# 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

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 $$R$$.

• As a localization $\mathbb{Q}$ is a faithfully flat $\mathbb{Z}$-module. Hence the complex $(R^{k_i})_i$ is exact iff $(R^{k_i}\otimes_\mathbb{Z}\mathbb{Q})_i=(R'^{k_i})_i$ is exact. Out of interest: In which area of physics is stuff from your question used?
– tj_
Apr 8, 2022 at 20:11
• @tj_ Doesn't flatness only give one implication: if $(R^{k_i})_i$ is exact, then $(R'^{k_i})_i$ is flat? The other direction requires faithful flatness, right? Is $\mathbb Q$ a faithfully flat $\mathbb Z$-module? About physics: I have seen such sequences where the coefficient ring is a finite field in some papers on fractons such as this. Apr 8, 2022 at 20:30
• @tj_ The $\mathbb Z$-module $\mathbb Q$ is not faithfully flat: $\mathbb F_p\otimes\mathbb Q=0$.
– Z. M
Apr 8, 2022 at 20:31
• TeX note: prefer $\operatorname{im} A$ \operatorname{im} A to \text{im}\,A, and similarly for \operatorname{rk}. I have edited accordingly. Also, if you are going to use such a command frequently, then you can start your post with $\DeclareMathOperator\im{im}$ (but don't include a trailing newline, or it will appear spuriously in the rendered post). Apr 9, 2022 at 18:16
• @LSpice Thanks for the edits! I will remember that. Apr 10, 2022 at 0:07

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 $$R'$$.
• Thanks for the counterexample! After seeing your answer, I came up with another (perhaps simpler) counterexample. Say $n=2$ and $A_0 = \begin{pmatrix} x_1 & x_2 \end{pmatrix}$, but $A_1 = \begin{pmatrix} 2x_2 \\ -2x_1 \end{pmatrix}$. (This is a minor modification of the example in the OP.) Then, over $R'$, this still gives an exact sequence but over $R$ it doesn't because $\begin{pmatrix} x_2 \\ -x_1 \end{pmatrix} \in \ker A_0$ over $R$ but it is not contained in $\operatorname{im}A_1$ over $R$. Apr 10, 2022 at 0:11
• While both of them are valid counterexamples to my question 1, we know that there is a choice of $A_1$ for which we do get sequences which are exact over both $R'$ and $R$. For instance, in your counterexample with $A_0 = \begin{pmatrix} x_1 - 2x_2 & x_1 - 2x_3 & x_1 \end{pmatrix}$, we can choose $A_1 = \begin{pmatrix} -x_1 + 2x_3 & x_3 & -x_1 + 2x_3 \\ x_1 - 2x_2 & -x_2 & -2x_2 \\ 0 & x_2 - x_3 & x_1 - 2x_3 \end{pmatrix}$ and $A_2 = \begin{pmatrix} x_1 \\ -x_1+2x_3 \\ x_1-2x_2 \end{pmatrix}$. (I found them on Macaulay2.) This gives an exact sequence over both $R'$ and $R$. Apr 10, 2022 at 0:18
• In both the counterexamples above, we could find a different choice of $A_1$'s that worked because $\mu_R(\ker A_0) = \mu_{R'}(\ker A_0)$, i.e., the minimal cardinalities of generating sets of $\ker A_0$ over $R$ and $R'$ are the same. I modified my question asking if they are always equal. Apr 10, 2022 at 0:35