$\DeclareMathOperator\Spec{Spec}$Let $R$ be a commutative ring with $1$ and $\Spec(R)$ be the set of all prime ideals of $R$ with the Zariski topology. It is well-known that if $P$ and $Q$ are two prime ideals of $R$ with $P\subseteq Q$, then there exists a path from $P$ to $Q$ in $\Spec(R)$. Now let $\frak{m}_1$ and $\frak{m}_2$ be two maximal ideals of $R$. I am looking for equivalent conditions under which there exists a path between $\frak{m}_1$ and $\frak{m}_2$ in $\Spec(R)$. Recall a path from a point $x$ to a point $y$ in a topological space $X$ is a continuous function $f$ from the unit interval $[0,1]$ to $X$ with $f(0) = x$ and $f(1) = y$.
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2$\begingroup$ As we can pass through the generic point, there exist a path between any two closed points of the same irreducible component. Moreover if two irreducible component share a point we can use it to pass from one to the other. Hence, at least if the number of irreducible components is finite (for example if R is noetherian), such a path exist iff the two closed points are in the same connected component. $\endgroup$– Antoine LabelleCommented Dec 19, 2020 at 20:52
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$\begingroup$ More generally, if we consider the graph whose vertices are the irreducible components and two components are connected by an edge iff they intersect, then two points of the spectrum are path connected iff their irreducible components are in the same connected component of the graph. $\endgroup$– Antoine LabelleCommented Dec 20, 2020 at 1:25
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