Let $A \to B$ be a homomorphism of commutative rings. I would like to find a criterion for the flatness of $A \to B$ which does not involve the notion of kernels; it should rather involve cokernels. This is partially motiviated by this question. I suggest the following: Consider for $A$-modules $M,N$ the canonical homomorphism of $B$-modules

$\alpha_{M,N} : \mathrm{Hom}_A(M,N) \otimes_A B \to \mathrm{Hom}_B(M \otimes_A B, N \otimes_A B) \cong \mathrm{Hom}_A(M,N \otimes_A B).$

For fixed $N$, the class $\{M : \alpha_{M,N} \text{ is an isomorphism}\}$ contains $A$ and is closed under finite direct sums. Now consider the following statement:

($\dagger$) For all $A$-modules $N$, the set $\{M : \alpha_{M,N} \text{ is an isomorphism}\}$ is closed under finite colimits, or equivalently, under cokernels.

Note that $\dagger$ implies, in particular, that $\alpha_{M,N}$ is an isomorphism whenever $M$ is finitely presented. But probably $\dagger$ is a stronger statement.

If $A \to B$ is flat, then $\dagger$ holds since then $\alpha_{-,N} : \mathrm{Hom}_A(-,N) \otimes_A B \to \mathrm{Hom}_A(-,N \otimes_A B)$ is a natural transformation between right exact functors $\mathrm{Mod}(A) \to \mathrm{Mod}(B)^{\mathrm{op}}$. What about the converse?

Assume that $\dagger$ holds. Then for all $N$ and all exact sequences $M'' \to M \to M' \to 0$ of modules over $A$ we have the following commutative diagram:

$$\begin{matrix} 0 & & 0 \\\ \downarrow & & \downarrow \\\ \mathrm{Hom}_A(M',N) \otimes_A B & \rightarrow & \mathrm{Hom}_A(M',N \otimes_A B) \\\ \downarrow & & \downarrow \\\ \mathrm{Hom}_A(M,N) \otimes_A B & \rightarrow & \mathrm{Hom}_A(M,N \otimes_A B) \\\ \downarrow & & \downarrow \\\ \mathrm{Hom}_A(M'',N) \otimes_A B & \rightarrow & \mathrm{Hom}_A(M'',N \otimes_A B) \end{matrix}$$

The right column is exact anyway. $\dagger$ says that we have an isomorphism in the first row as soon we have it in the two lower rows. This shows in particular that $- \otimes_A B$ preserves a large class of monics, namely those of the form $\mathrm{Hom}(M',N) \to \mathrm{Hom}(M,N)$ where $M \twoheadrightarrow M'$ and $M,M'$ belong to the class (for example, if they are finitely presented). Perhaps this is a first step in order to show that $A \to B$ is flat.

  • 1
    $\begingroup$ The array does not work properly. Does someone know how to repair this? With my latex dist it works, so it seems to be a problem of the latex implementation of MO. $\endgroup$ – Martin Brandenburg Nov 4 '11 at 10:05
  • $\begingroup$ I'm not sure if this is what you mean, but I once figured out how to do matrices in MO and (since you have plenty of rep) you can steal my code by "editing" my post: mathoverflow.net/questions/77094/… $\endgroup$ – David White Nov 4 '11 at 16:25
  • $\begingroup$ The key thing is begin{matrix} M & \to & A \\ f\downarrow & & \downarrow g\\ M' & \to & A' \end{matrix}, all in double dollar signs (if you use single, then you need \\\ rather than \) $\endgroup$ – David White Nov 4 '11 at 16:25

I cannot answer your question directly (though I suspect the criterion you give will be too weak to imply flatness). However, there are several points I can make.

  1. For the sort of question you are asking, you might as well let $B$ be an $A$-module, instead of restricting it to being an $A$-algebra.

  2. $\alpha_{M,N,B}$ (if I may call it that) is a functorial (in all three variables) map of $A$-modules, and when $B$ is flat and $M$ finitely generated (or when $B$ is arbitrary and $M$ is finitely-generated projective), it is an isomorphism. The map and these facts about it are well-studied, and can be souped-up in various ways involving derived categories and change-of-rings. It is called the "tensor-evaluation" map by Hans-Bjørn Foxby and his school.

  3. Yongwei Yao and I have written an article in Mathematische Zeitschrift on criteria for flatness (and for injectivity) of modules over a commutative Noetherian ring. Although the criteria in our article do not resemble what you discuss in the body of your question beyond the semicolon, they do satisfy your demand that they not involve kernels, at least not directly. For example, we show that if $A$ is reduced, then the following are equivalent for an $A$-module $M$.

(a) $M$ is flat.

(b) $P \otimes_R M$ is a torsion-free $R$-module for all $P \in \mathrm{Spec} A$.

(c) $\mathrm{Ass}_A (L \otimes_A M) \subseteq \mathrm{Ass}_A L$ for every $A$-module $L$.

  • $\begingroup$ 1. Yes. 2. Sure. 3. This goes into another direction as my question. Spectral data is even more 'complicated' than the use of monics. But you could help me if you indicate why you think that $\dagger$ is too weak for flatness. $\endgroup$ – Martin Brandenburg Nov 5 '11 at 21:45
  • $\begingroup$ To be honest, it's just a feeling arising from my work with flat and injective modules. Yao and I found that when things are restricted to Hom modules, conditions that look like they should be equivalent can fail to be so (e.g. with h-divisible modules). Perhaps your condition characterizes something weaker -- like torsionlessness (recall: an $A$-module $M$ is torsionless if the natural map $M \rightarrow \mathop{Hom}_A(\mathop{Hom}_A(M, A), A)$ is injective. If $A$ is a domain, this is the same as torsion-free). On the other hand, maybe you can characterize flatness this way. $\endgroup$ – Neil Epstein Nov 7 '11 at 12:10
  • $\begingroup$ Alright, thank you. I also suspect that my suggestion is too weak. Unfortunately, your characterizations all involve injections (implicitly). $\endgroup$ – Martin Brandenburg Nov 7 '11 at 20:00

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