Max-Spec vs Spec

This is an article on why using Max-Spec and Spec do not matter much if the ring $A$ is a Jacobson ring. 

Definition. A ring $A$ is a Jacobson ring if every prime ideal of $A$ is an intersection of maximal ideals. In other words,

\[ \mathrm{Nil}(A) = \bigcap_{\mathfrak{p}} \mathfrak{p} = \bigcap_{\mathfrak{m}} \mathfrak{m} = \mathrm{Jac}(A) \] 

Proposition. Let $X =\mathrm{Spec}(A)$ and $X_0 = \mathrm{MaxSpec}(A)$, i.e. the set of closed points of $X$. The $X_0$ is a very dense subset of $X$.

We show that some topological properties are preserved. Let's first describe what it means for $\mathrm{Spec}(A)$ to be disconnected. By this means that there exist disjoint closed subsets $V(I)$ and $V(J)$ that cover $\mathrm{Spec}(A)$. In other words,

1) $V((0))=\mathrm{Spec}(A) = V(I) \cup V(J) = V(I \cap J)$. 

2) $V((1)) = \varnothing = V(I) \cap V(J) = V(I + J)$.

This is same as saying that

1') $\sqrt{I \cap J} = \sqrt{(0)} = \mathrm{Nil}(A)$,

2') $\sqrt{I + J} = \sqrt{(1)} = (1)$

Again this is same as saying that

1'') $I \cap J \subset \mathrm{Nil}(A)$,

2'') $I+J = (1)$.

pf) Since $I \cap J \subset \sqrt{I \cap J}$,  1') implies 1''). Conversely, from 1''), we immediately get $\sqrt{I \cap J} \subset \mathrm{Nil}(A)$. If $a \in \mathrm{Nil}(A)$, then $a^n =0$ for some $n$. Then $a^n = 0 \in I \cap J \Rightarrow a \in \sqrt{I \cap J}$.

Similarly, 2'') implies 2') is clear. Suppose 2'). Then by definition $1^n = 1 \in I \cap J$. This shows that $I+J = (1)$.

We remark that 1) $\Leftrightarrow$ 1'') is actually easier. If $I \cap J \subset \mathrm{Nil}(A)$ if and only if $I \cap J \subset \mathfrak{p}$ for all $\mathfrak{p}$ if and only if $V(I \cap J) = \mathrm{Spec}(A)$.

From now, let $V_M(I)$ denote the set of maximal ideals containing $I$. Also, let $I_M(V)$ denote the ideal defined by $V$, i.e.

\[ I_M(V) = \bigcap_{\mathfrak{m} \in V} \mathfrak{m} \]

Similarly, let's think about what it means for $\mathrm{MaxSpec}(A)$ to be disconnected. We are not necessarily assuming that $A$ is a Jacobson ring yet. We have disjoint closed subsets $V_M(I)$ and $V_M(J)$ such that

1) $\mathrm{MaxSpec}(A) = V(I \cap J)$,

2) $V_M(I+J) = \varnothing$

This is same as

1') $I \cap J \subset \mathrm{Jac}(A)$,

2') $I+J = (1)$

For 1) $\Leftrightarrow$ 1') we can argue the same as 1) $\Leftrightarrow 1'')$ above. For 2), this is essentially Zorn's lemma. In particular, if $V_M(I+J) = \varnothing$ means that no maximal ideal contains $I+J$. The only ideal of $A$ satisfying this is the unit ideal $A$. Conversely, if $I+J=(1)$, then $V_M((1)) = \varnothing$.

Note that we assume that $A$ is a Jacobson ring. Then since $\mathrm{Nil}(A) = \mathrm{Rad}(A)$, we may conclude that $\mathrm{Spec}(A)$ is connected if and only if $\mathrm{MaxSpec}(A)$ is connected.

For irreducibility, we again describe what it means for $\mathrm{Spec}(A)$ to be reducible. There exist disjoint closed subsets $V(I)$ and $V(J)$ such that

1) $V(I) \cup V(J) = \mathrm{Spec}(A)$, i.e. $I \cap J \subset \mathrm{Nil}(A)$,

2) $V(I), V(J) \subsetneq \mathrm{Spec}(A)$, i.e. $I, J \subsetneq \mathrm{Nil}(A)$

Then again for a Jacobson ring $A$, $\mathrm{Spec}(A)$ is irreducible if and only if $\mathrm{MaxSpec}(A)$ is irreducible