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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