Spectrum of a ring is irreducible if and only if nilradical is prime (Atiyah-Macdonald, Exercise 1.19)

Question

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!

Answer 1

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.

Answer 2

We suppose that the nilradical is not prime and show that $\operatorname{Spec} A$ is reducible.

Let $\mathcal{N}$ be the nilradical of $A$. Suppose that $\mathcal{N}$ is not prime. Then there exist elements $a,b \in A$ such that $a,b \not\in \mathcal{N}$ but $ab \in \mathcal{N}$. Recall that $\operatorname{Spec} A = V(\mathcal{N})$ and $\mathcal{N} = \bigcap_{P \in \operatorname{Spec} A} P$. Next, define the sets $S_a = \left\{P \in \operatorname{Spec} A: \, a \in P \right\}$ and $S_b = \left\{P \in \operatorname{Spec} A: \, b \in P \right\}$. Notice that $S_a$ is non-empty (otherwise $b \in \mathcal{N}$) and proper subset of $\operatorname{Spec} A$ (otherwise $a \in \mathcal{N}$). Similarly for $S_b$. Writing $\mathcal{N} = \left(\bigcap_{P \in S_a} P \right) \cap \left(\bigcap_{P \in S_a^c} P\right)$, we have that $\operatorname{Spec}A = V(\mathcal{N}) = V\left(\bigcap_{P \in S_a} P \right) \cup V\left(\bigcap_{P \in S_a^c} P \right)$. It remains to show that $V\left(\bigcap_{P \in S_a} P \right)$ and $V\left(\bigcap_{P \in S_a^c} P \right)$ are proper subsets of $\operatorname{Spec}A$. Suppose that $\operatorname{Spec}A = V\left(\bigcap_{P \in S_a} P \right)$. Pick $P \in S_a^c$, then $P \in V\left(\bigcap_{P \in S_a} P \right)$ and so $a \in P$, contradiction. Next, suppose $\operatorname{Spec}A = V\left(\bigcap_{P \in S_a^c} P \right)$. Pick $Q \in S_b^c$. Then $Q \in V\left(\bigcap_{P \in S_a^c} P \right)$. But this is a contradiction, since $S_a^c \subset S_b$, and $b \in \bigcap_{P \in S_a^c} P$.

Answer 3

Let us denote the nilradical by $N$. $N= \cap P$. Let us assume N is not a prime ideal for X an irreducible space. So there must exist a and b in A, such that $a, b \not \in N$ but $ab\in N$. Then $ab \in P$ for all prime ideal P in A.

Now consider $X_{a} \cap X_{b}$, which is non-empty, by the property of irreducible space. Say some prime ideal, $P_{k} \in X_{a} \cap X_{b}$, $P_{k} \not \in (X_{a} \cap X_{b})^{c}= V(a) \cup V(b)$. That means $P_{k}$ is a prime ideal that do not contain a as well as b. That imply $ab \not \in P_{k}$. Which in turn shows $ab \not \in N$.

Answer 4

Here's another way to see this. Using the notation in Zach's answer, the points of $D(f)$ are in bijective correspondance with the prime ideals in the localization of $A$ at $f$, so we can identify $D(f)$ with $\textrm{Spec}(A_f)$.

By Zorn's lemma, any nonzero ring has a maximal ideal. Therefore $\textrm{Spec}(A_f)$ is empty if and only if $0\in \{f^n\}_{n\geq 0}$ if and only if $f\in\eta$.

We know that $\textrm{Spec}(A_f)\cap\textrm{Spec}(A_g)=\textrm{Spec}(A_{fg})$. By the previous paragraph, the two open sets $\textrm{Spec}(A_f)$ and $\textrm{Spec}(A_g)$ being nonempty is equivalent to $f,g\notin \eta$. Also, $\textrm{Spec}(A_{fg})$ being nonempty is equivalent to $fg\notin\eta$. This proves the equivalence because now we see that $\eta$ being prime is exactly the same statement as $\textrm{Spec}(A)$ being irreducible.