A sufficient condition for a morphism to be a closed immersion?

Question

Let $R$ be an integral $\bar{\mathbb{F}}_p$-algebra of finite type, let $V$ be an $R$-algebra. Consider a morphism $f \colon \mathrm{Spec}(V) \rightarrow \mathbb{A}^n_R$ that has the following properties:

• every fibre of $f$ is a closed immersion $f_K\colon \mathrm{Spec}(V)_K \hookrightarrow \mathbb{A}^n_K$
• the induced morphism of functors $\mathrm{Hom}_R(-, \mathrm{Spec}(V)) \rightarrow \mathrm{Hom}_R(-, \mathbb{A}^n_R)$ is injective
• $f$ is universally closed
• $V$ is at most countably generated as an $R$-algebra
• $\mathrm{Spec}(V)$ has a $\mathbb{G}_m$-action and $f$ is compatible with the $\mathbb{G}_m$-actions

Is it true that then $f$ is always a closed immersion? This should be clear if $f$ is of finite type, but in the particular example I am interested in, this is (a priori) not the case. If not, are there any easy to understand counterexamples? Some of the conditions above may be irrelevant (like $R$ living over $\bar{\mathbb{F}}_p$ or the group scheme action), but I listed them anyways on the off-chance that they could be useful.

Answer 1

Yes, it is true. More precisely any universally closed monomorphism of schemes (your second and third condition) with locally Noetherian target is a closed immersion. This is proven in Proposition 3.8 of Ferrand: Monomorphismes de schémas noethérian (http://www.numdam.org/article/SAC_1967-1968__2__A7_0.pdf).