Let $\phi:\mathcal F\rightarrow \mathcal G$ be a morphism between quasicoherent sheaves over a scheme $S$. Then is the zero locus of $\phi$ is a closed subscheme?
In the language of representable functor, let $F:(Sch/S)\rightarrow (Sets)$ be the functor $F(T) = \{f:T\rightarrow S|f^*(\phi)= 0\}$. It is equivalent to say $F$ is represented by a closed subscheme of $S$.
I already know it is true when $\mathcal G$ is finite locally free. I'm wondering if there is any mild(for example, $\mathcal G$ is flat, sheaves are coherent, $S$ is Noetherian, etc) condition guarantees this property.
Thank you.
Best Answer
Not in general. Let $Z\subset X$ be a closed subscheme, and consider the identity map $\mathcal{O}_Z\to\mathcal{O}_Z$. The zero locus is $X\setminus U$, which is open and not generally closed.
Off the top of my head, I do not know any condition more general than what you've already indicated which imply the result you seek. I suspect $\mathcal{G}$ flat is necessary if you're asking about all morphisms, but I doubt that's sufficient. It also seems likely that some mild finiteness hypothesis (either on the sheaves or on $S$) will be necessary, and you may be curious to know that this is already somewhat close to finite locally free, as seen in Stacks 053N.