Vtyjkyuk

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 $operatornameRad(J) = operatornameRad(I)$ and thus there is some $n in mathbbN$ 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 $J cap k[x] = I cap k[x]$.

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 $mathfrakm in operatornameSpec(R)$ relative to $I$ if $f in mathfrakmI$.

Then we obtain the following

Lemma:
Let $I subset R$ be an invertible ideal and let $f_1,ldots,f_n in I$. Then
$$langle f_1,ldots,f_nrangle = I quadquad Leftrightarrowquadquad f_1,ldots,f_n text do not share a common zero relative to I.$$

$ $Proof:
Let $langle f_1,ldots,f_nrangle = I$ and assume there is $mathfrakm$ such that $f_1,ldots,f_n in mathfrakmI$, then $R = langle f_1,ldots,f_nrangle I^-1 subset mathfrakm$; a contradiction!

Conversely, let $f_1,ldots,f_n$ don't have any common zero relative to $I$, that is for all $mathfrakm in operatornameSpec(R)$ we have $$langle f_1,ldots,f_nrangle not subset mathfrakmI quadRightarrowquad langle f_1,ldots,f_nrangle I^-1 not subset mathfrakm$$
and thus $langle f_1,ldots,f_nrangle I^-1$ is an integral ideal not contained in any prime ideal and therefore it must be equal to $R$, hence $langle f_1,ldots,f_nrangle = I$.