I am trying to prove that $R$ is a Jacobson ring iff for any $Y \subseteq Spec(R)$ closed in the Zariski topology, one has that the closure of $Spec_{max}(R) \cap Y$ is $Y$ itself.
I denote for any subset $S \subseteq R$, $V(S) := \left\{ P \in Spec(R): S \subseteq P \right\}$. In this notation $Y$ is closed iff $Y = V(S)$ for some $S \subseteq R$ by definition.
Firstly, to argue from left to right, I took $Z = V(T)$ and $Y = V(S)$ closed such that $Spec_{max}(R) \cap Y \subseteq Z \cap Y$. To show the implication, I would have to show that $Y \subseteq Z \cap Y$. Since $Spec_{max}(R) \cap Y \subseteq Z \cap Y$, we have that
\begin{equation}\label{lab} \bigcap\limits_{P \in Z \cap Y} P \subseteq \bigcap\limits_{P \in Spec_{max}(R) \cap Y} P \end{equation}
The RHS of the inclusion is just (denoting by $(\cdot)$ the ideal generated by $\cdot$ in $R$):
$$\bigcap\limits_{ \substack{\mathfrak{m} \in Spec_{max}(R) \\ (S) \subseteq \mathfrak{m} } } \mathfrak{m} \ ,$$
while the LHS of the inclusion is the intersection of primes in $V(S \cup T)$, which is just $\sqrt{(S \cup T)}$. By hypothesis $R$ is Jacobson, so:
$$\sqrt{(S \cup T)} = \bigcap\limits_{\substack{\mathfrak{m} \in Spec_{max}(R) \\ (S \cup T) \subseteq \mathfrak{m} }} \mathfrak{m} \ .$$
So $$ \bigcap\limits_{\substack{\mathfrak{m} \in Spec_{max}(R) \\ (S \cup T) \subseteq \mathfrak{m} }} \mathfrak{m} \subseteq \bigcap\limits_{ \substack{\mathfrak{m} \in Spec_{max}(R) \\ (S) \subseteq \mathfrak{m} } } \mathfrak{m} $$
The intersection on the right is over the same terms as the one on the left and possibly more, so it contains the one on the left. Hence the two intersections must be equal, so using again the hypothesis for $\sqrt{(S)}$, we get that $\sqrt{(S)} = \sqrt{(T \cup S)}$. Now an easy check shows that $V(\sqrt{I})=V(I)$ for any ideal $I$, hence we get that $Z \cap Y = V((T \cup S)) = V(\sqrt{(T \cup S)}) = V(\sqrt{(S)})= V((S)) = Y$ as desired. Is this correct? Is there a less "messy" proof (using only the very elementary properties of the Zariski topology, which are the only ones with which I'm familiar now)?
As for the second implication, I am a bit stuck: for $I$ an ideal in $R$ I have to show that it has the Jacobson property if $Spec_{max}(R) \cap V(I)$ is dense in $V(I)$. I think the problem is I can't figure out a way to take advantage of the density property. I tried deriving something from the density property expressed as:
$$ \bigcap\limits_{\substack{I \subseteq T \\ Spec_{max}(R) \cap V(I) \subseteq V(T)}} V(T) = V(I) = V(\sqrt{I}) \, $$
but this expression doesn't seem to help. So how is this implication proven?
I figured out the other implication too, it was almost there but I didn't realise it: if $Spec_{max}(R)\cap V(I)$ is dense in $V(I)$:
$$ \bigcap\limits_{\substack{I \subseteq T \\ Spec_{max}(R) \cap V(I) \subseteq V(T)}} V(T) = V(I) \, $$
The intersection on the LHS is necessarily contained in $Spec_{max}(R)\cap V(I)$, so we get that $V(I) \subseteq Spec_{max}(R) \cap V(I)$, so going to intersection of the primes in each set (which reverses inclusion) we get
$$ \bigcap\limits_{\substack{\mathfrak{m} \in Spec_{max}(R) \\ I \subseteq \mathfrak{m} }} \subseteq \sqrt{I} $$
Since we always have the other inclusion, we get equality.