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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?

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More generally, it is true if $\delta(\ker\sigma)=0$, for then $\ker\sigma$ is an ideal in the Ore extension. –  Mariano Suárez-Alvarez Jun 19 '11 at 21:26
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Mariano, $\ker\sigma \subseteq R$ is never an ideal of the Ore extension, but your idea is correct. It can be made more general: It is true if $\delta(\ker\sigma)\subseteq \ker\sigma$, for then (put $A:=R[x;\sigma,\delta]$) the set $(\ker\sigma) A$ is a proper ideal of the Ore extension. (I like to think of $R[x;\sigma,\delta]$ as a free left $R$-module with basis $\{1,x,x^2,...\}$.) –  Johan Öinert Jun 19 '11 at 23:11
    
Ups, that's what I meant, actually :) –  Mariano Suárez-Alvarez Jun 20 '11 at 19:32
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It is worth noting that $\ker \sigma$ does not have to be invariant under $\delta$. In fact there does not have to be any ideal that is invariant under both $\sigma$ and $\delta$. –  Johan Jun 23 '11 at 7:25
    
Johan, you obviously intended to write "non-trivial ideal" instead of "ideal". That was perhaps a stupid remark. While I'm at it, I might as well point out that in my previous comment I assumed that $\sigma$ was NOT injective. Otherwise $\ker\sigma = \{0\}$ is of course an ideal in the Ore extension, and the first sentense of my comment would be false. –  Johan Öinert Jun 23 '11 at 15:22
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2 Answers

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

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Just two Naive constructions i thought about and that do not give simple Ore extensions.

We start with an algebraically closed field $F$ of characteristic $0$ and consider the ring of polynomials $R_0=F[y]$. As a homomorphism $\sigma_0:R_0\rightarrow R_0$, we use the evaluation at $0$, i.e. $\sigma_0(\sum_{i=0}^n a_iy^i)=a_0$. The $\sigma_0$-derivation is then given by $\delta_0(\sum_{i=0}^n a_iy^i)=\sum_{i=1}^n a_iy^{i-1}$.

The Ore extension $R_0[x,\sigma_0,\delta_0]$ is not simple because $yx-1$ generates a non-trivial ideal. But we can do the following.

Define $R=R_0[(y-a)^{-1}\mid 0\not=a\in F]$ to be the localization of $R_0$ with respect to all polynomials $P\in R_0$ with $P(0)\not=0$. The homomorphism $\sigma_0$ and the derivation $\delta_0$ extend uniquely to this localization, i.e. we find a unique homomorphism $\sigma:R\rightarrow R$ such that $\sigma\vert_{R_0}=\sigma_0$, and a unique $\sigma$-derivation $\delta:R\rightarrow R$ such that $\delta\vert_{R_0}=\delta_0$.

Now the Ore extension $A=R[x,\sigma,\delta]$ is still not simple. Consider the set $I=\{b\in A\mid \exists k\in\mathbb{N}:by^k=0\}$. This set is clearly a left ideal, it does not contain $1$ and it contains $yx-1$. It is also immediate that $I\cdot R=I$. In order to show that $I$ is a non-trivial ideal, we only have to show that $Ix\subset I$. Suppose $b\in I$, so $by^k=0$ for some $k$. Then we see that $(bx)y^{k+1}=by^k=0$ so $bx\in I$.

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