Irreducible components in the spectrum of a ring

Question

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?

Answer 1

$(0)$ 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 (Exercise 19 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.

Just friendly advice, I would recommend you do most of the exercises in a book like Atiyah and Macdonald before seriously attempting to learn about schemes.