Question

Hi, I have the following question: why $Ext^1(\mathbb{G}_m,\mathbb{Z})=0$?

Answer 1

If ${\mathbb G}_m$ means the multiplicative group, then this is false over ${\mathbb C}$, where the exponential map ${\mathbb G}_a\rightarrow{\mathbb G}_m$ gives a nontrivial extension of ${\mathbb G}_m$ by ${\mathbb Z}$.

Answer 2

A proof can be found in Jantzen's book "Representations of Algebraic Groups", Part I, Lemma 4.3 (1st edition).

Edit: To be a little more precise here are some details of Jantzen's proof.

Let $H$ be an abstract abelian group, R a commuative ring and let the group scheme $G = Spec(R[H])$ where $R[H]$ denotes the group ring of $H$ with the usual Hopf algebra structure (in case of $\mathbb{G}_m$ take $H$ the additive group of rational integers).

We will show that $Ext_G^n(M,N) = 0$ for all $n>0$, $M$ a $R$-projective $G$-module and $N$ any $G$-module. With $M := R$ it follows $H^n(G;N) = Ext_G^n(R,N) = 0$ for all $n>0$ and all $G$-modules $N$.

The crucial step is to obtain a functorial decomposition $$Hom_G(M,N) = \Pi_{\lambda \in H}Hom_R(M_{\lambda},N_{\lambda})$$ of $R$-modules with $M = \oplus_{\lambda \in H}M_{\lambda}$ (as $R$-modules) since this implies $$Ext_G(M,N) = \Pi_{\lambda \in H}Ext_R^n(M_{\lambda},N_{\lambda}).$$ Then, if $M$ is $R$-projective, so is $M_{\lambda}$ as an $R$-direct summand of $M$ and therefore $Ext_R(M_{\lambda},N_{\lambda}) = 0$ for $n>0$, what was to be shown.

Let's construct $M_{\lambda}$: We know that the categories of $G$-modules and $R[H]$-comodules are equivalent (Jantzen 2.8). Therefore $M$ can be regarded as $R[H]$-comodule with a $R$-linear map $$\Delta: M \to M \otimes R[H],$$ making the usual diagrams commute (that's 2.8(2),(3) in Jantzen). Let $m \in M$. Using that $R[H]$ is a free $R$-module and tensor product is taken over $R$, we find $\rho_{\lambda}(m) \in M$ such that $$\Delta(m) = \sum_{\lambda \in H}\rho_{\lambda}(m) \otimes h.$$ It follows from the commutative diagrams of the comodule $M$ and the commutativity of $H$ that $\rho_{\lambda}: M \to M$ is $R$-linear and has the projection properties

$$id_M = \sum_{\lambda}\rho_{\lambda}, \hspace{10pt} \rho_{\lambda} \circ \rho_{\lambda} = \rho_{\lambda}, \hspace{10pt} \rho_{\lambda} \circ \rho_{\mu} = 0 \hspace{3pt} (\lambda \neq \mu).$$ Define $M_{\lambda} := \rho_{\lambda}(M)$. Then the properties of $\rho_{\lambda}$ directly yield the desired direct sum decomposition of $M$. Moreover they show $$M_{\lambda} = \lbrace m \in M \hspace{2pt} | \hspace{2pt} \Delta(m) = m \otimes \lambda \rbrace.$$ The equvalence of the categories of $G$-modules and $R[H]$-modules carries over to an isomorphism of $R$-modules $$Hom_G(M,N) \cong Hom_{R[H]-comod}(M,N)$$ (group of $R[H]$-comodule homomorphisms). Applying the projection properties of $\rho_{\lambda}$ once more it's easy to see that a $R[H]$-comodule homomorphism $f: M \to N$ has $f(M_{\lambda}) \subseteq N_{\lambda}$ and that $$Hom_{R[H]-comod}(M,N) \to \Pi_{\lambda \in H}Hom_R(M_{\lambda},N_{\lambda}), f \to (f|M_{\lambda})_{\lambda \in H}$$ is an isomorphism of $R$-modules that is natural in $M$ and $N$. This establishes the decompositon of $Hom_G(M,N)$ we were looking for.