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**Swlabr** If $\displaystyle b_1x^r+b_{r+1}x^{r+1} + \ldots + b_sx^s \in Z(R)$ then for all $\displaystyle a \in K, n \in \mathbb{Z}$ we have that $\displaystyle b_r \sigma^r(a)x^{n+r} + b_{r+1} \sigma^{r+1}(a)x^{n+r+1} + \ldots + b_s \sigma^s(a)x^{s+n} = $$\displaystyle a \sigma^n(b_r)x^{n+r} + a \sigma^{n}(b_{r+1})x^{n+r+1} + \ldots + a \sigma^n(b_{s})x^{s+n}$. As $\displaystyle <x> \cong C_{\infty}$ is a group we have that all the terms pair off, and so we have that $\displaystyle b_t \sigma^t(a) = a\sigma^n(b_t)$ for all $\displaystyle a \in K, n \in \mathbb{Z}$.

Let $\displaystyle P=\{b \in K : \sigma(b) = b\}$. If $\displaystyle o(\sigma) = m$ then let $\displaystyle P_m := \{a*x^{im}:a \in P, i \in \mathbb{Z} \}$. It is clear that both $\displaystyle P \leq Z(R)$ and $\displaystyle P_m \leq Z(R)$ as they are invarient under automorphisms and are subfields of the ring.

So, we wish to show that $\displaystyle Z(R) \leq P_{o(\sigma)}$ or if $\displaystyle o(\sigma) = \infty$, $\displaystyle Z(R) \leq P$. That is to say, $\displaystyle b_t \in P$ and either $\displaystyle n | o(\sigma)$ or $\displaystyle n=0$.

Rearranging $\displaystyle a \sigma^n(b_t) = b_t \sigma^t(a)$ we get that $\displaystyle \sigma(b_t)\sigma(a^{-1}) = \sigma^{-1}(b_t^{-1}a) = b_t^{-n}a^t$. As $\displaystyle a$ is arbitrary we have that $\displaystyle \sigma^{n} (b_t) = b_t $ for all $\displaystyle n \in K$ and $\displaystyle \sigma(a^{-1}) = a^{t}$. Thus, $\displaystyle b_t$ is invariant under every single automorphism and $\displaystyle a = \sigma^{t} (a)$ for all $\displaystyle a \in K$. Thus, $\displaystyle t=0$ or $\displaystyle t | o(\sigma)$ (as $\displaystyle \sigma^t=id_{Aut(R)}$), and $\displaystyle b_t \in P$, and we are done...almost.

It still needs to be shown what $\displaystyle P$ is. My initial thought was that it was the prime subfield (thus, I denoted it $\displaystyle P$). However, if $\displaystyle \sigma$ is trivial this is not the case. Thus, i suspect that the prime subfield is the intersection of all the $\displaystyle P$'s (it is certainly contained in the intersection). What $\displaystyle P$ actually is will depend on $\displaystyle \sigma$, and will either need someone with more knowledge of fields than I, or someone with a bigger brain...