Does $\operatorname{length}(M/xM) \leq \operatorname{rank}(M) \cdot \operatorname{length}(R/(x))$ hold over non-integral rings $R$?

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$\DeclareMathOperator{\len}{length} \DeclareMathOperator{\rk}{rank}$In Eisenbud's book Commutative Algebra with a View towards Algebraic Geometry he says:

The basic result of this section expresses the length of $M/aM$ for certain $R$-modules $M$ in terms of the length of $R/(a)$ and an invariant of $M$. [...] For simplicity, and because it suffices for the application, we shall assume here that $R$ is a domain.

He then defines the rank of $M$ over $R$ as $\rk(M) = \dim_{K} K \otimes_R M$ where $K$ is the field of fractions of $R$. After that he goes on and gives the following result:

Lemma 11.12. Let $R$ be a one-dimensional Noetherian domain. If $M$ is a torsion-free $R$-module, then $$\len(M/xM) \leq \rk(M) \cdot \len(R/(x))$$ with equality if $M$ is finitely generated as an $R$-module.

Now I am wondering if there holds a similar result in the case when $R$ has zero divisors. But then, if I recall correctly, the rank is only locally well-defined.

So, can anyone give me references where this is handled in the non-integral case? Thank you very much in advance!