Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

The absolute cohomological purity theorem in étale cohomology is as follows.

Let $X$ be a regular scheme over $\mathbb{Z}[1/n]$, and $i \colon Z \to X$ the inclusion of a regular closed subscheme everywhere of codimension $c$. Let $\Lambda$ be the constant sheaf $\mathbb{Z}/n\mathbb{Z}$ on $X$. Then $R^{2c} i^! \Lambda \cong \Lambda(-c)$, and $R^{p} i^! \Lambda$ is trivial for $p \neq 2c$.

Here $i^!$ is the functor taking a sheaf $\mathcal{F}$ on $X$ to $\mathrm{ker}(\mathcal{F} \to j_* j^* \mathcal{F})$, considered as a sheaf on $Z$, where $j$ is the inclusion of $X \setminus Z$. There are several other ways to state the result: chapter 16 of Milne's online Lectures on Etale Cohomology explains nicely how to go between them. As far as I understand, the theorem has been proved by Gabber in at least two ways.

My question is this:

What other sheaves $\mathcal{F}$ may replace $\Lambda$ in the purity theorem?

The problem seems to be in defining the morphism $R^{2c} i^! \mathcal{F} \to i^* \mathcal{F}(-c)$ in the first place: once that's done, proving that it's an isomorphism is purely local and ought to follow from the case of $\Lambda$, at least for $\mathcal{F}$ something nice like a flat sheaf of $\Lambda$-modules. Maybe one can define the morphism just by tensoring the original one with $\mathcal{F}$, but I'm not sufficiently happy with the formal properties of étale cohomology to understand how $R^p i^!$ works with tensor products.

share|improve this question
add comment

1 Answer

Having had a few up-votes but no answers, let me answer my own question in case anybody finds it useful in future. This is basically a formal calculation, similar to the proof of the projection formula, and I'm sure it's well known to experts in étale cohomology. But I couldn't find the details in the literature.

Everything happens in the category of sheaves of $\Lambda$-modules on $X$. Note that cohomology in this category is the same as cohomology in the category of sheaves of Abelian groups on $X$.$\DeclareMathOperator{\Hom}{Hom}\newcommand{\F}{\mathcal{F}}$

Step 1. Whenever $\F$ is flat, there is a "projection formula" $i^!(\mathcal{G} \otimes \F) \cong i^! \mathcal{G} \otimes i^* \F$ for any sheaf $\mathcal{G}$ on $X$. To see this, take the short exact sequence $0 \to i_* i^! \mathcal{G} \to \mathcal{G} \to j_* j^* \mathcal{G}$ and tensor with $\F$. The usual projection formula gives $(j_* j^* \mathcal{G}) \otimes \F \cong j_*(j^* \mathcal{G} \otimes j^* \F) \cong j_*j^*(\mathcal{G} \otimes \F)$ and $(i_* i^! \mathcal{G}) \otimes \F \cong i_*(i^! \mathcal{G} \otimes \F)$. We obtain

$0 \to i_* (i^! \mathcal{G} \otimes i^* \F) \to \mathcal{G} \otimes \F \to j_* j^* (\mathcal{G} \otimes \F)$.

Comparing this with the definition of $i^!(\mathcal{G} \otimes \F)$ gives the claimed isomorphism.

Step 2. If $\F$ is locally free of finite rank, then there is a natural isomorphism $- \otimes \F\to \Hom(\F^\vee,-)$, where $\F^\vee$ denotes $\mathrm{Hom}(\F,\Lambda)$; this can be checked on stalks and so follows from duality for $\mathbb{Z}/n\mathbb{Z}$-modules. It follows that

$\Hom(A, B\otimes\F) \cong \Hom(A,\Hom(\F^\vee,B)) \cong \Hom(A\otimes\F^\vee, B).$

So the functor $- \otimes \F$ admits an exact left adjoint $- \otimes \F^\vee$; therefore it is exact and takes injectives to injectives.

Step 3. Using these two facts, we can deduce that $R^p i^! \F \cong (R^p i^! \Lambda) \otimes \F$ for all $p$. To do so, consider an injective resolution of $\Lambda$. Tensoring with $\F$ gives an injective resolution of $\F$ by (2). Now apply $i^!$ and take homology; by the projection formula proved in (1), we get the claimed isomorphism.

Step 4. Now we can deduce the purity result for $\F$ by taking the isomorphism from the purity theorem and tensoring with $\F$.

So the purity theorem hold as least for $\Lambda$-modules $\F$ which are locally free of finite rank.

share|improve this answer
add comment

Your Answer

 
discard

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.