Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $R = \prod_{n\in\mathbf{N}}R_n$ be an infinite direct product of discrete valuation rings $R_n$. Why is $\mathrm{Pic}(R) = 0$?

share|improve this question
    
How did you know that this was true before Kestutis posted his answer? –  Georges Elencwajg May 8 at 7:29

1 Answer 1

up vote 7 down vote accepted

We aim to show that every $\mathbb{G}_m$-torsor over $R$ is trivial. By descent, such a torsor is represented by an affine $R$-scheme $X$. Due to affineness, $X(R) = \prod_n X(R_n)$, so it remains to (invoke the axiom of countable choice and) note that each $X(R_n)$ is nonempty because the pullback of $X$ to $\mathrm{Spec} R_n$ is the trivial $\mathbb{G}_m$-torsor as $R_n$ is local.

For the argument to work, it suffices that $\mathrm{Pic}(R_n) = 0$ for every $n$. Also, a similar argument (coupled with limit formalism) shows that there are no nontrivial vector bundles on $\mathrm{Spec } \mathbb{A}_K$ where $\mathbb{A}_K$ is the adele ring of a global field $K$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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