Simple ring without identity (Commutative Algebra by Zariski & Samuel p 133) 
Let $R$ be a commutative simple ring. If $R$ is not a field (equivalently, $R$ has no identity), there exists a nonzero element $a\in R$ such that $Ra\neq R$.

This statement appears in the page 133 of the book "Commutative Algebra" written by Oscar ZARISKI and Pierre SAMUEL. It seems not very hard to show it, but I am a bit confused. I shall be grateful if there is any expert who may share the ideas. Thank you!
 A: Unfortunately, I can't get a copy of the problem in the book fast, but I suspect you have misunderstood something. Here are the most relevant lines of thought.
If $R$ is commutative, then $aR$ is an ideal, and then there are only two possibilities for any given element $a\in R$: either $aR=\{0\}$ or $aR=R$.
If $R$ has identity, then $aR=\{0\}$ iff $a=0$ and $aR=R$ iff $a\neq 0$.
For a commutative simple ring$^\ast$ $R$ (not assuming identity anymore), if there is one single element $a$ such that $aR=R$, then $R$ has identity. For, there would exist $e$ such that $ae=a$, and for an arbitrary $b$, $as=b$. But then $be=ase=as=b$, showing $e$ is an identity.
So the only possibility is that $aR=\{0\}$ for every single $a\in R$, which does satisfy what you wrote, but in a rather extreme way. Under such conditions, every subgroup of $R$ is an ideal. It follows that $|R|=p$ for some prime $p$, and $ab=0$ for every $a,b\in R$.
$^\ast$ meaning a ring with only trivial ideals. Actually, simple rings without identity are usually additionally required to satisfy some nondegeneracy condition like $R^2\neq \{0\}$ or $R^2=R$. With that additional assumption, this problem statement becomes untenable.
