3
$\begingroup$

Let $R$ be a discrete valuation ring with residue field $k$. Denote by $\mathbb G_m:= R[x,1/x]$ the multiplicative group over $R$ and $\mathbb G_{m,k}:= k[x,1/x]$. If $B$ is a flat local $R$-algebra, do we have $\mathbb{G}_m(B)=\mathbb{G}_{m,k}(B)$?

It is equivalent to prove that $$ \operatorname{hom}(R[x,1/x],B) = \operatorname{hom}(k[x,1/x],B)?$$

Denote by $\mathfrak{m}$ the maximal ideal. We have $$ 0 \to \mathfrak{m} \to R[x,1/x] \to k[x,1/x] \to 0$$

Taking $\operatorname{hom}(-,B)$ we have $$ 0 \to \operatorname{hom}(k[x,1/x],B) \to \operatorname{hom}(R[x,1/x],B) \to \operatorname{hom}(\mathfrak{m},B) $$

Is it true that $\operatorname{hom}(\mathfrak{m},B)=0$?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
5
$\begingroup$

This is not true of any such $B$ unless $R=k$. Suppose the maximal ideal is generated by $y\in R$. For calculations if you think about the residue field as $k=R/y$, you see that for any scheme $X$ over $R$, the base change $X_k$ (rather, the fiber which is still based over $R$) has no $B$ points for any algebra $B$ which is not annihilated by $y$. In particular if $y\neq 0$, i.e. $R\neq k$, then taking $B=R$ is a counterexample for you, because $\mathbb{G}_m$ has nonempty $B$-points. But actually no flat algebra $B$ over $R$ is allowed to be annihilated by $y$ because that would imply $\mathrm{Tor_1^R(k, B)\neq 0}$.

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ Can you explain more about the last fact $Tor_1^R(k,B) \neq 0$? Why is that? And why we should have $Tor_1^R(k,B) = 0$? Thanks a lot for your kind guidance. $\endgroup$
    – MAY
    Commented Aug 7 at 23:14
  • 1
    $\begingroup$ I can elaborate without using derived functor if you like. What happens when you tensor the sequence $0\to\mathfrak{m} \to R\to k\to 0$ with an algebra $B$ which is annihilated by the element $y$ generating $\mathfrak{m}$? You get an isomorphism for the map on the right, but $B\otimes \mathfrak{m} $ you can show is nonzero when $y\neq 0$. So tensoring with $B$ isn't exact, by definition this makes it not flat. $\endgroup$
    – Justin Bloom
    Commented Aug 7 at 23:21
  • 1
    $\begingroup$ Then $R \otimes B \cong k \otimes B$. But why $Tor \neq 0$? $\endgroup$
    – MAY
    Commented Aug 7 at 23:24
  • 1
    $\begingroup$ Yes, so this can be used to argue that the sequence is not exact. See my edit (finger slipped before I was done with my comment) $\endgroup$
    – Justin Bloom
    Commented Aug 7 at 23:28
  • $\begingroup$ clear now, thanks a lot! $\endgroup$
    – MAY
    Commented Aug 7 at 23:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .