Lemma 56.6.2. Let $X$ be an affine scheme. Let $\mathcal{F}$ be a finitely presented $\mathcal{O}_ X$-module. Let $\mathcal{E}$ be a nonzero quasi-coherent $\mathcal{O}_ X$-module. If $\text{Supp}(\mathcal{E}) \subset \text{Supp}(\mathcal{F})$, then there exists a nonzero map $\mathcal{F} \to \mathcal{E}$.

Proof. Let us translate the statement into algebra. Let $A$ be a ring. Let $M$ be a finitely presented $A$-module. Let $N$ be a nonzero $A$-module. Assume $\text{Supp}(N) \subset \text{Supp}(M)$. To show: $\mathop{\mathrm{Hom}}\nolimits _ A(M, N)$ is nonzero. We may assume $N = A/I$ is cyclic (replace $N$ by any nonzero cyclic submodule). Choose a presentation

$A^{\oplus m} \xrightarrow {T} A^{\oplus n} \to M \to 0$

Recall that $\text{Supp}(M)$ is cut out by $\text{Fit}_0(M)$ which is the ideal generated by the $n \times n$ minors of the matrix $T$. See More on Algebra, Lemma 15.8.4. The assumption $\text{Supp}(N) \subset \text{Supp}(M)$ now means that the elements of $\text{Fit}_0(M)$ are nilpotent in $A/I$. Consider the exact sequence

$0 \to \mathop{\mathrm{Hom}}\nolimits _ A(M, A/I) \to (A/I)^{\oplus n} \xrightarrow {T^ t} (A/I)^{\oplus m}$

We have to show that $T^ t$ cannot be injective; we urge the reader to find their own proof of this using the nilpotency of elements of $\text{Fit}_0(M)$ in $A/I$. Here is our proof. Since $\text{Fit}_0(M)$ is finitely generated, the nilpotency means that the annihilator $J \subset A/I$ of $\text{Fit}_0(M)$ in $A/I$ is nonzero. To show the non-injectivity of $T^ t$ we may localize at a prime. Choosing a suitable prime we may assume $A$ is local and $J$ is still nonzero. Then $T^ t$ has a nonzero kernel by More on Algebra, Lemma 15.15.6. $\square$

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