Is this a right place to ask help for an exercise?
Let $n\geq 2$ be an integer and $D=\mathbb Z[1/n]$. Let $A$ be a complete commutative ring with unit for the $I$-adic topology, where $I$ is an ideal of $A$. Suppose that $n$ is invertible in $A$, and let $x\in I$. How can I show that there is a unique continuous homomorphism $\phi:D[[S]]\to A$ such that $\phi(S)=x$?
Starting from the unique ring homomorphism $\mathbb Z\rightarrow A$, it extends uniquely to a ring homomorphism $D = \mathbb Z[1/n]\rightarrow A$ since $n$ is invertible in $A$. By the universal property of $D[S]$ there exists a unique homomorphism $$ \varphi\colon D[S] \longrightarrow A,\quad \sum_{i=1}^md_iS^i\longmapsto \sum_{i=1}^md_i\cdot x^i. $$ Now, $\varphi$ is continuous (where $D[S]$ is equipped with the $(S)$-adic topology), because $x\in I$: For every $m\in \mathbb N$ we have $\varphi(S^m)\subseteq I^m$. Since $A$ is complete with respect to the $I$-adic topology this extends to a unique continuous homomorphism $$ \widehat \varphi\colon D[[S]]\longrightarrow A $$ (by the universal property of the completion of $D[S]$) and we are done.
