# Is the annihilator of a minimal prime ideal principal?

My setup is as follows: $$X$$ is a projective, reduced curve (which is not integral) with a finite morphism onto $$\mathbb{P}_k^1$$. $$\DeclareMathOperator{\Ann}{Ann}$$ Let $$R$$ be a coordinate ring of $$X$$ which is finite free over $$k[x]$$ (since $$X$$ has more than one irreducible component, $$R$$ has at least two minimal prime ideals). Let $$P$$ be a minimal prime of $$R$$ corresponding to an irreducible component of $$X$$.

Do we always have that $$\Ann(P)$$ is pincipal?

What I tried:

• Every example I was able to come up satisfied the above property. Hence I did not find any counter-example.

• There is some $$b \in R$$ such that $$P = \Ann(b)$$ and hence $$b \in \Ann(\Ann(b))$$. Thus a necessary condition is that such a generator $$a \in R$$ of $$\Ann(P)$$ must divide every $$b \in R$$ that satisfies $$\Ann(b) = P$$. Since $$\Ann(b) = \Ann(bf)$$ for all $$f \in k[x]$$ (since $$R$$ is torsion-free over $$k[x]$$) we may assume that no element $$f \in k[x]$$ divides $$b$$ in $$R$$. That's where I am stuck at.

I am grateful for any kind of help, counter-example or hints.

This is false. To see why, consider the following lemma.

Lemma. Let $$R$$ be a Noetherian ring with exactly two minimal primes $$\mathfrak p$$ and $$\mathfrak q$$ such that $$\mathfrak p \mathfrak q = 0$$. Then $$\operatorname{Ann}(\mathfrak p) = \mathfrak q$$.

The assumption is in particular satisfied if $$\mathfrak p \cap \mathfrak q = 0$$, which is equivalent to $$R$$ being reduced.

Proof. Clearly $$\mathfrak q \subseteq \operatorname{Ann}(\mathfrak p)$$, since $$\mathfrak p \mathfrak q = 0$$. Since $$\mathfrak p \not\subseteq \mathfrak q$$, we have $$\mathfrak p_{\mathfrak q} = R_{\mathfrak q}$$, so any element killing $$\mathfrak p$$ better be in $$\mathfrak q$$. (See also Tag 00L2.) $$\square$$

Thus, it suffices to construct such a ring $$R$$ where $$\mathfrak q$$ is not principal. Basically anything you write down will work.

Example. Let $$E$$ be an elliptic curve over an algebraically closed field $$k$$, and let $$p \in E$$ be a closed point. Glue two copies of $$E$$ at $$p$$, i.e. consider the union $$X = (E \times p) \cup (p \times E) \subseteq E \times E$$. This admits a finite flat map to $$\mathbb P^1$$ given by $$E \times E \to E \to \mathbb P^1$$ where the first map is $$(x,y) \mapsto x+y$$ and the second is any nonconstant map.

Removing any other point $$q$$ gives an affine open $$((E \setminus q) \times p) \cup (p \times (E \setminus q)) \subseteq X$$, and on its coordinate ring $$R$$ we have two minimal prime ideals $$\mathfrak p$$ and $$\mathfrak q$$ corresponding to the components $$(E \setminus q) \times p$$ and $$p \times (E \setminus q)$$ respectively.

If $$\mathfrak q$$ were principal, then the same is true for its restriction to $$R/\mathfrak p = \Gamma(E\setminus q,\mathcal O)$$. But the map \begin{align} \operatorname{Pic}(E \setminus q) &\to E(k) = \operatorname{Pic}^0(E)\\ r &\mapsto r-q \end{align} is an isomorphism, and if $$p \in E \setminus q$$ were principal this implies that $$p - q = 0$$, which is absurd. $$\square$$