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