# Minimal number of generators of kernel of a matrix of polynomials over $\mathbb Q$ and $\mathbb Z$

$$\DeclareMathOperator\im{im}$$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!