Ring extension and Jacobson rings

Question

If $R\subseteq S$ are commutative rings, is it a fact that $R$ is a Jacobson ring if and only if $S$ is so?

I guess the contraction of maximal and prime ideals of $S$ may be helpful in this regards. Thanks in advance!

Answer 1

Yes, it is true that if $R\subset S$ is an integral extenson of rings and if $S$ is Jacobson, then $R$ is Jacobson too. [For the converse, see here]
Indeed, let $\mathfrak p \subset R$ be a prime ideal and $\mathfrak P \subset S$ a prime lying over $\mathfrak p $ .
Since $S$ is Jacobson, we can write $\mathfrak P=\cap \mathfrak M_i$ for some family of maximal ideals $\mathfrak M_i \subset S$.
But then $$\mathfrak p =R\cap \mathfrak P=R\cap (\cap \mathfrak M_i)=\cap (R\cap \mathfrak M_i)$$
Now $\mathfrak m_i:=R\cap \mathfrak M_i$ is a maximal ideal of $R$ (by the integrality of $R\subset S$) and thus we have written an arbitrary prime ideal $\mathfrak p \subset R$ as the intersection $\mathfrak p=\cap (R\cap \mathfrak M_i)=\cap \mathfrak m_i$ of some maximal ideals $\mathfrak m_i\subset R$.
Thus $R$ is Jacobson.