# Ideals of a skew polynomial ring where no positive power of the automorphism is inner

The exercise I'm trying to answer is as follows:

Let $R$ be a ring, and $\alpha : R \rightarrow R$ an automorphism of $R$. Suppose that $R$ is simple and that no positive power of $\alpha$ is inner. Then the only nonzero ideals of $S = R[x;\alpha]$ are $S, Sx, Sx^2, \dots$

So far, I've made the following progress:

Let $I$ be a nonzero ideal of $S$, and let $J$ be the set of leading coefficients of elements of $I$, which is an ideal of $R$ (proof omitted). By simplicity of $R$, we see that $J = R$, and so $1 \in J$. Therefore there exists a monic polynomial $p$ of least degree in $I$, say $p = x^n + a_{n-1} x^{n-1} + \dots + a_k x^k$, for some $0 \le k \le n$, where we have assumed $a_k \neq 0$. For contradiction, assume that $k < n$. Let $r \in R$ be arbitrary. Then $pr - \alpha^n p \in I$ has degree at most $n-1$, so $pr - \alpha^n p = 0$. Therefore $a_k r = \alpha^n (r) a_k$ and so $a_k R \subseteq R a_k$. Similarly, $R a_k \subseteq a_k R$, and so $R a_k = a_k R$ is an ideal of $R$, which is therefore the entirety of $R$ by simplicity. Thus $a_k$ has an inverse in $R$, and so the equality $a_k r = \alpha^n (r) a_k$ implies that $\alpha^n$ is inner, a contradiction. Thus $k=n$ and so $p = x^n \in I$.

At this point I want to argue that $I = Sx^n$ but I can't seem to make it work. I feel like I'm missing something very obvious...

I've managed to answer this now, so I'll post my solution. My main issue was that I chose $n$ at the wrong time.
Let $I$ be a nonzero ideal of $S$, and let $n$ be the minimum degree of nonzero elements of $I$. In particular, we have $I \subseteq S x^n$. Let $J$ be the set of leading coefficients of elements of $I$, which is an ideal of $R$ (proof omitted). By simplicity of $R$, we see that $J = R$. In particular, $1 \in J$, so choose a monic $p = x^n + a_{n-1} x^{n-1} + \dots + a_k x^k$ in $I$, where $0 \le k \le n$, and where $a_k \neq 0$. For contradiction, assume that $0 \le k < n$. Let $r \in R$ be arbitrary. Then $pr - \alpha^n p \in I$ has degree at most $n-1$, so $pr - \alpha^n p = 0$. Therefore $a_k r = \alpha^n (r) a_k$ and so $a_k R \subseteq R a_k$. Similarly, $R a_k \subseteq a_k R$, and so $R a_k = a_k R$ is an ideal of $R$, which is therefore the entirety of $R$ by simplicity. Thus $a_k$ has an inverse in $R$, and so the equality $a_k r = \alpha^n (r) a_k$ implies that $\alpha^n$ is inner, a contradiction. Thus $k=n$ and so $p = x^n \in I$. Therefore $Sx^n \subseteq I$, and so $I = Sx^n$ for some $n \in \mathbb{N}$.