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.


  • $A$ 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!

  • 3
    $\begingroup$ I seem to be making this comment a lot, but if a book has two authors, then it is not appropriate to refer to it just by using one of their names, even if one is more familiar to you than the other. In this case, the book should be referred to as Atiyah--Macdonald. $\endgroup$
    – user64687
    Oct 27, 2013 at 20:36

3 Answers 3


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.


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 $b \in \bigcap_{P \in S_a^c} P$.


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 $ab \not \in N$.


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