= Coleman norm contraction
{c}
{title2=$N(1+p^kR)\subseteq1+p^{k+1}R$}
For odd $p$, the <Coleman norm operator> satisfies the displayed inclusion for every $k\geq1$. Write $M_h$ for multiplication by $h$ on the <finite free module> $R$ over $\phi(R)$. Its <trace> is $p\phi(S(h))$, where $S$ is the <Coleman trace operator>. In $\det(I+p^kM_h)$ 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 $1+p^kR$ for every $k$ and hence equals $1$.
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