Lemma 30.13.1. Let $f : Y \to X$ be a morphism of schemes. Assume $f$ is finite, surjective and $X$ locally Noetherian. Let $Z \subset X$ be an integral closed subscheme with generic point $\xi $. Then there exists a coherent sheaf $\mathcal{F}$ on $Y$ such that the support of $f_*\mathcal{F}$ is equal to $Z$ and $(f_*\mathcal{F})_\xi $ is annihilated by $\mathfrak m_\xi $.

**Proof.**
Note that $Y$ is locally Noetherian by Morphisms, Lemma 29.15.6. Because $f$ is surjective the fibre $Y_\xi $ is not empty. Pick $\xi ' \in Y$ mapping to $\xi $. Let $Z' = \overline{\{ \xi '\} }$. We may think of $Z' \subset Y$ as a reduced closed subscheme, see Schemes, Lemma 26.12.4. Hence the sheaf $\mathcal{F} = (Z' \to Y)_*\mathcal{O}_{Z'}$ is a coherent sheaf on $Y$ (see Lemma 30.9.9). Look at the commutative diagram

We see that $f_*\mathcal{F} = i_*f'_*\mathcal{O}_{Z'}$. Hence the stalk of $f_*\mathcal{F}$ at $\xi $ is the stalk of $f'_*\mathcal{O}_{Z'}$ at $\xi $. Note that since $Z'$ is integral with generic point $\xi '$ we have that $\xi '$ is the only point of $Z'$ lying over $\xi $, see Algebra, Lemmas 10.36.3 and 10.36.20. Hence the stalk of $f'_*\mathcal{O}_{Z'}$ at $\xi $ equal $\mathcal{O}_{Z', \xi '} = \kappa (\xi ')$. In particular the stalk of $f_*\mathcal{F}$ at $\xi $ is not zero. This combined with the fact that $f_*\mathcal{F}$ is of the form $i_*f'_*(\text{something})$ implies the lemma. $\square$

## 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Your email address will not be published.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.

## Comments (0)

There are also: