Is it true that $\mathbb Z[G]\otimes_\mathbb Z \mathbb Q$ is isomorphic to $\mathbb Q[G]$ as a $\mathbb Q$-algebra?

Context: If I have the Jacobian of a smooth curve $X$, call it $J_X$, then I've read the claim that any subgroup $G$ of Aut$(X)$ will induce a canonical map of $\mathbb Q$-algebras $\mathbb Q[G]\to$ End$(J_X)\otimes_\mathbb Z\mathbb Q$. I'm trying to come up with this map

My Interpretation: Since $\mathbb Z[G]$ and End$(J)$ are both $\mathbb Z$- modules, I can get a map of $\mathbb Z$-modules $$\phi:\mathbb Z[G] \to \text{End}(J)$$ just by extending the $G$ action on $X$ to $Jac(X)$ (Since $Jac(X)$ is just the free abelian group on points of $X$) linearly. Now if I tensor this map with the identity map id:$\mathbb Q \to \mathbb Q$, I get the map $$\phi\otimes_\mathbb Z \text{id}: \mathbb Z[G]\otimes_\mathbb Z \mathbb Q \to \text{End}(J_X)\otimes_\mathbb Z\mathbb Q.$$ So if the left side is $\mathbb Q[G]$, I will be done! (Well, I'd still have to check that this is actually a map of $\mathbb Q $ algebras but baby steps.) :D

  • $\begingroup$ It is a bit confusing for me that you ask for such a basic algebra isomorphism, and at the same time, study the Jacobian of a smooth algebraic curve. $\endgroup$ – Martin Brandenburg Feb 14 '16 at 11:18

If $S$ is any commutative $R$-algebra and $M$ is any monoid, then $$R[M] \otimes_R S \cong S[M]$$ in $S\mathsf{-Alg}$. In fact, if $T$ is an $S$-algebra, then (denoting the evident forgetful functors $C \to D$ here by $(-)|_D$) $$\hom(R[M] \otimes_R S,T) \cong \hom(R[M],T|_{R\mathsf{-Alg}}) \cong \hom(M,T|_{R\mathsf{-Alg}}|_{\mathsf{Mon}})$$ $$ = \hom(M,T|_{\mathsf{Mon}}) \cong \hom(S[M],T).$$ This is natural in $T$, so that Yoneda Lemma implies $R[M] \otimes_R S \cong S[M]$.

  • $\begingroup$ Thanks! I'll have to study some category theory to understand your answer. $\endgroup$ – Ravi Mar 6 '16 at 13:28

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