Points of multiplicative groups

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 $$\operatorname{hom}(\mathfrak{m},B)=0$$?

1 Answer

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 $$\mathrm{Tor_1^R(k, B)\neq 0}$$.

• 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.
– MAY
Commented Aug 7 at 23:14
• 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. Commented Aug 7 at 23:21
• Then $R \otimes B \cong k \otimes B$. But why $Tor \neq 0$?
– MAY
Commented Aug 7 at 23:24
• 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) Commented Aug 7 at 23:28
• clear now, thanks a lot!
– MAY
Commented Aug 7 at 23:29