I'm following David Goss's book Basic structures of function field arithmetic. Let $L$ be an extension field of $L_0$ and $\sigma$ an automorphism of infinite order which fixes $L_0$. Let $L\{\sigma\}$ be finite expressions of the form $\Sigma a_n \sigma^n$, $a_n \in L$. Let $R \subset L\{\sigma\}$ be a commutative subring, with the following statements:
- $L_0$ is strictly contained in $R$ as $L_0 \cdot \sigma^0$.
- $R\cap L=L_0 \cdot \sigma^0$.
Let $t$ the greatest common divisor of $\{\deg(r): r \in R\}$ (we think $L\{\sigma\}$ as polynomials in $\sigma$, so the degree is defined as the maximum of powers in $\sigma$), $A_1$ and $A_2$ of degree $mt$ in $R$. We write \begin{equation} A_i=a_i\sigma^{mt}+\{ \mbox{lower terms} \} \end{equation} The texts says: In the quotient field of $R$ we can find an element $\alpha$ of degree $t$. Using the commutativity of $R$ and $\alpha$ we deduced that \begin{equation} \sigma^t\left(\dfrac{a_1}{a_2}\right)= \dfrac{a_1}{a_2}. \end{equation}
How can I verify this? My attempt was try to divide $\dfrac{A_1}{A_2}$ to get an expression of the form \begin{equation} \dfrac{A_1}{A_2}=\dfrac{\alpha}{\sigma^t}, \end{equation},
where $\alpha = \dfrac{a_1}{a_2} \sigma^t +\sum_{k\geq 0} \dfrac{b_k}{\sigma^{kt}}$. Then, evaluate $\sigma (\dfrac{A_1}{A_2})$ and, anyway, get the statement. By they way, Can I choose $\alpha$ in that way?
Thanks in advance!
