The Stacks project

Lemma 37.67.2 (Derived Chow's lemma). Let $A$ be a ring. Let $X$ be a separated scheme of finite presentation over $A$. Let $x \in X$. Then there exist an open neighbourhood $U \subset X$ of $x$, an $n \geq 0$, an open $V \subset \mathbf{P}^ n_ A$, a closed subscheme $Z \subset X \times _ A \mathbf{P}^ n_ A$, a point $z \in Z$, and an object $E$ in $D(\mathcal{O}_{X \times _ A \mathbf{P}^ n_ A})$ such that

  1. $Z \to X \times _ A \mathbf{P}^ n_ A$ is of finite presentation,

  2. $b : Z \to X$ is an isomorphism over $U$ and $b(z) = x$,

  3. $c : Z \to \mathbf{P}^ n_ A$ is a closed immersion over $V$,

  4. $b^{-1}(U) = c^{-1}(V)$, in particular $c(z) \in V$,

  5. $E|_{X \times _ A V} \cong (b, c)_*\mathcal{O}_ Z|_{X \times _ A V}$,

  6. $E$ is pseudo-coherent and supported on $Z$.

Proof. We can find a finite type $\mathbf{Z}$-subalgebra $A' \subset A$ and a scheme $X'$ separated and of finite presentation over $A'$ whose base change to $A$ is $X$. See Limits, Lemmas 32.10.1 and 32.8.6. Let $x' \in X'$ be the image of $x$. If we can prove the lemma for $x' \in X'/A'$, then the lemma follows for $x \in X/A$. Namely, if $U', n', V', Z', z', E'$ provide the solution for $x' \in X'/A'$, then we can let $U \subset X$ be the inverse image of $U'$, let $n = n'$, let $V \subset \mathbf{P}^ n_ A$ be the inverse image of $V'$, let $Z \subset X \times \mathbf{P}^ n$ be the scheme theoretic inverse image of $Z'$, let $z \in Z$ be the unique point mapping to $x$, and let $E$ be the derived pullback of $E'$. Observe that $E$ is pseudo-coherent by Cohomology, Lemma 20.45.3. It only remains to check (5). To see this set $W = b^{-1}(U) = c^{-1}(V)$ and $W' = (b')^{-1}(U) = (c')^{-1}(V')$ and consider the cartesian square

\[ \xymatrix{ W \ar[d]_{(b, c)} \ar[r] & W' \ar[d]^{(b', c')} \\ X \times _ A V \ar[r] & X' \times _{A'} V' } \]

By Lemma 37.67.1 the schemes $X \times _ A V$ and $W'$ are Tor independent over $X' \times _{A'} V'$. Hence the derived pullback of $(b', c')_*\mathcal{O}_{W'}$ to $X \times _ A V$ is $(b, c)_*\mathcal{O}_ W$ by Derived Categories of Schemes, Lemma 36.22.5. This also uses that $R(b', c')_*\mathcal{O}_{Z'} = (b', c')_*\mathcal{O}_{Z'}$ because $(b', c')$ is a closed immersion and simiarly for $(b, c)_*\mathcal{O}_ Z$. Since $E'|_{U' \times _{A'} V'} = (b', c')_*\mathcal{O}_{W'}$ we obtain $E|_{U \times _ A V} = (b, c)_*\mathcal{O}_ W$ and (5) holds. This reduces us to the situation described in the next paragraph.

Assume $A$ is of finite type over $\mathbf{Z}$. Choose an affine open neighbourhood $U \subset X$ of $x$. Then $U$ is of finite type over $A$. Choose a closed immersion $U \to \mathbf{A}^ n_ A$ and denote $j : U \to \mathbf{P}^ n_ A$ the immersion we get by composing with the open immersion $\mathbf{A}^ n_ A \to \mathbf{P}^ n_ A$. Let $Z$ be the scheme theoretic closure of

\[ (\text{id}_ U, j) : U \longrightarrow X \times _ A \mathbf{P}^ n_ A \]

Since the projection $X \times \mathbf{P}^ n \to X$ is separated, we conclude from Morphisms, Lemma 29.6.8 that $b : Z \to X$ is an isomorphism over $U$. Let $z \in Z$ be the unique point lying over $x$.

Let $Y \subset \mathbf{P}^ n_ A$ be the scheme theoretic closure of $j$. Then it is clear that $Z \subset X \times _ A Y$ is the scheme theoretic closure of $(\text{id}_ U, j) : U \to X \times _ A Y$. As $X$ is separated, the morphism $X \times _ A Y \to Y$ is separated as well. Hence we see that $Z \to Y$ is an isomorphism over the open subscheme $j(U) \subset Y$ by the same lemma we used above. Choose $V \subset \mathbf{P}^ n_ A$ open with $V \cap Y = j(U)$. Then we see that (3) and (4) hold.

Because $A$ is Noetherian we see that $X$ and $X \times _ A \mathbf{P}^ n_ A$ are Noetherian schemes. Hence we can take $E = (b, c)_*\mathcal{O}_ Z$ in this case, see Derived Categories of Schemes, Lemma 36.10.3. This finishes the proof. $\square$


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 0CSJ. Beware of the difference between the letter 'O' and the digit '0'.