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.



Is the annihilator of a minimal prime ideal principal?
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](https://stacks.math.columbia.edu/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. $\square$


