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.