Inverse image of maximal ideals under finite type ring maps.

Question

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.

Answer 1

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.