$\DeclareMathOperator\im{im}$I have asked some related questions before on math.SE here and on MathOverflow here (answered here). This post is self-contained.
Let $R' = \mathbb Q[x_1,\dotsc,x_n]$, and let $M'$ be a submodule of a free $R'$-module, $R'^k$. Since $R'$ is Noetherian, $M'$ is finitely generated. Choose a minimal generating set of $M'$ with least cardinality, say $l$, and let $A$ be the $k\times l$ matrix whose columns are these generators of $M'$. Without loss of generality, we can choose these generators to have polynomials with only integer coefficients. Let $\varphi':R'^l \rightarrow R'^k$ be the map associated with the matrix $A$.
Let $R = \mathbb Z[x_1,\dotsc,x_n]$, and consider the map $\varphi:R^l \rightarrow R^k$ associated with the same matrix $A$.
Question: Is $\mu_R(\ker \varphi) = \mu_{R'}(\ker \varphi')$?
Here $\mu_S(N)$ denotes the minimal number of generators of $N$, an $S$-module. Since both $R$ and $R'$ are Noetherian, the above numbers are finite.
Attempt: When $n\le 1$, $R'$ is a PID, so $M'$ is free, which means $\ker \varphi' = 0$. Then $\ker \varphi = 0$ because both $\varphi$ and $\varphi'$ are associated with the same matrix $A$. Hence, when $n\le 1$, the answer is yes. I am not sure how to proceed for $n\ge 2$. In all the examples I have seen so far, both kernels had the same minimal number of generators (some examples can be found in the posts linked above).
Motivation: My original motivation was to construct a free resolution of $M:=\im \varphi$ using the free resolution of $M' = \im \varphi'$. If the answer to the above question is yes, then we can construct a matrix $B$ whose columns are minimal number of generators of $\ker \varphi$, and hence also of $\ker \varphi'$. By induction, we can construct a free resolution of both $M$ and $M'$ using the same matrices!
