I have asked a related question on math.SE [here](https://math.stackexchange.com/questions/4417531/finite-free-resolution-of-a-finitely-generated-mathbb-zx-1-ldots-x-n-modul), 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](https://arxiv.org/abs/1905.08117)).


**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?](https://doi.org/10.1016/0021-8693(73)90044-6) 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](https://mathoverflow.net/a/420023), 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 $R$.



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
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](https://math.stackexchange.com/questions/630564/projective-dimension-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 $R'$.


