0
$\begingroup$

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...

$\endgroup$
0
$\begingroup$

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 $n \in \mathbb{N}$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.