# Condition for equality of modules generated by columns of matrices

Let $$R$$ be a commutative ring with unit. Let $$M_A$$ denote the submodule of $$R^m$$ generated by columns of a matrix $$A$$ with entries in $$R$$. Suppose we are given two matrices $$A,B \in R^{m \times k}$$. I want to know if the following statement is true:

1. $$M_A = M_B$$ if and only if there exists an invertible matrix $$P \in R^{k \times k}$$ such that $$A = BP$$.

Note that one direction is super-easy: if $$A = BP$$ for invertible $$P$$, then the columns of both $$A$$ and $$B$$ generate the same submodule of $$R^m$$. So really the question is about the other direction. This is what I have tried for the "only if" direction: since $$M_A = M_B$$, we know that one can find matrices $$P, Q \in R^{k \times k}$$ such that $$A = BP$$ and $$B = AQ$$. From this we deduce that $$A(I - QP)= B(I - PQ) = 0$$. From here on, I don't really know what to do, but most likely this approach is just too naive.

However, I think the answer is true for a ring $$R$$ when $$R$$ is a principal ideal ring. Does anybody have a reference to this result (the particular case I want is when $$R = \mathbb{Z}_d$$, so a reference to this will also suffice)? More generally, I'd like to know for which rings does this property hold, and where can I find a proof of this result.

EDIT: Actually now I'm beginning to think that the result may not be true in general for PIRs. However, some special cases are true. I know the following facts:

(i) If you replace PIR by PID (principal integral domain), then the result is true.

(ii) If you have a PIR, and both $$A,B$$ have one row, then the result is true. Proof goes like this: Since a PIR is a Hermite ring, there exists an invertible matrices $$K,K'$$ such that $$AK = (s,0,\dots,0)$$ and $$BK' = (s',0,\dots,0)$$. Thus $$s,s'$$ generate the same ideal, and thus they must be associates (which is a property of PIRs by a result of I.Kaplansky). The result now follows.

(iii) If you are allowed to add zero columns to both $$A,B$$, then also the result is true, because of the existence of the Howell normal form. (see this and the references therein).

I'd like to stress that I'm interested in the case when the ring is $$\mathbb{Z}_d$$.

• Please consider the case k = 1. Dec 27, 2022 at 14:25
• This seems relevant (and I naively assumed the opposite when first thinking about it): math.stackexchange.com/questions/355994/… Dec 27, 2022 at 16:34
• @MartinBrandenburg At least for principal ideal rings, the case k=1 is true. Dec 27, 2022 at 18:49
• Yes ok, but it shows that your general question has to be answered with "No". It also shows that any general proof attempt will not work. Regarding your proof for principal rings, it would be very helpful for the community if you share the proof, or at least summarize it. And by the way, I believe that it might then also work with Bezout rings (every f.g. ideal is principal), right? Dec 27, 2022 at 23:06
• @JeremyRickard thanks for catching the typo. i meant submodules of $R^m$, and I believe I have fixed the typos. Let me know if there is still something that is not clear or confusing. Jan 6 at 10:07

Interpreting the various matrices as maps between free modules in the usual way, the question becomes:

If $$M$$ is a submodule of $$R^m$$, and $$\alpha,\beta: R^k\to M$$ are epimorphisms, then must $$\alpha$$ and $$\beta$$ be equivalent, in the sense that there is an automorphism $$f:R^k\to R^k$$ with $$\alpha = \beta f$$?

For $$R=\mathbb{Z}/d\mathbb{Z}$$, or indeed any artinian $$R$$, the answer is yes, since then for finitely generated modules we have projective covers and a Krull-Schmidt theorem, so both $$\alpha$$ and $$\beta$$ are equivalent to the same epimorphism $$\gamma:R^k=P(M)\oplus Q\to M$$, where $$P(M)$$ is the projective cover of $$M$$, and $$\gamma$$ restricts to the usual map $$P(M)\to M$$ on $$P(M)$$ and the zero map on $$Q$$.

However, the answer is no if $$R$$ has a stably free module $$P$$ that is not free. For then, suppose $$P\oplus R^n\cong R^k$$. Then we have two epimorphisms $$R^k\to R^n$$, one with kernel $$P$$, and one with kernel $$R^{k-n}$$. Since they have non-isomorphic kernels, these epimorphisms can't be equivalent.

I want to elaborate on the answer that I accepted above by Jeremy Rickard. The facts in the answer were not entirely obvious to me. But after some reading, I managed to put together some of the missing details. Hopefully, these details are useful to someone else too.

We record a few definitons that I found non-trivial; however I won't define everything (such as an artinian ring, and short exact sequences). We assume a commutative ring $$R$$ with unit. I will use the following references: [these lecture notes], [this Stacks project page], [this Stacks project page].

Definition 1: A $$R$$-module $$P$$ is called a projective module if there exists a $$R$$-module $$Q$$ such that $$P \oplus Q$$ is a free $$R$$-module. (Note: any free $$R$$-module is a projective $$R$$-module).

Definition 2: A $$R$$-module epimorphism $$f: P \rightarrow M$$ is called an essential surjection if for every strictly smaller submodule $$Q \subset P$$, we have $$f(Q) \neq M$$.

Definition 3: A $$R$$-module epimorphism $$f: P \rightarrow M$$ is called a projective cover if $$P$$ is a projective $$R$$-module, and $$f$$ is an essential surjection.

Next we record some facts that go into the proof.

Fact 1: Any finitely generated module over an artinian ring $$R$$ admits a projective cover. [see here]

Fact 2: Suppose $$P$$ is a projective $$R$$-module, $$f: P \rightarrow M$$ is a $$R$$-module homomorphism, and $$g: N \rightarrow M$$ is a $$R$$-module epimorphism. Then there is a $$R$$-module homomorphism $$h: P \rightarrow N$$ such $$f = gh$$. [Theorem 43.9]

Fact 3: [Theorem 43.9] Suppose $$P$$ is a projective $$R$$-module, and we have a short exact sequence of $$R$$-module homomorphisms $$\require{AMScd}$$ $$\begin{CD} 0 @>>> N @>u>> M @>v>> P @>>> 0. \end{CD}$$ Then the sequence is a split exact sequence, meaning that there exists a $$R$$-module isomorphism $$w: M \rightarrow N \oplus P$$ such that the following diagram commutes $$\require{AMScd}$$ $$\begin{CD} 0 @>>> N @>u>> M @>v>> P @>>> 0 \\ @. @V=VV @VwVV @V=VV @. \\ 0 @>>> N @>>> N \oplus P @>>> P @>>> 0. \end{CD}$$

Now we may start the proof. We restate the theorem.

Theorem: For an artinian ring $$R$$, let $$M$$ be a finitely generated $$R$$-module, $$N$$ be a finitely generated free $$R$$-module, and $$\alpha, \beta: N \rightarrow M$$ are $$R$$-module epimorphisms. Then there exists an $$R$$-module isomorphism $$f: N \rightarrow N$$ such that $$\alpha = \beta f$$.

Proof: Since $$R$$ is an artinian ring, by Fact 1, there exists a projective cover $$\mu: P \rightarrow M$$. Now consider the maps $$\alpha, \mu$$ both of which are $$R$$-module epimorphisms. By Fact 2, noting that $$N$$ is a free $$R$$-module implying it is also a projective $$R$$-module, we get a homomorphism $$h_{\alpha} : N \rightarrow P$$ such that $$\mu h_{\alpha} = \alpha$$. Now, since $$\mu$$ is an essential surjection, $$h_{\alpha}$$ must be an epimorphism too. Thus we have the short exact sequence $$\require{AMScd}$$ $$\begin{CD} 0 @>>> \ker(h_{\alpha}) @>>> N @>h_{\alpha}>> P @>>> 0. \end{CD}$$

Then Fact 3 implies that there is a $$R$$-module isomorphism $$\lambda_{\alpha}: N \rightarrow \ker(h_{\alpha}) \oplus P$$ such that $$h_{\alpha} = \pi_{\alpha} \lambda_{\alpha}$$, where $$\pi_{\alpha}: \ker(h_{\alpha}) \oplus P \rightarrow P$$ is the projection map. Thus so far we have $$\alpha = \mu \pi_{\alpha} \lambda_{\alpha}$$, or equivalently $$\alpha \lambda^{-1}_{\alpha} = \mu \pi_{\alpha}: \ker(h_{\alpha}) \oplus P \rightarrow M$$. The map $$\mu \pi_{\alpha}$$ restricts to the projective cover $$\mu$$ on $$P$$.

We can repeat the same argument now for the map $$\beta$$ to get the corresponding maps $$h_{\beta}, \lambda_{\beta}, \pi_{\beta}$$. The end result is that we have the following isomorphisms between modules: $$N \cong \ker(h_{\alpha}) \oplus P \cong \ker(h_{\beta}) \oplus P$$. We would be done if we can show that we have an isomorphism $$\ker(h_{\alpha}) \cong \ker(h_{\beta})$$. But this follows by applying the Krull-Schmidt theorem, using the fact that $$N$$ is finitely generated (see discussion in comments below).

• You've proved that (up to isomorphism) any epimorphism $N\to M$ from a free module to $M$ is of the form $N'\oplus P\to M$, where the restriction of the map to $N'$ is zero (since $N'=\operatorname{ker}(h)$) and the restriction to $P$ is a projective cover $P\to M$ of $M$. By Lemma 47.4.2 of your first Stacks Project reference, projective covers are unique up to isomorphism (meaning that there is an isomorphism between any two projective covers that makes the obvious triangle commute). Finally, by the Krull-Schmidt theorem, $N'$ is also uniquely determined up to isomorphism. Jan 7 at 9:50
• When writing the previous comment, I noticed that I only need the existence of projective covers for finitely generated modules, and the Krull-Schmidt theorem for finitely generated projectives, which means that the argument works not only for artninan rings, but for any semiperfect ring. Jan 7 at 9:57
• @JeremyRickard seems like I can avoid invoking Lemma 47.4.2 from the Stacks project reference if I just choose $P$ to be same for both $\alpha, \beta$ in the argument. What I'm still unsure of is the last step, where we apply the Krull-Schmidt theorem, to conclude that the kernels must be isomorphic as well. Can you state the theorem precisely? Seems most references I check, require the direct summands to be indecomposable modules. How do I know that the kernels are indecomposable? I'm guessing there is a way around this? Jan 8 at 0:35
• (1) Yes, that works. (2) The kernels may not be indecomposable, but it follows from Krull-Schmidt that if $P\oplus X\cong P\oplus Y$ then $X\cong Y$. Proof: write $P=P_1\oplus\dots \oplus P_k$, $X=X_1\oplus\dots\oplus X_m$ and $Y=Y_1\oplus\dots\oplus Y_n$ as direct sums of indecomposable modules. Then $P_1\oplus\dots \oplus P_k\oplus X_1\oplus\dots\oplus X_m\cong P_1\oplus\dots \oplus P_k\oplus Y_1\oplus\dots\oplus Y_n$, and so $m=n$ and $X_i\cong Y_i$ up to some permutation, so $X\cong Y$. Jan 8 at 10:55
• The Krull-Schmidt theorem, when it applies, states that things are finite direct sums of indecomposable things, and that these indecomposable things are uniquely determined up to isomorphism and reordering. Example where it holds include finitely generated modules for an artinian ring, and finitely generated projective modules for a semiperfect ring. I'm sure there are similar statements that hold for infinite direct sums in certain situations, but these will necessarily involve infinitely generated modules, since a finitely generated module is never an infinite direct sum. Jan 9 at 8:11