In the book "Rings and Categories of Modules" by Anderson & Fuller, this problem is given: If $V^A$, i.e. the direct product of the module $V$ by the index set $A$, is flat for all sets $A$, then $(V^{(B)})^A$ is flat for all sets $A$and $B$. My try: the module $V$is itself flat by adopting a one-element set $A$ in the hypothesis. Hence, any direct sum $V^{(B)}$ is also flat (for all sets $B$). Also, by Exercise 4.2 of Lam's book "Exercises in Modules and Rings", if every finitely generated submodule of a module $P$ is contained in a flat submodule of $P$, then $P$ itself is a flat module. Any help would be appreciated.
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2 Here's the way I'd prove this fact.
Monomorphism $f: S \hookrightarrow M$ in the category of (say, left) $R$-modules is called pure, if one of those equivalent conditions are met:
a) $N \otimes f$ is injective for any right R-module $N$;
b) $f_*: \operatorname{Hom}(C, M) \to \operatorname{Hom}(C, M/S)$ is surjective for any finitely presented left module $C$;
c) $f^+: M^+ \to S^+$ is a split epimorphism, where $(-)^+$ is the duality functor from left to right modules $M \mapsto \operatorname{Hom}_{\Bbb Z}(M, \Bbb Q / \Bbb Z)$.
Now we need three observations, which follow pretty clearly from definitions above.
Pure submodule of a flat module is flat. This is because quotient of a flat module by pure submodule is flat by a), and long exact sequence of Tors gives you this.
$\bigoplus_I A_i$ is a pure submodule of $\prod_I A_i$ via natural inclusion by b) — every morphism from a finitely presented factors through the direct sum.
Product of a pure monomorphisms is a pure monomorphism, by c).
There are a few good excercises on pure submodules in the book "Rings of quotients" by Bo Stenström. For a more detailed exposition of "pure homological algebra" one may consult monograph "Purity, spectra and localisation" my Mike Prest.
- $\begingroup$ If I well understand your reasoning, the direct sum $V^{(B)}$ is a pure submodule of $V^A$. Now, if $(V^{(B)})^A$ is a pure submodule of $(V^B)^A$, then by your argument and the hypothesis of the problem, we are done. But, how do we get the pureness of the direct product $(V^{(B)})^A$ in the direct product $(V^B)^A$? If I am not right, would you please explain more. $\endgroup$karparvar– karparvar2024-08-23 19:52:21 +00:00Commented Aug 23, 2024 at 19:52
- 1$\begingroup$ Product of pure monomorphisms is again pure. This can be deduced directly, or by noticing that monomorphism $f: M \to N$ is pure iff $f^+: N^+ \to M^+$ is a split epi, where $M^+$ is $Hom_{\Bbb Z}(M, \Bbb Q / \Bbb Z)$. $\endgroup$Denis T– Denis T2024-08-23 21:03:11 +00:00Commented Aug 23, 2024 at 21:03