Conceptual (homological) interpretation of a theorem about flat modules

Question

I just learned the following beautiful result about flat modules from J. S. Milne's book Étale Cohomology:

Lemma 2.10(b'): Let $M$ be any flat $A$-module. If $$\sum_i a_ix_i=0,$$ $a_i\in A$, $x_i\in M$, then there are equations $$x_i = \sum_j a_{ij}x_j'$$ with $x_j'\in M$, $a_{ij}\in A$, such that $$\sum_i a_ia_{ij} = 0$$ for all $j$.

It is stated under the hypothesis that $A$ is a commutative, noetherian ring, but as far as I can tell, the proof does not use noetherianity. (Actually I am not sure it even uses commutativity, but I will leave that matter alone for the purposes of this question.)

The proof given in Milne is essentially computational. However, the result struck me as almost a statement that something is exact (I'll make this precise in a moment), and of course flatness is all about exactness, so I wondered, is there a conceptual argument, or at least some reasonably satisfying conceptual handwaving, that gives us the claimed result in terms of an assertion that something is exact, or some (co)homology vanishes, in a way that is straightforwardly connected to $M$'s flatness?

(I'm using the soft-question tag because, while there could easily be a fully precise answer [which would be best possible], I will probably be happy with what I'm calling "satisfying conceptual handwaving," and what I mean by "straightforwardly" isn't precise either. Apologies in advance about this vagueness.)

Here is what I mean when I say the result struck me as "almost a statement that something is exact". The tuple $(a_i)_{i=1,\dots,r}$ defines a map $\varphi: M^r\rightarrow M$ by

$$(y_i)_{i=1,\dots,r} \mapsto \sum_i a_iy_i.$$

If $A$ is commutative, this is a map of $A$-modules. Then the equation

$$\sum_i a_ix_i=0$$

asserts that $(x_1,\dots,x_r)\in M^r$ lies in the kernel of this map. Similarly, the matrix $(a_{ij})$ defines a map $\psi: M^s\rightarrow M^r$ by

$$(y'_j)_{j=1,\dots,s}\mapsto \left(\sum_j a_{ij}y'_j\right)_{i=1,\dots,r},$$

whereupon the equations

$$\sum_i a_ia_{ij} = 0$$

assert that the composed map $\psi\circ\varphi = 0$, and the equations

$$x_i = \sum_j a_{ij}x_j'$$

assert that $(x_1,\dots,x_r)$ lies in the image of $\psi$. Thus the lemma "comes close" to asserting that we have a complex $M^s\xrightarrow{\psi} M^r \xrightarrow{\varphi} M$ which is exact in the middle.

It is not actually asserting this because the matrix $(a_{ij})$ is allowed to depend on the tuple $(x_i)$ (and the proof indeed requires the data of the $(x_i)$ for the construction of the $(a_{ij})$). So the real statement is: given an $A$-linear map $\varphi: M^r\rightarrow M$ defined by a tuple $(a_1,\dots,a_r)$, and given an element $(x_1,\dots,x_r)$ of the kernel of this map, there is an $A$-linear map $\varphi: M^s\rightarrow M^r$ defined by a matrix $(a_{ij})$, such that $M^s\rightarrow M^r\rightarrow M$ forms a complex and the particular given $(x_1,\dots,x_r)\in M^r$ lies in the image of $\psi$. There is no assertion that the entire kernel of $\varphi$ is exhausted by the image of $\psi$.

Nonetheless, these musings are sufficiently suggestive to me that I wanted to ask here: can you explain Milne's Lemma 2.10(b') in terms of the exactness of something, or the vanishing of some homology or cohomology, in a way that is evidently linked to $M$'s flatness over $A$?

Answer 1

Maybe you already know this and are looking for something more conceptual, but here's a way to connect this statement to a more homologically flavoured one.

A module $M$ over some commutative ring $R$ is flat precisely if $\mathbf{Tor}_1^R(R/I,M)$ vanishes for all finitely generated ideals $I \subset R$.

A handwavy proof: $M$ is flat if $\mathbf{Tor}_1^R(N,M)$ vanishes for all $N$. By cocontinuity using that every module is the colimit of its finitely generated submodules, flatness is equivalent to the vanishing of $\mathbf{Tor}_1^R(N,M)$ for all $N$ finitely generated. Such a module sits in a filtration with cyclic quotients and allows us to assume $N$ to be cyclic, hence of the form $R/I$. "$\square$"

Moreover, in the same proposition it is shown that $M$ is flat if for any finitely generated ideal $I \subset R$ tensoring the inclusion gives an injective map $I \otimes M \to M$. This is a matter of using the aforementioned equivalence and the long exact sequence for $\mathbf{Tor}$ applied to the short exact sequence

$$ 0 \to I \to R \to R/I \to 0. $$

Indeed, since $\mathbf{Tor}_1^R(R,M) = 0$ we have

$$ 0 \to \mathbf{Tor}_1^R(R/I,M) \to I \otimes M \to R \otimes M \simeq M $$ so the vanishing of $\mathbf{Tor}$ is exactly characterized the injectivity of $I \otimes M \to M$.

Therefore, flatness is intrinsically related to all the kernels of inclusion-induced maps $\langle a_1, \ldots, a_n\rangle \otimes M \to M$. Concretely, for $M$ to be flat we need that whenever

$$ \sum_{i}b_i a_i m_i = 0 $$ in $M$, $$ \sum_{i} b_i a_i \otimes m_i = 0 $$ in $\langle a_1, \ldots, a_n\rangle \otimes M$. This already looks connected to the equational criterion, and in fact the connection comes from the following lemma on zero elements of certain tensor products,

Lemma. Let $R$ be a commutative ring, $M, N$ two $R$-modules and $\{n_i\}_{i \in I}$ a finite generating set for $N$. Given $\{m_i\}_i \subset M$, we have $\sum_i m_i \otimes n_i = 0$ if and only if there exists a finite set $J$ and elements $r_{ij} \in R, \widetilde{m}_j \in M$ such that $$ \sum_{j} r_{ij}\widetilde{m}_j = m_i, \quad \sum_{i \in I}r_{ij}n_i = 0 \quad (i \in I, j \in J). $$

All of this can be found in section $6.3$ of Eisenbud's Commutative Algebra with a View Toward Algebraic Geometry.