Irreducible components in the spectrum of a ring

I have a question concerning page 43 of this book. In Corollary 2.7 it says that the map $\mathfrak{p}\mapsto \overline{\{\mathfrak{p}\}}$ is a bijection from Spec($A$) onto the sets of closed irreducible subsets of Spec($A$). It further says that minimal prime ideals of $A$ correspond to the irreducible components of Spec($A$). I don't understand this point. By the first part $\overline{\{0\}}=Spec(A)$ should be irreducible. So $Spec(A)$ has only one irreducible component. And this component corresponds to the zero ideal and not the minimal prime ideals. So what did I get wrong?

• $(0)$ is not always in $\operatorname {Spec} A$ ... – PVAL-inactive Jul 7 '14 at 21:15
• Thx. I forgot that (0) is not always prime. So if (0) is prime then I'm right and Spec(A) is irreducible and has therefore only one irreducible component. Otherwise there might be more such components. – principal-ideal-domain Jul 7 '14 at 21:24
• @PVAL: This is an answer. Please add it as an answer so that the question can be removed from the list of unanswered questions ;). – Martin Brandenburg Jul 8 '14 at 6:11

$(0)$ is not always in $\operatorname {Spec} A$ is where your reasoning went wrong. One can show rather straightforwardly that $\operatorname {Spec} A$ is irreducible iff the the nilradical is prime (an exercise in Ch. 1 of Atiyah and Macdonald's commutative algebra book) Note there is a natural homeomorphism induced by the projection map between $\operatorname {Spec} A/I$ and the closed subset of $\operatorname {Spec} A$, $V(I)= \{p\in \operatorname {Spec} A\ | \ I\subseteq p\}$, so this deals with the irreducible components question as well.