Let $X \subset \mathbb{C}^n$ be an irreducible affine algebraic curve with coordinate ring $$\mathbb{C}[X] = \mathbb{C}[z_1, \ldots, z_n] / (f_1, \ldots, f_m ) $$ with each $f_i \in \mathbb{Z}[z_1, \ldots, z_n]$. Define $R[X] = R[z_1, \ldots, z_n] / (f_1, \ldots, f_m )$ for any subring $R \subseteq \mathbb{C}$ containing $\mathbb{Z}$.

Now, if $f\in \mathbb{C}[X]$ can be represented by a polynomial in with integer coefficients, then $\mathbb{C}[X]$ (resp. $R[X]$) can be regarded as a module over $\mathbb{C}[f]$ (resp. $R[f]$), where $\mathbb{C}[f]$ is the $\mathbb{C}$-algebra generated by $f$ in $\mathbb{C}[X]$. Note that $\mathbb{C}[f]$ is the image of the induced homomorphims $f^\ast \colon \mathbb{C}[\mathbb{A}^1] \to \mathbb{C}[X]$.

Suppose $\mathbb{C}[X]$ and $\mathbb{Z}[X]$ are finitely generated over $\mathbb{C}[f]$ and $\mathbb{Z}[f]$. For a ring $R$ as above, let $\mu_R(f)$ denote the minimal number of generators for $R[X]$ over $R[f]$. Then $$\mu_\mathbb{Z}(f) \geq \mu_\mathbb{Q} ( f) = \mathrm{rank}_{\mathbb{Q}[f]} \mathbb{Q}[X]$$ where the equality follows from the fact that $\mathbb{Q}[X]$ is a finitely generated torsion free module over a PID. The inequality can be strict, however.

What is the appropriate geometric interpretation of the case when $\mu_\mathbb{Z} ( f) > \mu_\mathbb{Q} (f)$? What does this say about $f$ when thinking of it as a regular function on $X$? Does this inequality tell us anything geometric about $X$?

From a few Overflow posts (like this and this), Corollary A3.3 in Eisenbud's Commutative Algebra, and glancing through Hartshorne, it appears that this has to do with algebraic vector bundles on $\mathbb{A}^1$ and/or coherent sheaves, but I can't figure out how the regular function $f$ fits into the picture or what intersecting down to $\mathbb{Z}$ does specifically.

Note: this is my first Overflow question and I previously asked this question on StackExchange where it was upvoted a few times, but had no responses. If it is an inappropriate question for this site, please let me know and I will delete it.