Proper mapping theorem

Question

Let $Z\to X$ be a closed immersion of schemes. Assume $\mathcal{O}_Z$ and $\mathcal{O}_X$ both are coherent sheaves of $\mathcal{O}_Z$, resp. $\mathcal{O}_X$-modules.

In particular, the coherent sheaves on $Z$, resp. $X$, are precisely the finitely presented $\mathcal{O}_Z$, resp. $\mathcal{O}_X$-modules.

It is true that if $\mathcal{F}$ is a quasi-coherent $\mathcal{O}_Z$-module of finite type, then $i_*\mathcal{F}$ is a quasi-coherent $\mathcal{O}_X$-module of finite type (in general. No need of coherence of the structure sheaves).

Is it true, under the assumptions explained up above, that if $\mathcal{F}$ is a finitely presented quasi-coherent $\mathcal{O}_Z$-module, then $i_*\mathcal{F}$ is also finitely presented?

Answer 1

The answer is no. Let $i_0 : Z_0\to X_0$ be a closed immersion of smooth varieties over a finite field. Upon passing to the perfection, denoted $(\cdot)^{\rm perf}$, i.e. the limit along the absolute Frobenius morphisms, we call:

$$Z := Z_0^{\rm perf}, X := X_0^{\rm perf}\ \ \text{and}\ \ i := i^{\rm perf} : Z\to X.$$

By regularity of $Z_0$ and $X_0$, their absolute Frobenii are flat. $\mathcal{O}_{Z_0}$ and $\mathcal{O}_{X_0}$ are coherent, hence so are $\mathcal{O}_{Z}$ and $\mathcal{O}_{X}$ (since a filtered colimit of coherent rings with flat transition maps is coherent).

$\mathcal{F} = \mathcal{O}_Z$ is already a counterexample.