Let $R=\mathbb Z[x_1,\ldots,x_n]$. Say $M = \text{im}\, \varphi$, where the homomorphism $\varphi:R^k \rightarrow R^l$ is given by an $l\times k$ matrix $A$ with elements in $R$. So $M$ is a finitely generated $R$-module. I am interested in the length of a minimal free resolution of $M$.
Here's what I thought might work. Say $R' = \mathbb Q[x_1,\ldots,x_n]$, and $M'= \text{im}\, \varphi'$, where the homomorphism $\varphi':R'^k \rightarrow R'^l$ is given by the same matrix $A$ as above (i.e., all the entries of $A$ are polynomials with integer coefficients). By the Hilbert's syzygy theorem, a minimal free resolution of $M'$ is of length at most $n$. Let $A_i$, for $i=0,\ldots,n$, be the matrices associated with the maps $\varphi'_i$ in the minimal free resolution of $M'$, with $A_0 = A$ and $\varphi'_0 = \varphi'$.
Here are my questions:
Are the entries of $A_i$ also polynomials with integer coefficients for all $i$? I think, if they are not integers, we can scale them appropriately to make them integers.
Can we use the matrices $A_i$ (with integer coefficients) to construct a minimal free resolution of $M$? It seems to make sense intuitively, and it seems to be true in several examples I have worked out on Macaulay2.
If the answer to question 2 is "no" in general, is there a characterisation of $A$ for which the answer is "yes"?
Since $\mathbb Z$ has global (homological) dimension $1$, $R$ has global dimension $n+1$. However, this is projective dimension, and I am not sure if a projective resolution is the same as a free resolution for $R$-modules. Moreover, $n+1>n$, where $n$ is the global dimension of $R'$. So it is possible that the above procedure might be wrong. If so, I would like to know where it goes wrong.
Update 1:
As an example, consider $n=0$. Then, $M$ is a free $\mathbb Z$-module because any submodule of a free $\mathbb Z$-module is free. Therefore, the free resolution of $M$ has length $0$. Of course, this is not true if $M$ has torsion (which is why global dimension of $\mathbb Z$ is $1$ instead of $0$). So I think the above procedure implies that whenever $M$ is the image of a homomorphism between free $R$-modules of finite rank, its minimal free resolution is of length at most $n$.
Update 2:
There are some trivial counterexamples when the matrix $A$ is such that its columns do not form a minimal generating set of $M$. For example, consider $n=1$, i.e., $R=\mathbb Z[x]$, and the $1\times 2$ matrix $A = \begin{pmatrix} x &1 \end{pmatrix}$. Then $\ker A$ is generated by $\begin{pmatrix} 1 \\ -x \end{pmatrix}$, which means the length of the free resolution is $1$. However, the module $M$ is free because $\text{im}\,A \cong R$, so the length of its minimal free resolution is $0$. So, I am adding the assumption that the columns of $A$ form a minimal generating set of $M$.
Update 3:
There are matrices $A$ such that the columns of $A$ form a minimal generating set of $M$ but not of $M'$. For example, the columns of $A = \begin{pmatrix} 2 & x \end{pmatrix}$ form a minimal generating set of $M$ but not of $M'$. Consequently, the minimal free resolutions of $M$ and $M'$ do not have the same length. So, I am modifying the assumption to the following stronger version: the columns of $A$ form a minimal generating set of $M'$. It follows that they form a minimal generating set of $M$ as well.
Update 4:
First, the answer to question 1 is clearly yes because this won't affect the free resolution of $M'$.
After the discussion in comments below, I see that the notion of minimal free resolution in my question is not standard. So let me define what I mean by a minimal free resolution: it is a free resolution with minimal length. By the syzygy theorems for $R$ (see here) and $R'$ (Hilbert's syzygy theorem), both $M$ and $M'$ have finite lengths, so this notion gives a finite answer. With this definition, question 2 can be restated as follows:
- Given a minimal free resolution of $M'$ with matrices $A_i$ chosen so that they have only integer coefficients, can the same matrices be used to construct a free resolution of $M$? (Note that, now, I don't require the latter to be a minimal free resolution.)
I think question 2 then boils down to the following:
- Say $A_1$ is a matrix of polynomials with integer coefficients such that $\text{im}\,A_1 = \ker A$ over $R'$. Then, does $A_1$ satisfy $\text{im}\,A_1 = \ker A$ over $R$? If this is true for $A_1$, then by induction, all $A_i$ satisfy $\text{im}\,A_{i+1} = \ker A_i$ over both $R'$ and $R$, as long as we choose $A_i$ such that its columns form a minimal generating set with least cardinality of $\text{im}\,A_i = \ker A_{i-1}$ over $R'$.
Let me give an example of what I mean by this. Consider $n=2$, and $A = \begin{pmatrix} x & y \end{pmatrix}$. Then, in a minimal free resolution of $M'$, we have $A_1 = \begin{pmatrix} y\\ -x\end{pmatrix}$ and $A_2 = 0$. Now, $\text{im}\,A_1 = \ker A$ and $\text{im}\,A_2 = \ker A_1$ over both $R'$ and $R$. Therefore, $A_1$ and $A_2$ give a (probably minimal) free resolution of $M$ as well.