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
\[ \xymatrix{ Z' \ar[r]_{i'} \ar[d]_{f'} & Y \ar[d]^ f \\ Z \ar[r]^ i & X } \]
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. $\square$
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