The Stacks project

Lemma 36.31.4. Let $X$ be a scheme. Let $E \in D(\mathcal{O}_ X)$ be perfect of tor-amplitude in $[a, b]$ for some $a, b \in \mathbf{Z}$. Let $r \geq 0$. Then there exists a locally closed subscheme $j : Z \to X$ characterized by the following

  1. $H^ a(Lj^*E)$ is a locally free $\mathcal{O}_ Z$-module of rank $r$, and

  2. a morphism $f : Y \to X$ factors through $Z$ if and only if for all morphisms $g : Y' \to Y$ the $\mathcal{O}_{Y'}$-module $H^ a(L(f \circ g)^*E)$ is locally free of rank $r$.

Moreover, $j : Z \to X$ is of finite presentation and we have

  1. if $f : Y \to X$ factors as $Y \xrightarrow {g} Z \to X$, then $H^ a(Lf^*E) = g^*H^ a(Lj^*E)$,

  2. if $\beta _ a(x) \leq r$ for all $x \in X$, then $j$ is a closed immersion and given $f : Y \to X$ the following are equivalent

    1. $f : Y \to X$ factors through $Z$,

    2. $H^0(Lf^*E)$ is a locally free $\mathcal{O}_ Y$-module of rank $r$,

    and if $r = 1$ these are also equivalent to

    1. $\mathcal{O}_ Y \to \mathop{\mathcal{H}\! \mathit{om}}\nolimits _{\mathcal{O}_ Y}(H^0(Lf^*E), H^0(Lf^*E))$ is injective.

Proof. First, let $U \subset X$ be the locally constructible open subscheme where the function $\beta _ a$ of Lemma 36.31.1 has values $\leq r$. Let $f : Y \to X$ be as in (2). Then for any $y \in Y$ we have $\beta _ a(Lf^*E) = r$ hence $y$ maps into $U$ by Lemma 36.31.1. Hence $f$ as in (2) factors through $U$. Thus we may replace $X$ by $U$ and assume that $\beta _ a(x) \in \{ 0, 1, \ldots , r\} $ for all $x \in X$. We will show that in this case there is a closed subscheme $Z \subset X$ cut out by a finite type quasi-coherent ideal characterized by the equivalence of (4) (a), (b) and (4)(c) if $r = 1$ and that (3) holds. This will finish the proof because it will a fortiori show that morphisms as in (2) factor through $Z$.

If $x \in X$ and $\beta _ a(x) < r$, then there is an open neighbourhood of $x$ where $\beta _ a < r$ (Lemma 36.31.1). In this way we see that set theoretically at least $Z$ is a closed subset.

To get a scheme theoretic structure, consider a point $x \in X$ with $\beta _ a(x) = r$. Set $\beta = \beta _{a + 1}(x)$. By More on Algebra, Lemma 15.75.6 there exists an affine open neighbourhood $U$ of $x$ such that $K|_ U$ is represented by a complex

\[ \ldots \to 0 \to \mathcal{O}_ U^{\oplus r} \xrightarrow {(f_{ij})} \mathcal{O}_ U^{\oplus \beta } \to \ldots \to \mathcal{O}_ U^{\oplus \beta _{b - 1}(x)} \to \mathcal{O}_ U^{\oplus \beta _ b(x)} \to 0 \to \ldots \]

(This also uses earlier results to turn the problem into algebra, for example Lemmas 36.3.5 and 36.10.7.) Now, if $g : Y \to U$ is any morphism of schemes such that $g^\sharp (f_{ij})$ is nonzero for some pair $i, j$, then $H^0(Lg^*E)$ is not a locally free $\mathcal{O}_ Y$-module of rank $r$. See More on Algebra, Lemma 15.15.7. Trivially $H^0(Lg^*E)$ is a locally free $\mathcal{O}_ Y$-module if $g^\sharp (f_{ij}) = 0$ for all $i, j$. Thus we see that over $U$ the closed subscheme cut out by all $f_{ij}$ satisfies (3) and we have the equivalence of (4)(a) and (b). The characterization of $Z$ shows that the locally constructed patches glue (details omitted). Finally, if $r = 1$ then (4)(c) is equivalent to (4)(b) because in this case locally $H^0(Lg^*E) \subset \mathcal{O}_ Y$ is the annihilator of the ideal generated by the elements $g^\sharp (f_{ij})$. $\square$


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