Ring with no identity (that has a subring with identity) has zero divisors.

Question

Let $L$ be a non-trivial subring with identity of a ring $R$. Prove that if $R$ has no identity, then $R$ has zero divisors.

So I assumed that there $\exists$ $e \in L$, such that $ex=xe=x$, $\forall$ $x\in L$ and there $\exists$ $x' \in R/L$ such that $ex'\neq x'$ and $x'e\neq x'$. I then show (we obtain it from the last two inequalities), that we have $y=ey=ye,$ where $y=ex', y\neq x'$, which is a contradiction because $y\notin L.$ Is there a way to show from this that $R$ has zero divisors? Because in one of the steps I have $e(x'-y)=0$, but I can't seem to figure out if this can directly lead to the desired result.

Since $L$ is non-trivial, you have $e \neq 0$, which makes $x' - y$ a zero divisor, once you rule out $x' = y$. — David Wheeler, May 24 '15 at 16:43

score 2 · Accepted Answer · answered May 24 '15 at 16:49

Your proof essentially works with $x'-y$ being the zero divisor, as long as you justify why you can pick $x'$ such that $ex'\neq x'$. If $R$ is not commutative you can't guarantee this; $e$ is not an identity, but it may be a left identity. Assume this is the case. Then $e$ is not a right identity, so instead we can pick $x'$ so that $x'e\neq x'$ and argue analogously, with $x'e-x'$ being the zero divisor.

score 0 · Answer 2 · edited Apr 13 '17 at 12:20

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Actually you can prove something even better: a nonzero idempotent in a rng without nonzero zero divisors is actually an identity for the rng.. This applies to your case since the identity of the subrng would be a nonzero idempotent.

The linked solution proves the contrapositive of your statement very simply, and along analogous lines to yours.

edited Apr 13 '17 at 12:20

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answered May 26 '15 at 02:53

rschwieb

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Ring with no identity (that has a subring with identity) has zero divisors.

2 Answers2

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