The Stacks project

Lemma 51.22.4. With $q_0 = q(S)$ and $d = d(S)$ as above, suppose we have integers $n \geq c \geq 0$, an $(A, n, c)$-module $M$, an index $i \in \{ 0, 1, \ldots , d\} $, and an integer $q$. Then we distinguish the following cases

  1. In the case $i = d \geq 1$ and $q \geq q_0$ we have $H^ d(X, p^*M(q)) = 0$.

  2. In the case $i = d - 1 \geq 1$ and $q \geq q_0$ we have $H^{d - 1}(X, p^*M(q)) = 0$.

  3. In the case $d - 1 > i > 0$ and $q \geq q_0 + (d - 1 - i)c$ the map $H^ i(X, p^*M(q)) \to H^ i(X, p^*M(q - (d - 1 - i)c))$ is zero.

  4. In the case $i = 0$, $d \in \{ 0, 1\} $, and $q \geq q_0$, there is a surjection

    \[ I^ qM \longrightarrow H^0(X, p^*M(q)) \]

  5. In the case $i = 0$, $d > 1$, and $q \geq q_0 + (d - 1)c$ the map

    \[ H^0(X, p^*M(q)) \to H^0(X, p^*M(q - (d - 1)c)) \]

    has image contained in the image of the canonical map $I^{q - (d - 1)c}M \to H^0(X, p^*M(q - (d - 1)c))$.

Proof. Let $M$ be an $(A, n, c)$-module. Choose a short exact sequence

\[ 0 \to K \to (A/I^ n)^{\oplus r} \to M \to 0 \]

We will use below that $K$ is an $(A, n, c)$-module, see More on Algebra, Lemma 15.70.6. Consider the corresponding exact sequence

\[ p^*K \to (\mathcal{O}_{Y_ n})^{\oplus r} \to p^*M \to 0 \]

We split this into short exact sequences

\[ 0 \to \mathcal{F} \to p^*K \to \mathcal{G} \to 0 \quad \text{and}\quad 0 \to \mathcal{G} \to (\mathcal{O}_{Y_ n})^{\oplus r} \to p^*M \to 0 \]

By Lemma 51.22.2 the coherent module $\mathcal{F}$ is scheme theoretically supported on $Y_ c$.

Proof of (1). Assume $d > 0$. We have to prove $H^ d(X, p^*M(q)) = 0$ for $q \geq q_0$. By the vanishing of the cohomology of twists of $\mathcal{G}$ in degrees $> d$ and the long exact cohomology sequence associated to the second short exact sequence above, it suffices to prove that $H^ d(X, \mathcal{O}_{Y_ n}(q)) = 0$. This is true by Lemma 51.22.1.

Proof of (2). Assume $d > 1$. We have to prove $H^{d - 1}(X, p^*M(q)) = 0$ for $q \geq q_0$. Arguing as in the previous paragraph, we see that it suffices to show that $H^ d(X, \mathcal{G}(q)) = 0$. Using the first short exact sequence and the vanishing of the cohomology of twists of $\mathcal{F}$ in degrees $> d$ we see that it suffices to show $H^ d(X, p^*K(q))$ is zero which is true by (1) and the fact that $K$ is an $(A, n, c)$-module (see above).

Proof of (3). Let $0 < i < d - 1$ and assume the statement holds for $i + 1$ except in the case $i = d - 2$ we have statement (2). Using the long exact sequence of cohomology associated to the second short exact sequence above we find an injection

\[ H^ i(X, p^*M(q - (d - 1 - i)c)) \subset H^{i + 1}(X, \mathcal{G}(q - (d - 1 - i)c)) \]

as $q - (d - 1 - i)c \geq q_0$ gives the vanishing of $H^ i(X, \mathcal{O}_{Y_ n}(q - (d - 1 - i)c))$ (see above). Thus it suffices to show that the map $H^{i + 1}(X, \mathcal{G}(q)) \to H^{i + 1}(X, \mathcal{G}(q - (d - 1 - i)c))$ is zero. To study this, we consider the maps of exact sequences

\[ \xymatrix{ H^{i + 1}(X, p^*K(q)) \ar[r] \ar[d] & H^{i + 1}(X, \mathcal{G}(q)) \ar[r] \ar[d] \ar@{..>}[ld] & H^{i + 2}(X, \mathcal{F}(q)) \ar[d] \\ H^{i + 1}(X, p^*K(q - c)) \ar[r] \ar[d] & H^{i + 1}(X, \mathcal{G}(q - c)) \ar[r] \ar[d] & H^{i + 2}(X, \mathcal{F}(q - c)) \\ H^{i + 1}(X, p^*K(q - (d - 1 - i)c)) \ar[r] & H^{i + 1}(X, \mathcal{G}(q - (d - 1 - i)c)) } \]

Since $\mathcal{F}$ is scheme theoretically supported on $Y_ c$ we see that the canonical map $\mathcal{G}(q) \to \mathcal{G}(q - c)$ factors through $p^*K(q - c)$ by Lemma 51.22.3. This gives the dotted arrow in the diagram. (In fact, for the proof it suffices to observe that the vertical arrow on the extreme right is zero in order to get the dotted arrow as a map of sets.) Thus it suffices to show that $H^{i + 1}(X, p^*K(q - c)) \to H^{i + 1}(X, p^*K(q - (d - 1 - i)c))$ is zero. If $i = d - 2$, then the source of this arrow is zero by (2) as $q - c \geq q_0$ and $K$ is an $(A, n, c)$-module. If $i < d - 2$, then as $K$ is an $(A, n, c)$-module, we get from the induction hypothesis that the map is indeed zero since $q - c - (q - (d - 1 - i)c) = (d - 2 - i)c = (d - 1 - (i + 1))c$ and since $q - c \geq q_0 + (d - 1 - (i + 1))c$. In this way we conclude the proof of (3).

Proof of (4). Assume $d \in \{ 0, 1\} $ and $q \geq q_0$. Then the first short exact sequence gives a surjection $H^1(X, p^*K(q)) \to H^1(X, \mathcal{G}(q))$ and the source of this arrow is zero by case (1). Hence for all $q \in \mathbf{Z}$ we see that the map

\[ H^0(X, (\mathcal{O}_{Y_ n})^{\oplus r}(q)) \longrightarrow H^0(X, p^*M(q)) \]

is surjective. For $q \geq q_0$ the source is equal to $(I^ q/I^{q + n})^{\oplus r}$ by Lemma 51.22.1 and this easily proves the statement.

Proof of (5). Assume $d > 1$. Arguing as in the proof of (4) we see that it suffices to show that the image of

\[ H^0(X, p^*M(q)) \longrightarrow H^0(X, p^*M(q - (d - 1)c)) \]

is contained in the image of

\[ H^0(X, (\mathcal{O}_{Y_ n})^{\oplus r}(q - (d - 1)c)) \longrightarrow H^0(X, p^*M(q - (d - 1)c)) \]

To show the inclusion above, it suffices to show that for $\sigma \in H^0(X, p^*M(q))$ with boundary $\xi \in H^1(X, \mathcal{G}(q))$ the image of $\xi $ in $H^1(X, \mathcal{G}(q - (d - 1)c))$ is zero. This follows by the exact same arguments as in the proof of (3). $\square$

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



View Lemma 51.22.4 as pdf