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Let $A$ be a $\mathbb{Z}_p$-algebra which as a $\mathbb{Z}_p$-module is free of finite rank. Let $x$ be in $A$. Suppose that $A \otimes_{\mathbb{Z}_p}\mathbb{Q}_p$ is semi-simple, i.e. it has no nilpotent element. Now my book says “then it is a product of finite extension of $\mathbb{Q}_p$ and the image of $x$ in each simple factor is contained in the $p$-adic ring of integer of the factor” without explanation. But why is this true?

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  • $\begingroup$ $A \otimes \mathbb{Q}_p$ has two natural $\mathbb{Q}_p$-topologies, which are equivalent (one is given by $A$ being a free $\mathbb{Z}_p$-module and the other one is “product of $p$-adic fields”). It follows that the image of $A$ in each factor of the product of fields is a compact $\mathbb{Z}_p$-subalgebra of this $p$-adic field – so must be contained in the ring of integers. $\endgroup$
    – Aphelli
    Jul 8, 2022 at 18:15

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By the Artin-Wedderburn Theorem we have an isomorphism $$ A \otimes_{\mathbb{Z}_p} \mathbb{Q}_{p} \cong \prod_{i=1}^{k}\operatorname{Mat}_{n_i}(D_i) $$ for some division algebras $D_i$ over $\mathbb{Q}_p$ and for some $n_i \in \mathbb{N}$. Because you have assumed your algebra $A$ to be commutative, each division algebra $D_i$ can be taken to be a field extension $F_i$ of $\mathbb{Q}_p$ and each $n_i = 1$ so that you have an isomorphism $$ A \cong \prod_{i=1}^{k} \operatorname{Mat}_{n_i}(D_i) \cong \prod_{i=1}^{k} F_i. $$ Each field extension $F_i/\mathbb{Q}_p$ must be finite; this can be seen from the fact that $A \otimes_{\mathbb{Z}_p} \mathbb{Q}_p$ has finite dimension as a $\mathbb{Q}_p$-vector space (if even a single $F_i$ was an infinite field extension then $A \otimes_{\mathbb{Z}_p} \mathbb{Q}_p$ would have infinite dimension as a $\mathbb{Q}_p$-vector space).

To see the statement about the element $x \in A$, first write the isomorphism of $\mathbb{Z}_p$-modules $$ A \cong \bigoplus_{j=1}^{\ell} \mathbb{Z}_p $$ and chase $x$ through the isomorphism to associate $x \mapsto (x_j)_{j=1}^{\ell}$ for $x_i \in \mathbb{Z}_p$. For each $1 \leq i \leq k$ let $n_i = [F_i:\mathbb{Q}_p]$ and note that we have the identity $$ \sum_{i=1}^{k} n_i = \ell. $$ Rewrite the description of $A$ to be of the form $$ A \cong \bigoplus_{j=1}^{\ell} \mathbb{Z}_p \cong \bigoplus_{i=1}^{k}\left( \bigoplus_{j=1}^{n_i}\mathbb{Z}_p\right) $$ and note that as $\mathbb{Z}_p$-modules each free module $$ \bigoplus_{i=1}^{n_i}\mathbb{Z}_p \cong \mathcal{O}_{F_i} $$ where $\mathcal{O}_{F_i}$ is the ring of integers in $F_i$. Finally once you check that the induced map $A \to A \otimes_{\mathbb{Z}_p} \mathbb{Q}_p \cong \prod_{i=1}^{k} F_i$ factors as $$ A \cong \bigoplus_{i = 1}^{k} \mathcal{O}_{F_i} \to \prod_{i=1}^{k} F_i \cong \mathbb{Q}_p \otimes_{\mathbb{Z}_p} \left(\bigoplus_{i=1}^{k}\mathcal{O}_{F_i}\right) $$ checking that the image of $x$ in each component is an $F_i$-integer is routine.

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