I am currently reading the book Cyclotomic fields and zeta values by J. Coates and Sujatha. There is a statement(Page 16) that is not clear immediately.
Given $f$ in $R$, define $h(T)=\prod_{\zeta\in\mu_{p}}f(\zeta(1+T)−1)$, which is clearly also in R.
Here $R$ is the ring $\mathbb{Z}_{p}[[T]]$. I don't get why $h(T)$ has to be in $R$. Am I missing some trivial check?
This is Galois theory: the factors $f(\zeta(1+T)-1) = f(\zeta - 1 + \zeta{T})$ are typically not in $R$ (except when $\zeta = 1$), but their product is guaranteed to be in $R$.
Let the group $G = {\rm Gal}(\mathbf Q_p(\zeta_p)/\mathbf Q_p)$ act on $\mathbf Q_p(\zeta_p)[[T]]$ by acting on coefficients: $\sigma(\sum a_n T^n) := \sum \sigma(a_n)T^n$. The fixed set is trivially $\mathbf Q_p[[T]]$ by Galois theory. Show the action of $G$ on $\mathbf Q_p(\zeta_p)[[T]]$ is by ring automorphisms and then see what $\sigma(f(\zeta(1+T)-1))$ is when $\zeta$ is a $p$th root of unity.