MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
What is ${\mathbb G}_m$? – Steven Landsburg Mar 30 '11 at 14:55
$\mathbb{G}_m = GL_1$. – Mike Skirvin Mar 30 '11 at 15:01
According to the title, I guess $\mathbb{G}_m$ means the multiplicative group scheme (defined over $\mathbb{Z}$ ?). – Ralph Mar 30 '11 at 15:27
Unknown: Could you please explain over what base scheme you are working? – Bisi Agboola Mar 30 '11 at 15:50
@Bisi Agboola does happens that answer depends on the base? – unknown Mar 30 '11 at 16:26

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 ${\mathbb Z}$.

share|cite|improve this answer
@Steven: I touched up your answer just a little bit by adding the word "nontrivial" (but I resisted the temptation to explain that it is nontrivial because, e.g., $\mathbb{G}_m$ is connected). I hope you don't mind. (If you do, please feel free to roll it back...) – Pete L. Clark Mar 30 '11 at 16:12
@Steven Landsburd exponential map is not algebraic – unknown Mar 30 '11 at 16:23
Indeed, that is not a map of group schemes. – Ben Webster Mar 30 '11 at 16:31
Ben Webster and unknown: Point well taken. – Steven Landsburg Mar 30 '11 at 16:39
+1: This is partly the fault of "unknown" for not adequately specifying the category in which Ext is being taken. For example, you can take Ext in the category of abelian sheaves on the analytic site, where your example is a perfectly valid nontrivial extension. – S. Carnahan Mar 31 '11 at 3:28

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.

share|cite|improve this answer
@Ralph it seems to me that the lemma says that $Ext_{\mathbb{G}_m}^1(k,\mathbb{Z})$, it is not clear to me how $\mathbb{Z}$ is a $\mathbb{G}_m$-module and how to relate this group with my question – unknown Mar 30 '11 at 16:38
In Jantzen's lemma $k$ can be any commutative ground ring (I'll elaborate his argument later on in case $\mathbb{G}_m$). What's the defining ground ring for your $\mathbb{G}_m$ ? – Ralph Mar 30 '11 at 16:47
@Ralph DVR is it fine, I don't understand why this depends on the ground ring. Are u looking at $\mathbb{Z}$ as the trivial module? – unknown Mar 30 '11 at 17:03
The ground ring is an intrinsic part of the definition of an affine scheme and thus a group scheme. It's similar to polynomial rings: You can't just talk from a "polynomial ring". You also need to tell from what ring the coefficients are taken from. For the vanishing of the cohomology of $\mathbb{G}$ it doesn't matter if the action on the coefficients is trivial or not. The relevant fact here is that in $Ext_{\mathbb{G}_m}^n(M,N)$ the $\mathbb{G}_m$-module $M$ is projective as module over the ground ring. – Ralph Mar 31 '11 at 0:36
@Ralph: The OP's question seems to be different: it is concerned with the vanishing of the $Ext^1$ group computed in either the category of commutative group schemes (base is a field) or the category of etale/Zariski sheaves (base is arbitrary, but perhaps Spec Z is the important case). – SGP Mar 31 '11 at 2:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.