# Basis for the completion of a free module

This (or similar) question might have been asked before- apologies for any duplication.

I've got a Dedekind domain $R$, a non-zero prime ideal $P$ of $R$ and the completion $\widehat{R}$ of $R$ wrt the valuation associated with $P$.

Let $M$ be a finitely generated, free $R$-module with basis $\{v_i\}_{i=1}^n$. Then would $\{1\otimes v_i\}$ be an $\widehat{R}$-basis of $\widehat{R}\otimes_R M$?

More generally, if $f:R\to S$ is a ring homomorphism and $M$ is a free left $R$-module with basis $\{m_i\}_{i\in I}$ then is $S\otimes_R M$ a free left $S$-module with basis $\{1\otimes m_i\}_{i\in I}$?

• Yes to the first, no to the second. Imagine $f$ is the zero map, then the $R$-module structure defined on $S$ is trivial, hence it is the $0$ R-module, and tensoring against $0$ give the zero module. But then a basis for that is the empty set, and if $M$ was non-trivial, the set you describe is not empty, hence not a basis. If you demand your ring homomorphisms send 1 to 1, it can be done. Commented Jul 10, 2014 at 19:38
• I think it's pretty standard to require ring maps to send 1 to 1, especially in commutative algebra. ("Ring means commutative with unity" is practically a mantra...) Commented Jul 11, 2014 at 3:52

Tensor products commute with direct sums, so $S\otimes_R \bigoplus_i R \cong \bigoplus_i (S\otimes_R R) \cong \bigoplus_i S$. If you write down these isomorphisms explicitly, you will get the conclusion you want.
EDIT: Adam Hughes has made the important observation that our ring morphism must send $1$ to $1$. This is usually an assumption in commutative algebra, but I thought I'd mention it.