I stumbled across the following assertion:
Let $A$ be a commutative ring and Spec($A$) given with the Zariski topology. In this topology all closed sets are of the form:
$V(\mathfrak{a}):=$ {$\mathfrak{p} \in$ Spec($A$) | $\mathfrak{a} \subset \mathfrak{p}$}.
What was claimed then is that
$V(\mathfrak{a})= \emptyset $ $\Rightarrow$ $\mathfrak{a}=$ ($1$) and
$V(\mathfrak{a})=$Spec($A$) $\Rightarrow $ $\mathfrak{a} \subset \mathfrak{N}$ where $\mathfrak{N}$ is the nil radical
How can one prove that?
Best Answer
For the first part use contra positive: Assume that $a \neq (1)$ i.e. that $a$ is a proper ideal of $A$, then you know that $a$ is contained in a maximal ideal $m \subseteq A$. Since maximal ideals are prime, $m \in V(a)$ which is thus non-empty.
For the second one, note that $V(a) = \text{Spec}(A)$ means that $a$ is contained in every prime ideal. Hence we need that $a \subseteq \cap_{p \subseteq A, \text{prime}} p$. Now recall that the right side is precisely the definition of the nil radical.