$ is generated by a single element. Let us see that we can pick an idempotent generator. Since $R=Ra$, there is $e\in R$ such that $a=ea$, whence $ea=e^2a$, i.e., $(e-e^2)a=0$. Suppose $e$ is not idempotent; then $e^2-e\neq0$, so $R=R(e^2-e)$ and there is $b\in R$ such that $e=b(e^2-e)$, so $a=ea=b(e^2-e)a=0$, a contradiction. Hence $e^2=e$. Now let us see that $e$ is actually the identity element of $R$. Let $x\in R$ and suppose $ex-x\neq0$. Then again $R=(ex-x)R$, so there is $y$ (depending on $x$) such that $e=(ex-x)y$, so $e=e^2=e(ex-x)y=(ex-ex)y=0$, a contradiction. Therefore $ex=x$ for all $x\in R$.

Remark: A generalization of the previous proposition is: Let $R$ be a (not necessarily unital nor commutative) semiprime ring and $I$ be a minimal left ideal of $R$. Then there is an idempotent $e\in I$ such that $I=Ie=Re$.

Jose Brox
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What about the two-element ring with zero multiplication?

Hagen von Eitzen
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  • But by definition in the nonunital setting, a ring $R$ is simple if $R^2\neq0$ and it has no nontrivial ideals. – Jose Brox Nov 12 '20 at 09:04