# Proof regarding $\operatorname{Spec}A$ being irreducible (Atiyah-MacDonald).

I'm working on exercise 19 on chapter 1 of Atiyah-MacDonald's Commutative Algebra. The question asks to prove that $$X=\operatorname{Spec}A$$ is irreducible (any intersection of open sets is non-empty) if and only if the nilradical $$\mathfrak{N}$$ is a prime ideal.

To prove the forward implication, suppose $$X$$ is irreducible, and that $$ab\in\mathfrak{N}$$. Since $$X$$ is irreducible, $$(X-V(a))\cap(X-V(b))\neq\varnothing$$ where $$V(E) = \{\mathfrak{p}\in X : E\subset\mathfrak{p}\}$$, thus there is some prime ideal $$\mathfrak{p}$$ containing neither $$a$$ nor $$b$$ (since $$\mathfrak{p}\in X-V(a)\iff\mathfrak{p}\not\ni a$$), but then $$ab \in \mathfrak{N}\subset\mathfrak{p}$$, but $$\mathfrak{p}$$ is prime, so we can't have $$ab\in\mathfrak{p}$$ while neither $$a$$ nor $$b$$ belong to $$\mathfrak{p}$$.

What went wrong? The use of the nilradical here isn't important either. I believe the assertion is true, but why am I running into an evident contradiction before the proof even starts? I didn't start by assuming anything contradictory to start.

Edit: I think I see the error, I erroneously assumed that $$X-V(a)$$ is necessarily non-empty. Assuming that both $$X-V(a)$$ and $$X-V(b)$$ are non-empty yields a contradiction. WLOG, say $$X-V(a)$$ is empty, that is $$X=V(a)$$ meaning every prime ideal must contain $$a$$, thus $$a$$ is in the nilradical which proves the nilradical is prime. Is this right?

• I don't know the answer for what you're asking (I haven't much looked at this topic since Fall 1980, when I took a course using Atiyah/Macdonald), but you might be interested in Errata for Atiyah-Macdonald. (Note: The second author's name uses a lower-case 'd'.) Commented Jul 30, 2019 at 6:08

The nilradical $$\mathfrak{N}$$ of a commutative ring is the intersection of all the prime ideals in the ring.
Suppose first that $$\mathrm{Spec}(A)$$ is irreducible. Now let $$ab \in \mathfrak{N}$$. This means that every prime ideal of $$A$$ contains $$ab$$. Therefore, $$\mathrm{Spec}(A) = V(ab) = V(a) \cup V(b)$$. Since $$\mathrm{Spec}(A)$$ is irreducbile, this means either $$\mathrm{Spec}(A) \subset V(a)$$ or $$\mathrm{Spec}(A) \subset V(b)$$. Therefore, either $$a$$ or $$b$$ is contained in all the primes ideals of $$\mathrm{Spec}(A)$$, that is, either $$a \in \mathfrak{N}$$ or $$b \in \mathfrak{N}$$. I think it's more 'natural' to work with closed subsets as the Zariski topology is defined in terms of closed subsets.
Suppose $$\mathfrak{N}$$ is a prime ideal of $$A$$. Now $$\mathrm{Spec}(A) = V(\mathfrak{N})$$. Any closed subset $$V(I)$$, where $$I$$ is an ideal of $$A$$, contains $$\mathfrak{N}$$ if and only if $$I$$ is contained in all the prime ideals of $$A$$, that is, $$I \subset \mathfrak{N}$$, which implies $$V(\mathfrak{N})\subset V(I)$$. So $$V(\mathfrak{N})$$ is the smallest closed subset containing $$\mathfrak{N}$$. So $$\mathrm{Spec}(A) = \overline{ \{ \mathfrak{N} \}}$$ is the closure of a single point, therefore it is irreducible.
More generally, for any subset $$Y$$ of $$\mathrm{Spec}(A)$$, you can define $$I(Y) := \bigcap_{\mathfrak{p} \in Y} \mathfrak{p}$$. Similar arguments as above also show that $$Y$$ is irreducible if and only if $$I(Y)$$ is a prime ideal.