# Semi-simple product of $p$-adic numbers

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?

• $A \otimes \mathbb{Q}_p$ 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. Jul 8, 2022 at 18:15

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.