Can anyone help me with this exercise, please?
A topological space $X$ is said to be irreducible if $X\neq\emptyset$ and if every pair of non-empty open sets in $X$ intersect, or equivalently, if every non-empty open set is dense in $X$. Show that $\text{Spec}(A)$ is irreducible if and only if the nilradical of $A$ is a prime ideal.
Notation:
- $A$ is a commutative ring with $1$ (not necessarily $1\ne0$)
- $\eta= \text{nilradical of $A$ }= \bigcap\limits_{\mathscr{p}\text{ prime}}\mathscr{p}=\{a\in A:\text{$a$ is nilpotent}\}$
- $\text{Spec}(A)=\{p\subset A:\text{$p$ prime}\}$, and the topology is such that $V(E)=\{p\subset A\text{ prime}:E\subset A\}$ is a basis for closed sets, for all subset $E\subset A$ (we can show that the complementar of these sets form a basis for open sets)
If the nilradical $\eta=\mathscr{p}$ is prime, then every non-empty closed set $V(E)$ satisfy: "$p\in V(E)\implies V(E)=\text{Spec}(A)$" (since every prime contains $\eta=p$), hence, every non-empty open set contains $p$, so $\text{Spec}(A)$ is irreducible.
The conversely is the problem…
A previous exercise showed that there exists minimal prime ideals in every ring $A$.
I assumed that $\eta$ is not a prime ideal, hence there exists at least two distinct minimal prime ideals. So, let $p$ be a minimal prime ideal and $E=\bigcap\{q\subset A:\text{$q$ is prime minimal, $q\ne p$}\}$. If there are a finite number of minimal prime ideals (for example, if $A$ is Noetherian), then the complementar of $V(E)$ is contained in $V(p)$ (since if a finite intersection of prime ideals is contained in any ideal $I$, then at least one of these prime ideal is contained in $I$), hence, $\text{Spec}(A)$ is not irreducible.
But this argument seems not to work for general rings…
Any help will be appreciated!
Thanks!
Best Answer
I think the open sets definition of irreducibility is easier to work with. You should show these useful facts about $Spec(A)$: the sets $D(f) = \{\mathfrak{p} \in Spec(A) : f \notin \mathfrak{p}\}$ form a basis of the topology of the spectrum, and $D(f) \cap D(g) = D(fg)$. Then we suppose that $f \notin Nil(A)$ and $g \notin Nil(A)$. This means that $D(f)$ and $D(g)$ are then nonempty open sets, and so if $Spec(A)$ is irreducible, their intersection $D(fg)$ is nonempty.