How does one show that the prime spectrum of a domain is irreducible in the Zariski topology?
Why is the prime spectrum of a domain irreducible in the Zariski topology
6
$\begingroup$
algebraic-geometry
commutative-algebra
-
3$0$ is the only minimal prime of $R$. Suppose, $R=V(I)\cup V(J)$. If $I,J$ are nonzero, then the zero ideal is in neither closed set. So, one of them must be zero, in which case, the corresponding closed set is the entire spectrum. – 2011-02-06
-
0A little typo: it's $\text{Spec } R=V(I)\cup V(J)$. – 2011-02-07
-
0The set $\{x\}$ is closed (we say that $x$ is a "closed point") in $\mathrm{Spec} (A) \Leftrightarrow $x is maximal. – 2014-06-08