Along a finite ring extension, the restriction of invertible ideals cannot be zero

Question

Let $S := k[x] \subseteq R$ be a finite ring extension where $k$ is a field. Moreover, let $R$ be one-dimensional, noetherian and reduced. Let $I$ be an invertible ideal of $R$.

Do we have $I \cap S \neq \{0\}$?

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My attempt:

The answer is yes if $I$ contains (properly) a prime ideal $P \subsetneq I$ since then by incomparability we have $0 \subseteq P \cap S \subsetneq I \cap S$.

But do we know that invertible ideals always contain a prime?

Answer 1

Since $I$ is invertible, it contains at least one regular element. That is $I \not\subseteq P$ for any minimal prime ideal $P$ of $R$.

Since $R$ is of dimension one, any prime ideal chain $P_0 \subsetneq P_1$ provides that $P_0$ is minimal. Thus $$\dim R/I = 0.$$

Now I use the following theorem:

Theorem: [A.8 in Bruns-Herzog, Cohen-Macaulay Rings]

Let $S \subseteq R$ be an integral ring extension of noetherian rings and $I \subsetneq R$ be a proper ideal. Then $$\dim R/I = \dim S/(I \cap S).$$

Now this implies $\dim S/(I \cap S) = 0$, in particular $I \cap S \neq 0$ (since $\dim S = 1$).