# Inverse image of maximal ideals under finite type ring maps.

All rings are commutative with $$1$$.

I am trying to find a name of such ring $$R$$ with the following property:

For any finite type ring map $$f: R\to A$$, inverse image $$f^{-1}(\mathfrak m)$$ of any maximal ideal $$\mathfrak m\subset A$$ is also maximal in $$R$$.

In other words, $$f^*:\operatorname{Spec}(A)\to\operatorname{Spec}(R)$$ maps closed points to closed points.

By the proposition here any Jacobson ring $$R$$ satisfies this property but not conversely. Is there a name for such ring $$R$$? I am trying to find a reference.

The property you claimed above is equivalent to "$$R$$ is a Jacobson ring", i.e. we have
$$R$$ is Jacobson if and only if for any finite type ring map $$R\to A$$, inverse image of maximal ideals are maximal as well.
As you cited in the question, if $$R$$ is Jacobson, then any finite type ring map $$R\to A$$ satisfies the property.
Conversely, assume $$R$$ is not Jacobson and use this lemma with the finite type ring morphism $$h:R\to (R/\mathfrak p)_f$$, where $$\mathfrak p$$ is a non-maximal prime ideal in $$R$$ and $$(R/\mathfrak p)_f$$ is a field. Now $$h^{-1}((0))=\mathfrak p$$ gives a contradiction since $$\mathfrak p$$ is not maximal.