Coleman norm contraction ($N(1+p^{k}R)\subseteq 1+p^{k+1}R$)

For odd $p$, the Coleman norm operator satisfies the displayed inclusion for every $k\ge 1$. Write $M_h$ for multiplication by $h$ on the finite free module $R$ over $\varphi (R)$. Its trace is $p\varphi (S(h))$, where $S$ is the Coleman trace operator. In $\det (I+p^k{M}_h)$ the linear term is therefore divisible by $p^{k+1}$, and all higher terms are divisible by $p^{2k}$. Congruence reflection for Frobenius substitution gives the assertion. A norm-fixed element of $1+pR$ lies in $1+p^{k}R$ for every $k$ and hence equals $1$.