# Global to local principle for f.g. $\mathbb{Z}[x]$ modules

In graduate school, while I was working on the maximal subgroup growth of certain metabelian groups, I discovered and proved a lemma which gave me the impression that it was already known. Do you know of any reference for this result? Does it follow easily from a known result? Here is the result:

Let $$N$$ be a finitely generated $$\mathbb{Z}[x]$$-module. Suppose (by using the fundamental theorem of f.g. modules over PIDs) we have the following isomorphism of $$\mathbb{Q}[x]$$-modules:

$$\mathbb{Q} \otimes_\mathbb{Z} N \cong \left( \bigoplus_{j=1}^{s_1} \mathbb{Q}[x]/(a_j) \right) \oplus \mathbb{Q}[x]^{s_2}$$ for some $$a_j \in \mathbb{Q}[x]$$ that are not units and such that $$a_1 \mid a_2 \mid \ldots \mid a_{s_1}$$.

Then for all large primes $$p$$, we have the following isomorphism of $$\mathbb{F}_p[x]$$-modules:

$$N/pN \cong \left( \bigoplus_{j=1}^{s_1} \mathbb{F}_p[x]/(\overline{a_j}) \right) \oplus \mathbb{F}_p[x]^{s_2}.$$

While my proof of the above result is about three pages long, I do have a short heuristic argument:

When doing the computation required in finding the decomposition of $$\mathbb{Q} \otimes N$$ into a direct sum of cyclic modules, the only thing keeping us from doing this computation to $$N$$ itself (as a $$\mathbb{Z}[x]$$-module) is that we may need to divide by finitely many integers.

So if $$p$$ is large enough, then in $$\mathbb{F}_p$$ we can divide by all those integers (i.e. their residues mod $$p$$). For such $$p$$, the steps of the algorithm would be the same for $$N/pN$$ as for $$\mathbb{Q} \otimes N$$.

In case you are interested, this is Lemma 41 in my arXiv paper Maximal subgroup growth of some metabelian groups. I actually proved a slight generalization, concerning finitely generated $$D^{-1}\mathbb{Z}[x]$$-modules, where $$D$$ is the multiplicative closure of a finite set of primes.

• I guess generic flatness is hiding behind this (I used it in a similar context, although in Krull dimension greater than 2). – YCor Apr 24 at 18:11
• @YCor: May I ask why you've added the algebraic-geometry tag? – Todd Leason Apr 25 at 10:49
• @ToddLeason because algebraic geometry has the right point of view to think of such issues; a standard reference for generic flatness is SGA. (I don't claim it's the only approach or point of view.) – YCor Apr 25 at 15:04

Nothing in the sequel should be new to you. But it is shorter and it also makes clear that your lemma generalizes immediately when replacing $$\mathbb{Z}$$ by any integral domain with infinitely many maximal ideals ("for all prime $$p$$ large enough" becomes "for all but finitely many maximal ideals").
Given a rational prime number $$p$$, let $$\mathbb{Z}_{(p)} = \left\{ \frac{r}{s} \, \vert \, r, s \in \mathbb{Z},\, \mathbb{Z}s + \mathbb{Z}p = \mathbb{Z} \right\}$$ and let $$\phi_p: \mathbb{Z}_{(p)} \rightarrow \mathbb{Z}_{(p)} / p\mathbb{Z}_{(p)} \simeq \mathbb{F}_p$$ be the natural epimorphism. Restricted to $$\mathbb{Z}$$, the map $$\phi_p$$ is the reduction modulo $$p$$. Given a commutative and unital ring $$R$$, we denote by $$\text{M}_{m,n}(R)$$ the $$R$$-module of the $$m$$-by-$$n$$ matrices over $$R$$. Abusing notation, we denote also by $$\phi_p$$ the epimorphisms $$\mathbb{Z}_{(p)}[x] \rightarrow \mathbb{F}_p[x]$$ and $$\text{M}_{m,n}(\mathbb{Z}_{(p)}[x]) \rightarrow \text{M}_{m,n}(\mathbb{F}_p[x])$$ induced by the reduction of coefficients.
Proof of OP's Lemma. Let $$N$$ be a finitely generated module over $$\mathbb{Z}[x]$$ and let $$$$\label{EqSeq} \mathbb{Z}[x]^n \xrightarrow[]{f} \mathbb{Z}[x]^m \rightarrow N \rightarrow 0 \quad\quad (1)$$$$ be an exact sequence that provides us with a presentation of $$N$$ where $$m$$ is the minimal number of generators of $$N$$. Let $$M(f) \in \text{M}_{m, n}(\mathbb{Z}[x])$$ be the matrix of $$f$$ with respect to the canonical bases. Applying the tensor functors $$-\otimes_{\mathbb{Z}[x]} \mathbb{Q}[x]$$ and $$-\otimes_{\mathbb{Z}[x]} \mathbb{F}_p[x]$$ to $$(1)$$, we obtain two other exact sequences. The exactness of the first is immediate as $$-\otimes_{\mathbb{Z}[x]} \mathbb{Q}[x]$$ is exact since we just inverted non-zero rational integers. For the second, some easy diagram chasing involving also $$(1)$$ settles the claim. Thus we get a presentation of $$N \otimes_{\mathbb{Z}[x]}\mathbb{Q}[x]$$ as the cokernel of the matrix $$M(f\otimes_{\mathbb{Z}[x]}\mathbb{Q}[x]) = M(f)$$ and a presentation of $$N \otimes_{\mathbb{Z}[x]}\mathbb{F}_p[x] \simeq N/pN$$ as the cokernel of $$M(f\otimes_{\mathbb{Z}[x]}\mathbb{F}_p[x]) = \phi_p(M(f)) \in \text{M}_{m,n}(\mathbb{F}_p[x]))$$. Since $$\mathbb{Q}[x]$$ is a principal ideal ring, the Smith Normal Form Theorem applies. Therefore we can find $$A \in \text{GL}_m(\mathbb{Q}[x]), B \in \text{GL}_n(\mathbb{Q}[x])$$ such that $$AM(f)B = \text{diag}(a_1, \dots, a_{s_1}, \underbrace{0, \dots, 0}_{s_2 \text{ zeroes}}) \quad (2)$$ with $$a_j \in \mathbb{Q}[x]$$ for all $$j$$, $$a_j \vert a_{j + 1}$$ and $$a_1 \notin \mathbb{Q} \setminus \{0\}$$. For all $$p$$ sufficiently large, we have $$A \in \text{GL}_m(\mathbb{Z}_{(p)}[x]), B \in \text{GL}_n(\mathbb{Z}_{(p)}[x])$$ and hence $$a_j \in \mathbb{Z}_{(p)}[x]$$ for all $$j$$. Applying $$\phi_p$$ to both sides of $$(2)$$, we get $$\phi_p(A) \phi_p(M(f)) \phi_p(B) = \text{diag}(\phi_p(a_1), \dots, \phi_p(a_{s_1}), 0, \dots, 0).$$ The proof is then complete.