Sufficient conditions for two ideals with the same zero set to be equal Let $R$ be a one-dimensional noetherian ring finite over $k[x]$. Let $I$, $J$ be two integral ideals of $R$ with $J \subseteq I$ and $I$ invertible. Moreover, assume that $V(I) = V(J)$.

What are sufficient conditions for $I = J$?


What I tried:
From $V(I) = V(J)$ one concludes $\operatorname{Rad}(J) = \operatorname{Rad}(I)$ and thus there is some $n \in \mathbb{N}$ such that $$I^n \subseteq J \subseteq I.$$
Until now I did not use the 'invertible'-assumption on $I$. From that it follows at least that $V(I)$ and a fortiori $V(J)$ is finite. It also follows that $I \cap k[x] = fk[x] \neq 0$.
I appreciate any kind of help or idea.

Edit: You may also assume that $f \in J$ and hence $J \cap k[x] = I \cap k[x]$.
 A: There is a possibility to characterize elements that generate $I$ in terms of having zeros relative to $I$ as follows:
We say that $f \in I$ has a zero $\mathfrak{m} \in \operatorname{Spec}(R)$ relative to $I$ if $f \in \mathfrak{m}I$.
Then we obtain the following 

Lemma:
  Let $I \subset R$ be an invertible ideal and let $f_1,\ldots,f_n \in I$. Then
  $$\langle f_1,\ldots,f_n\rangle = I \quad\quad \Leftrightarrow\quad\quad f_1,\ldots,f_n\ \ \text{ do not share a common zero relative to }I.$$

$\ \ $Proof:
Let $\langle f_1,\ldots,f_n\rangle = I$ and assume there is $\mathfrak{m}$ such that $f_1,\ldots,f_n \in \mathfrak{m}I$, then $R = \langle f_1,\ldots,f_n\rangle I^{-1} \subset \mathfrak{m}$; a contradiction!
Conversely, let $f_1,\ldots,f_n$ don't have any common zero relative to $I$, that is for all $\mathfrak{m} \in \operatorname{Spec}(R)$ we have $$\langle f_1,\ldots,f_n\rangle \not \subset \mathfrak{m}I \quad\Rightarrow\quad \langle f_1,\ldots,f_n\rangle I^{-1} \not \subset \mathfrak{m}$$
and thus $\langle f_1,\ldots,f_n\rangle I^{-1}$ is an integral ideal not contained in any prime ideal and therefore it must be equal to $R$, hence $\langle f_1,\ldots,f_n\rangle = I$.
