Geometric interpretation of the annihilator of a zero divisor

Let me recap the setup here just so all notations are on the same page. Let $R$ be a reduced Noetherian ring and $f$ a zero divisor (also assume $f\ne 0$ because this is trivial). Because $R$ is Noetherian, it has finitely many minimal prime ideals, say $p_1, p_2, \dots, p_n$. Geometrically, the irreducible components of $\mathrm{Spec} R$ are exactly the closed sets $Z_i = V(p_i)$. A neat observation here is that since $R$ is reduced, the intersection of all these minimal primes is $(0)$ (why?).

Anyways, let us do some commutative algebra first. Partition the minimal primes into two sets - $S$ which are the minimal primes containing $f$, and $T$ the rest of them. Note that since $f$ is a zero divisor, $S$ is non-empty (why? Hint: you will need $f$ being a zero divisor crucially here. Try to prove that the set of zero divisors is the union of minimal prime ideals). The claim is now that in any reduced Noetherian ring

$$\mathrm{Ann}(f) = \bigcap_{p_j \in T} p_j$$

The proof is simple. Suppose $g \in \mathrm{Ann}(f)$. Then $fg = 0$. Since $0$ is contained in every prime ideal, $fg \in p_i$ for all $i = 1, \dots, n$. For the primes $p_j \in T$, we know $f \notin p_j$. Because $p_j$ is a prime ideal, $fg \in p_j$ implies $g \in p_j$. Therefore, $g$ is in the intersection of all primes in $T$. Conversely, suppose $g \in \bigcap_{p_j \in T} p_j$. We want to show $fg = 0$. To do this, it suffices to show $fg$ is in every minimal prime $p_1, \dots, p_n$, because their intersection is $(0)$. If $p_i \in S$, then $f \in p_i$, so $fg \in p_i$; if $p_j \in T$, then $g \in p_j$ by our assumption, so $fg \in p_j$. Since $fg$ is in every minimal prime, $fg \in (0)$, meaning $fg = 0$. Thus, $g \in \mathrm{Ann}(f)$.

You might notice that the assumption that $f$ is a zero divisor was only used to guarantee that $S$ is non-empty (meaning $f$ actually vanishes on at least one component). The algebraic result, however, doesn't actually require it. If $f$ is a non-zero-divisor, $S$ is simply empty, hence $T$ contains all minimal primes, and thus $\mathrm{Ann}(f) = (0)$. In fact we can get rid of the condition of being Noetherian too, that just would mean we may not have finitely many minimal primes (but that's all okay).

As a very nice corollary of this, we get the following. In a reduced ring, the annihilator of any element is always a radical ideal because it is an intersection of prime ideals.

What does this really mean geometrically? Well, let's assume again that $R$ is Noetherian reduced so we have finitely many components, and let $X = \mathrm{Spec} R$. Since $R$ is reduced, we can view elements of $R$ as functions on $X$, where the space decomposes into its irreducible components $X = \bigcup_{i=1}^n Z_i$. The algebraic condition $fg = 0$ translates to the geometric statement that $V(f) \cup V(g) = X$. Since $X$ is the union of its irreducible components, the vanishing loci of $f$ and $g$ must cover every $Z_i$. By definition, $f$ vanishes identically on the components $Z_i$ for $p_i \in S$, and fails to vanish identically on the components $Z_j$ (that is, $f$ is non-zero on a dense open subset of $Z_j$) for $p_j \in T$. To satisfy $V(f) \cup V(g) = X$, the function $g$ is forced to cover the gap and hence must vanish entirely on the remaining components. Geometrically, this means $V(g) \supseteq \bigcup_{p_j \in T} Z_j$. Translating this back to algebra, $g$ must lie in the radical ideal defining this union. Therefore, $\mathrm{Ann}(f)$ is precisely the intersection $\bigcap_{p_j \in T} p_j$, representing the ideal of functions forced to vanish on all irreducible components of $\mathrm{Spec} R$ that are not contained in $V(f)$.