TITLE: Simple Ore extensions
QUESTION [6 upvotes]: Let $R[x;\sigma,\delta]$ be an Ore extension, where $R$ is an associative and unital ring and $\sigma : R\to R$ is a (not necessarily injective!) ring endomorphism. (In the literature it is often assumed to be injective).
My question is the following:
If $R[x;\sigma,\delta]$ is a simple ring, is $\sigma$ necessarily an injective map?
Note that the answer is affirmative in the case when the maps $\sigma$ and $\delta$ commute. Does anyone know of an example of a simple Ore extension where $\sigma$ is NOT injective?

REPLY [2 votes]: Actually, the argument in my earlier answer can be refined to show the following.
Proposition: Suppose $R$ is a reduced abelian ring, $\sigma:R\rightarrow R$ is a ring homomorphism and $\delta:R\rightarrow R$ is a $\sigma$-derivation. If the Ore extension $A=R[x,\sigma,\delta]$ is simple, then $\sigma$ has to be injective.
Proof: Suppose $\sigma$ is not injective. Then we find a non-zero element $b\in\ker\sigma$. Because $R$ is reduced we know that all powers of $b$ are non-zero. Define $I=\{P\in A\mid \exists k\in \mathbb{N}:P\;b^k=0\}$. Then it is clear that $I$ is a left ideal of $A$, that it is a right $R$-submodule of $A$, that it does not contain $1$ and that $bx-\delta(b)\in I$. To show that $I$ is a non-trivial ideal, we only have to show that $Ix\subset I$. Take any $P\in I$, so $P\;b^k=0$ for some $k$. Now we see that $(P\; x)b^{k+1}=P\;b^k\delta(b)=0$ and hence $P\;x\in I$. QED