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This question has two parts, the first part will be to obtain the desired exact sequence while the second will be to study it in the corresponding derived category and try to obtain it from there.

Let $X$ be a smooth geometrically integral variety over a number field $k$ with canonical morphism $\pi:X \rightarrow \mathrm{Spec}\,k$, I want to obtain the following exact sequence of $\mathrm{Gal}(\bar{k}/k)$-modules (all tensor products are over $\mathbb{Z}$):

$$\mu_\infty \rightarrow \bar{k}[X]^* \rightarrow \bar{k}[X]^* \otimes \mathbb{Q} \rightarrow H^1(\bar{X},\mu_\infty) \rightarrow \mathrm{Pic}(\bar{X}) \rightarrow \mathrm{Pic}(\bar{X})_{free} \rightarrow 0,$$

where $\bar{X} := X \times _k \bar{k}$, $\mu_\infty = \mathrm{colim}_n\mu_n$, and $\mathrm{Pic}(\bar{X})_{free}$ denotes the maximal free quotient of $\mathrm{Pic}(\bar{X})$.

A very natural approach will be to apply the etale cohomology functor $H^i(\bar{X},-)$ to the exact sequence

$$1 \rightarrow \mu_\infty \rightarrow \mathbb{G}_m \rightarrow \mathbb{G}_m \otimes \mathbb{Q} \rightarrow 1.$$

To see why this sequence is exact refer to the answer of my post A Kummer exact sequence involving $\mu_\infty$.

So everything is fine except for the surjective map $\mathrm{Pic}(\bar{X}) \rightarrow \mathrm{Pic}(\bar{X})_{free}$, which can be rewritten as

$$\mathrm{Pic}(\bar{X}) \rightarrow \mathrm{Ker}(H^1(\bar{X},\mathbb{G}_m) \otimes \mathbb{Q} \rightarrow H^2(\bar{X},\mu_\infty)).$$

Question 1. How do I show that $\mathrm{Pic}(\bar{X})_{free}$ is precisely the kernel written above?

One thought I have is that perhaps we can try to show that $H^2(\bar{X},\mu_\infty) = 0$, and then we would need to prove that $\mathrm{Pic}(\bar{X}) \otimes \mathbb{Q} \cong \mathrm{Pic}(\bar{X})_{free}$, which would not be surprising since we are sort of 'killing off' the torsion parts of the Picard group.

Now on to the next part of the question, we have that the inclusion $\mu_\infty \rightarrow \mathbb{G}_m$ induces the map

$$\varphi: \tau_{\leq 1}R\pi_*\mu_\infty \rightarrow \tau_{\leq 1}R\pi_*\mathbb{G}_m$$

in the category of bounded complexes of sheaves of discrete Galois modules. Let $D := \mathrm{Cone}(\varphi)$, thus we have a distinguished triangle

$$\tau_{\leq 1}R\pi_*\mu_\infty \rightarrow \tau_{\leq 1}R\pi_*\mathbb{G}_m \rightarrow D \rightarrow (\tau_{\leq 1}R\pi_*\mu_\infty)[1].$$

This would give rise to the exact sequence

$$0 \rightarrow \tau_{\leq 1}R\pi_*\mathbb{G}_m \rightarrow D \rightarrow (\tau_{\leq 1}R\pi_*\mu_\infty)[1] \rightarrow 0.$$

Let $h^i(A)$ denote the $i$-th cohomology group of the complex $A$, then we have a long exact sequence of cohomology groups

$$h^{-1}(D) \rightarrow h^{-1}((\tau_{\leq 1}R\pi_*\mu_\infty)[1]) \rightarrow h^0(\tau_{\leq 1}R\pi_*\mathbb{G}_m) \rightarrow h^0(D) \rightarrow h^0((\tau_{\leq 1}R\pi_*\mu_\infty)[1])$$ $$\rightarrow h^1(\tau_{\leq 1}R\pi_*\mathbb{G}_m) \rightarrow h^1(D) \rightarrow h^1((\tau_{\leq 1}R\pi_*\mu_\infty)[1]).$$

Here we only consider $h^i(D)$ for $i = -1,0,1$ because by definition, the cohomology is zero outside these degrees. It is well-known that since $X$ is smooth over $k$, $\tau_{\leq 1}R\pi_*\mathbb{G}_m$ is quasi-isomorphic to the complex $[\bar{k}(X)^* \rightarrow \mathrm{Div}(\bar{X})]$ in degrees 0 and 1. Thus one easily computes that $h^0(\tau_{\leq 1}R\pi_*\mathbb{G}_m) \cong \bar{k}[X]^*$ and $h^1(\tau_{\leq 1}R\pi_*\mathbb{G}_m) \cong \mathrm{Pic}(\bar{X})$. Also, we have

$$h^i((\tau_{\leq 1}R\pi_*\mu_\infty)[1]) = h^{i+1}(\tau_{\leq 1}R\pi_*\mu_\infty)$$

and so the last term of the above long exact sequence is 0.

Question 2. We want to show that this long exact sequence is precisely the one mentioned in the first part of the question. All we need to do is find an injective resolution for $\mu_\infty$, this will enable us to get an explicit presentation of $R\pi_*\mu_\infty$, but this is where I have no idea how to proceed.

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    $\begingroup$ I'm a little bit confused, this seems quite straightforward, maybe I'm missing something. You want to identify the image of the morphism induced on the first cohomology group, by morphism of sheaves $\mathbb{G}_m\rightarrow \mathbb{G}_m\otimes \mathbb{Q}$. Note that tensoring with $\mathbb{Q}$ commutes with cohomology. So you are asking what is the image of $Pic(X)\rightarrow Pic(X)\otimes \mathbb{Q}$. It is image is going to be free and you can see it is also the maximal one. $\endgroup$ – user127776 Jul 22 at 4:24
  • $\begingroup$ @user127776 So what you're saying is that we have $\mathrm{Pic}(\bar{X})_{free} \subset \mathrm{Pic}(\bar{X}) \otimes \mathbb{Q}$ right? Do you know if this inclusion is strict? I'm also interested in the group $H^2(\bar{X},\mu_\infty)$, do you know any convenient ways to compute it for, say, a curve $\bar{X}$? For example, if $X$ is a smooth proper connected curve over $k$ then $H^2(X,\mu_n) \cong \mathbb{Z}/n\mathbb{Z}$. $\endgroup$ – Kelvin Lian Jul 22 at 4:53
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    $\begingroup$ Yes by definition maximal free quotient is just the natural image of group in its rationalization. It depends for example if you are working with smooth projective curves over number fields, it is going to be finitely generated and the inclusion is going to be strict. I don't think there is a general formula for $H^2(X, \mu_n)$, it depends on the Brauer group of $X$. $\endgroup$ – user127776 Jul 22 at 5:14

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