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 $R$.

  • $\begingroup$ 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? $\endgroup$
    – tj_
    Apr 8, 2022 at 20:11
  • $\begingroup$ @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. $\endgroup$ Apr 8, 2022 at 20:30
  • 1
    $\begingroup$ @tj_ The $\mathbb Z$-module $\mathbb Q$ is not faithfully flat: $\mathbb F_p\otimes\mathbb Q=0$. $\endgroup$
    – Z. M
    Apr 8, 2022 at 20:31
  • 1
    $\begingroup$ 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). $\endgroup$
    – LSpice
    Apr 9, 2022 at 18:16
  • 1
    $\begingroup$ @LSpice Thanks for the edits! I will remember that. $\endgroup$ Apr 10, 2022 at 0:07

1 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 $R'$.

  • $\begingroup$ 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 $R$. $\endgroup$ Apr 10, 2022 at 0:11
  • $\begingroup$ 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 $R$. $\endgroup$ Apr 10, 2022 at 0:18
  • $\begingroup$ 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. $\endgroup$ Apr 10, 2022 at 0:35
  • $\begingroup$ I upvoted your answer but haven't accepted it yet. Technically, you answered my original question and it is still relevant to the new question. If you think it's better for the new question (which is a weaker version of the old question) to be a new post, I can accept your answer and create a new post for the new question. $\endgroup$ Apr 10, 2022 at 0:59
  • $\begingroup$ I modified my answer upon your new question. $\endgroup$
    – Aurora
    Apr 10, 2022 at 1:56

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