Constructing a free resolution of a $\mathbb Z[x_1,\dotsc,x_n]$-module using a related free resolution of a $\mathbb Q[x_1,\dotsc,x_n]$-module

Question

I have asked a related question on math.SE here, but the notation is a bit different.

As the title says, I am interested in constructing a finite free resolution of a $\mathbb Z[x_1,\dotsc,x_n]$-module using a related finite free resolution of a $\mathbb Q[x_1,\dotsc,x_n]$-module. Let $R=\mathbb Z[x_1,\dotsc,x_n]$ and $R' = \mathbb Q[x_1,\dotsc,x_n]$. Let $M'$ be a submodule of $R'^k$. Since $R'$ is Noetherian, $M'$ is finitely generated. Consider a finite free resolution of $M'$: $$ 0 \longrightarrow R'^{k_l} \overset{A_l}\longrightarrow R'^{k_{l-1}} \overset{A_{l-1}}\longrightarrow \cdots \overset{A_1}\longrightarrow R'^{k_0}\overset{A_0}\longrightarrow R'^k~. $$ Here, $l \le n$ by Hilbert's syzygy theorem, $M' = \operatorname{im} A_0$, and each matrix $A_i$ can be chosen such that its elements are polynomials with integer coefficients. Moreover, at every step in constructing this free resolution, I choose the minimal generating set of least cardinality.

Let $M$ be the submodule of $R^k$ generated by the columns of $A_0$. Since $R$ is also Noetherian, $M$ is also finitely generated and has a finite free resolution of length at most $n+1$ (this is proved in Gamanda, Lombardi, Neuwirth, and Yengui - The syzygy theorem for Bézout rings).

Goal: I want to construct a free resolution of $M$ using the above free resolution of $M'$.

Consider the complex $$ 0 \longrightarrow R^{k_l} \overset{A_l}\longrightarrow R^{k_{l-1}} \overset{A_{l-1}}\longrightarrow \cdots \overset{A_1}\longrightarrow R^{k_0}\overset{A_0}\longrightarrow R^k~. $$

Question 1: Is this complex exact? In other words, is it a free resolution of $M$?

For example, say $n=2$, and $M'$ is the ideal $(x_1,x_2)$ in $R'$. Then, the matrices $A_0 = \begin{pmatrix} x_1 & x_2 \end{pmatrix}$, and $A_1 = \begin{pmatrix} x_2 \\ -x_1 \end{pmatrix}$ give a free resolution of $M'$. In fact, they also give a free resolution of $M$, which is the ideal $(x_1,x_2)$ in $R$. I have tried several other examples and it always worked in the same way. (Most of the examples I worked out come from my research in physics. And for several reasons associated with my research, I believe that the answer to the above question is yes.)

Attempt 1: To prove that the above complex is exact, I thought I would use the following result of Buchsbaum and Eisenbud - What makes a complex exact? for commutative Noetherian rings. It says the above complex is exact if and only if, for $i=0,\dotsc,l$,

1. $r_i + r_{i+1} = k_i$, where $r_i = \operatorname{rk} A_i$ and $A_{l+1} = 0$, and
2. depth of the ideal $I(A_i)$ generated by the $r_i \times r_i$ minors of $A_i$ is at least $i+1$ (not $i$, because of the way I indexed the complex).

The first condition is easy because a minor of $A_i$ is nonzero over $R'$ if and only if it is nonzero over $R$. However, I am stuck at the second condition. Since the first complex over $R'$ is exact, for each $i$, there is an $R'$-regular sequence $(f_1,\dotsc,f_{i+1})$ of length $i+1$ in $I(A_i;R')$ (here, $I(A_i;R')$ is the ideal generated by $r_i\times r_i$ minors of $A_i$ in $R'$). Without loss of generality, we can assume that each $f_a$, for $a=1,\dotsc,i+1$, is a polynomial with integer coefficients. Then, it is clear that $(f_1,\dotsc,f_{i+1})$ is an $R$-regular sequence, but

Question 2: is $(f_1,\ldots,f_{i+1})$ an $R$-regular sequence in $I(A_i;R)$?

While each $f_a$ is an $R'$-linear combination of $r_i\times r_i$ minors of $A_i$, it may not be an $R$-linear combination. In fact, there may not be any $R$-linear combinations of these minors generating the above regular sequence. This is where I am stuck.

It is clear that there is a least positive integer $m_a$ such that $m_a f_a$ is an $R$-linear combination of the $r_i \times r_i$ minors of $A_i$. If $\gcd(m_a,m_b) = 1$ for all $a\ne b$, then $(m_1 f_1,\dotsc,m_{i+1} f_{i+1})$ is an $R$-regular sequence in $I(A_i;R)$. However, I am not sure how to show that $\gcd(m_a,m_b) = 1$ in general.

I would like to know if the answer to question 1 is known, or if there is a different approach to settle it. I am also interested in any counterexamples (I tried constructing some counterexamples but failed so far). In particular, a counterexample where $\operatorname{im} A_1 = \ker A_0$ over $R'$ but $\operatorname{im} A_1 \subsetneq \ker A_0$ over $R$ is enough. Note that the columns of $A_1$ and $A_0$ should be minimal generating sets with least cardinality of "$\ker A_0$ over $R'$" and $M'$ respectively.

Update 1:

The answer to question 1 is yes when $n\le 1$. This is because, for $n\le 1$, $R'$ is a PID, so any submodule $M'$ of a free module $R'^k$ is also free. Choosing the columns of $A_0$ to be a basis of $M'$, we have $\ker A_0 = 0$ over $R'$. Therefore, $\ker A_0 = 0$ over $R$ as well.

Attempt 2: Let $\mu_S(N)$ denote the infimum of cardinalities of generating sets of $N$, an $S$-module, where $S$ is commutative Noetherian ring. Then, if $\mu_R(\ker A_0) = \mu_{R'}(\ker A_0)$ for any $A_0$ defined as above, then by induction, the answer to question 1 is yes. Conversely, a counterexample can be obtained by finding an $A_0$, defined as above, such that $\mu_R(\ker A_0) > \mu_{R'}(\ker A_0)$. I have been unsuccessful in constructing such a counterexample so far.

Note that $\operatorname{im} A_i$ is always a torsion-free module over both $R$ and $R'$ because it is a submodule of a free module. I am not sure if this is helpful.

Update 2:

After looking at Aurora's answer, which finds a counterexample to question 1, I am modifying the question to the following:

Question 1': Given an $A_0$ associated with a minimal generating set of $M'$ with least cardinality, is it always the case that $\mu_{R'}(\ker A_0) = \mu_R(\ker A_0)$?

If yes, then there is always a choice of $A_1$ such that $\operatorname{im} A_1 = \ker A_0$ over both $R'$ and $R$ (this is what I mentioned in Attempt 2 above). We can then proceed by induction to show that there is a choice of $A_i$ for $i>0$ such that the complex is exact over both $R'$ and $R$.

Answer 1

Let $\mathfrak{a}:=(X_1-2X_2,X_1-2X_3,X_1)$ as ideal of $R$. Then the Koszul complex of the mentioned generating set of $\mathfrak{a}$ is not acyclic, because $X_1-2X_2,X_1-2X_3,X_1$ is not a regular sequence. However, $X_1-2X_2,X_1-2X_3,X_1$ forms a regular sequence in $R'$, thus the Koszul complex $K_\bullet(\mathbf{a};R)\otimes_RR'=K_\bullet(\mathbf{a};R')$ is acyclic. To see the regular sequence property in $R'$ one can compute by Macaulay; or by hand $(X_1-2X_2,X_1-2X_3,X_1)R'=(X_1,X_2,X_3)R'$! To see the non-regular sequence property in $R$:

We have $X_1(X_2-X_3)=(X_1-2X_2)(-X_3)+(X_1-2X_3)(X_2)$, thus $X_2-X_3\in (X_1-2X_2,X_1-2X_3):X_1$, while $X_2-X_3\notin (X_1-2X_2,X_1-2X_3)$. Note that $X_2-X_3$ will be in the $2$-generated ideal after inverting $2$.

For your new question, set $$M:=\mathbb{Z}[X_1,X_2,X_3]/(X_1-2X_2,X_1-2X_3,X_1,X_2^2,X_3^2).$$ Then $M$ has projective dimension $4$ over $$R=\mathbb{Z}[X_1,X_2,X_3];$$ because after localizing at $(2,X_1,X_2,X_3)$ the module $M$ has depth $0$ and then apply the Auslander-Buchsbaum Formula as well as Formula of Pdim and Localization. However, $M\otimes_RR'=\mathbb{Q}$ has projective dimension 3 over $R'$. Thus it is impossible to obtain a minimal resolution over $R'$ for $R'\otimes_RM$ (minimal in the sense that the length agrees with the projective dimension), by tensoring a free resolution of $M$ over $R$. It is also impossible to obtain a resolution of $M$ over $R$ from, simply, lifting to $R$ a minimal free resolution of $M\otimes_RR'$ over $R'$.