The Stacks project

Lemma 59.97.5. Let $K$ be a field. Let $j : U \to X$ be an open immersion of schemes of finite type over $K$. Let $Y$ be a scheme of finite type over $K$. Consider the diagram

\[ \xymatrix{ Y \times _{\mathop{\mathrm{Spec}}(K)} X \ar[d]_ q & Y \times _{\mathop{\mathrm{Spec}}(K)} U \ar[l]^ h \ar[d]^ p \\ X & U \ar[l]_ j } \]

Then the base change map $q^{-1}Rj_*\mathcal{F} \to Rh_*p^{-1}\mathcal{F}$ is an isomorphism for $\mathcal{F}$ an abelian sheaf on $U_{\acute{e}tale}$ whose stalks are torsion of orders invertible in $K$.

Proof. Write $\mathcal{F} = \mathop{\mathrm{colim}}\nolimits \mathcal{F}[n]$ where the colimit is over the multiplicative system of integers invertible in $K$. Since cohomology commutes with filtered colimits in our situation (for a precise reference see Lemma 59.86.3), it suffices to prove the lemma for $\mathcal{F}[n]$. Thus we may assume $\mathcal{F}$ is a sheaf of $\mathbf{Z}/n\mathbf{Z}$-modules for some $n$ invertible in $K$ (we will use this at the very end of the proof). In the proof we use the short hand $X \times _ K Y$ for the fibre product over $\mathop{\mathrm{Spec}}(K)$. We will prove the lemma by induction on $\dim (X) + \dim (Y)$. The lemma is trivial if $\dim (X) \leq 0$, since in this case $U$ is an open and closed subscheme of $X$. Choose a point $z \in X \times _ K Y$. We will show the stalk at $\overline{z}$ is an isomorphism.

Suppose that $z \mapsto x \in X$ and assume $\text{trdeg}_ K(\kappa (x)) > 0$. Set $X' = \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}^{sh})$ and denote $U' \subset X'$ the inverse image of $U$. Consider the base change

\[ \xymatrix{ Y \times _ K X' \ar[d]_{q'} & Y \times _ K U' \ar[l]^{h'} \ar[d]^{p'} \\ X' & U' \ar[l]_{j'} } \]

of our diagram by $X' \to X$. Observe that $X' \to X$ is a filtered colimit of étale morphisms. By smooth base change in the form of Lemma 59.89.3 the pullback of $q^{-1}Rj_*\mathcal{F} \to Rh_*p^{-1}\mathcal{F}$ to $X'$ to $Y \times _ K X'$ is the map $(q')^{-1}Rj'_*\mathcal{F}' \to Rj'_*(p')^{-1}\mathcal{F}'$ where $\mathcal{F}'$ is the pullback of $\mathcal{F}$ to $U'$. (In this step it would suffice to use étale base change which is an essentially trivial result.) So it suffices to show that $(q')^{-1}Rj'_*\mathcal{F}' \to Rj'_*(p')^{-1}\mathcal{F}'$ is an isomorphism in order to prove that our original map is an isomorphism on stalks at $\overline{z}$. By Lemma 59.95.5 there is a separably closed field $L/K$ such that $X' = \mathop{\mathrm{lim}}\nolimits X_ i$ with $X_ i$ affine of finite type over $L$ and $\dim (X_ i) < \dim (X)$. For $i$ large enough there exists an open $U_ i \subset X_ i$ restricting to $U'$ in $X'$. We may apply the induction hypothesis to the diagram

\[ \vcenter { \xymatrix{ Y \times _ K X_ i \ar[d]_{q_ i} & Y \times _ K U_ i \ar[l]^{h_ i} \ar[d]^{p_ i} \\ X_ i & U_ i \ar[l]_{j_ i} } } \quad \text{equal to}\quad \vcenter { \xymatrix{ Y_ L \times _ L X_ i \ar[d]_{q_ i} & Y_ L \times _ L U_ i \ar[l]^{h_ i} \ar[d]^{p_ i} \\ X_ i & U_ i \ar[l]_{j_ i} } } \]

over the field $L$ and the pullback of $\mathcal{F}$ to these diagrams. By Lemma 59.86.3 we conclude that the map $(q')^{-1}Rj'_*\mathcal{F}' \to Rj'_*(p')^{-1}\mathcal{F}$ is an isomorphism.

Suppose that $z \mapsto y \in Y$ and assume $\text{trdeg}_ K(\kappa (y)) > 0$. Let $Y' = \mathop{\mathrm{Spec}}(\mathcal{O}_{X, x}^{sh})$. By Lemma 59.95.5 there is a separably closed field $L/K$ such that $Y' = \mathop{\mathrm{lim}}\nolimits Y_ i$ with $Y_ i$ affine of finite type over $L$ and $\dim (Y_ i) < \dim (Y)$. In particular $Y'$ is a scheme over $L$. Denote with a subscript $L$ the base change from schemes over $K$ to schemes over $L$. Consider the commutative diagrams

\[ \vcenter { \xymatrix{ Y' \times _ K X \ar[d]_ f & Y' \times _ K U \ar[l]^{h'} \ar[d]^{f'} \\ Y \times _ K X \ar[d]_ q & Y \times _ K U \ar[l]^ h \ar[d]^ p \\ X & U \ar[l]_ j } } \quad \text{and}\quad \vcenter { \xymatrix{ Y' \times _ L X_ L \ar[d]_{q'} & Y' \times _ L U_ L \ar[l]^{h'} \ar[d]^{p'} \\ X_ L \ar[d] & U_ L \ar[l]^{j_ L} \ar[d] \\ X & U \ar[l]_ j } } \]

and observe the top and bottom rows are the same on the left and the right. By smooth base change we see that $f^{-1}Rh_*p^{-1}\mathcal{F} = Rh'_*(f')^{-1}p^{-1}\mathcal{F}$ (similarly to the previous paragraph). By smooth base change for $\mathop{\mathrm{Spec}}(L) \to \mathop{\mathrm{Spec}}(K)$ (Lemma 59.90.1) we see that $Rj_{L, *}\mathcal{F}_ L$ is the pullback of $Rj_*\mathcal{F}$ to $X_ L$. Combining these two observations, we conclude that it suffices to prove the base change map for the upper square in the diagram on the right is an isomorphism in order to prove that our original map is an isomorphism on stalks at $\overline{z}$1. Then using that $Y' = \mathop{\mathrm{lim}}\nolimits Y_ i$ and argueing exactly as in the previous paragraph we see that the induction hypothesis forces our map over $Y' \times _ K X$ to be an isomorphism.

Thus any counter example with $\dim (X) + \dim (Y)$ minimal would only have nonisomorphisms $q^{-1}Rj_*\mathcal{F} \to Rh_*p^{-1}\mathcal{F}$ on stalks at closed points of $X \times _ K Y$ (because a point $z$ of $X \times _ K Y$ is a closed point if and only if both the image of $z$ in $X$ and in $Y$ are closed). Since it is enough to prove the isomorphism locally, we may assume $X$ and $Y$ are affine. However, then we can choose an open dense immersion $Y \to Y'$ with $Y'$ projective. (Choose a closed immersion $Y \to \mathbf{A}^ n_ K$ and let $Y'$ be the scheme theoretic closure of $Y$ in $\mathbf{P}^ n_ K$.) Then $\dim (Y') = \dim (Y)$ and hence we get a “minimal” counter example with $Y$ projective over $K$. In the next paragraph we show that this can't happen.

Consider a diagram as in the statement of the lemma such that $q^{-1}Rj_*\mathcal{F} \to Rh_*p^{-1}\mathcal{F}$ is an isomorphism at all non-closed points of $X \times _ K Y$ and such that $Y$ is projective. The restriction of the map to $(X \times _ K Y)_{K^{sep}}$ is the corresponding map for the diagram of the lemma base changed to $K^{sep}$. Thus we may and do assume $K$ is separably algebraically closed. Choose a distinguished triangle

\[ q^{-1}Rj_*\mathcal{F} \to Rh_*p^{-1}\mathcal{F} \to Q \to (q^{-1}Rj_*\mathcal{F})[1] \]

in $D((X \times _ K Y)_{\acute{e}tale})$. Since $Q$ is supported in closed points we see that it suffices to prove $H^ i(X \times _ K Y, Q) = 0$ for all $i$, see Lemma 59.97.4. Thus it suffices to prove that $q^{-1}Rj_*\mathcal{F} \to Rh_*p^{-1}\mathcal{F}$ induces an isomorphism on cohomology. Recall that $\mathcal{F}$ is annihilated by $n$ invertible in $K$. By the Künneth formula of Lemma 59.97.2 we have

\begin{align*} R\Gamma (X \times _ K Y, q^{-1}Rj_*\mathcal{F}) & = R\Gamma (X, Rj_*\mathcal{F}) \otimes _{\mathbf{Z}/n\mathbf{Z}}^\mathbf {L} R\Gamma (Y, \mathbf{Z}/n\mathbf{Z}) \\ & = R\Gamma (U, \mathcal{F}) \otimes _{\mathbf{Z}/n\mathbf{Z}}^\mathbf {L} R\Gamma (Y, \mathbf{Z}/n\mathbf{Z}) \end{align*}


\[ R\Gamma (X \times _ K Y, Rh_*p^{-1}\mathcal{F}) = R\Gamma (U \times _ K Y, p^{-1}\mathcal{F}) = R\Gamma (U, \mathcal{F}) \otimes _{\mathbf{Z}/n\mathbf{Z}}^\mathbf {L} R\Gamma (Y, \mathbf{Z}/n\mathbf{Z}) \]

This finishes the proof. $\square$

[1] Here we use that a “vertical composition” of base change maps is a base change map as explained in Cohomology on Sites, Remark 21.19.4.

Comments (0)

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0F1H. Beware of the difference between the letter 'O' and the digit '0'.