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 $\{1\otimes m_i\}_{i\in I}$?

  • 2
    $\begingroup$ 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. $\endgroup$ – Adam Hughes Jul 10 '14 at 19:38
  • $\begingroup$ 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...) $\endgroup$ – Jake Levinson Jul 11 '14 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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.