9
$\begingroup$

Let $A$ be a PID and $M$ a flat (i.e., torsion-free) $A$-module. Then, for every $A$-module $N$, $\text{Ext}_A^1(M, N)$ is injective in $A\text{-}\mathbf{Mod}$.

It is easy when $M$ is finitely generated, since it is free (in particular, projective); thus $\text{Ext}_A^\bullet$-acyclic. Does the general case follow by a filtered colimit argument?

$\endgroup$
1
  • $\begingroup$ $\text{Ext}_A^\bullet(-,-)$ denotes the Ext funtor [en.wikipedia.org/wiki/Ext_functor], that is both the right-derived funtor of $\text{Hom}_{A\text{-}\mathbf{Mod}}(M,-)$ and $\text{Hom}_{(A\text{-}\mathbf{Mod})^{\text{op}}}(-, N)$. @YACP: I think you are right. I get confused with $\text{Tor}_\bullet^A$, which actually does. $\endgroup$ Jan 28 '13 at 21:38
11
$\begingroup$

You have to prove that $\operatorname{Ext}_A^1(M,N)$ is divisible. Take $a\in A$, $a\neq 0$. Since $M$ is torsion-free we have a short exact sequence $0\to M\stackrel{a\cdot}\to M\to M/aM\to 0$ and this gives rise to a long exact sequence of homology: $\operatorname{Ext}_A^1(M,N)\stackrel{a\cdot}\to\operatorname{Ext}_A^1(M,N)\to\operatorname{Ext}_A^2(M/aM,N)=0$ (the last $\operatorname{Ext}$ is $0$ since a PID has global dimension $\le 1$), and we are done.

$\endgroup$
4
  • $\begingroup$ Usefull! I will try to think this way. $\endgroup$ Jan 28 '13 at 21:46
  • $\begingroup$ I'm back to this answer after a long time. I can't see why the morphism $\text{Ext}^1_A(a\cdot, N)$ has to be multiplication by $a$. I'd thought to use the same resolution $P_\bullet$ for both the $M$'s, in such a way the induced morphism has to be $a\cdot$, but clearly it can't work, since $\text{coker}(a\cdot\colon P_0 \to P_0)$ is torsion, thus not projective (hence, cannot be the 0-term for a projective resolution $Q_\bullet$ of $M/aM$). Actually, looking at the horseshoe lemma, I have even suspected it is false (because we have an immersion map $P_0 \to P_0\otimes Q_0$). $\endgroup$ May 31 '13 at 23:58
  • 1
    $\begingroup$ @AndreaGagna books.google.ro/… $\endgroup$
    – user26857
    Jun 1 '13 at 0:26
  • $\begingroup$ Sure, balacing Ext! Great, thank you. $\endgroup$ Jun 1 '13 at 0:37
5
$\begingroup$

Does the general case follow by a filtered colimit argument? $\newcommand{\Z}{\mathbb{Z}}$ $\newcommand{\Q}{\mathbb{Q}}$ $\newcommand{\R}{\mathbb{R}}$ $\newcommand{\Hom}{\operatorname{Hom}}$ $\newcommand{Ext}{\operatorname{Ext}}$

No. I spoke about precisely this in class yesterday. In light of the corresponding properties for the Tor functors and since $\Ext^{\bullet}( \bigoplus_{i \in I} M_i,N) \cong \prod_{i \in I} \Ext^{\bullet}(M_i,N)$, it is natural to wonder about

$\Ext^{\bullet} (\varinjlim M_i,N) \stackrel{?}{\cong} \varprojlim \Ext^{\bullet}(M_i,N)$.

Since for all modules $N$, projective modules are acyclic for $\Hom(\cdot,N)$, if the above "continuity" property of $\Ext$ held true, then all (I mean filtered here) direct limits of projective modules would also be acyclic for $\Hom(\cdot,N)$. By a famous result of Govorov-Lazard, these are precisely the flat modules, so we'd have $\Ext^n_R(M,N) = 0$ for all flat $M$, all $N$, and all $n > 0$.

However, there are plenty of counterexamples to this. My favorite at the moment (as a relative newcomer to this material) is

$\Ext^1_{\Z}(\Q,\Z) \cong \R$.

(Writing $\R$ is a bit "lyrical": we mean of course the additive group of $\mathbb{R}$, which is characterized by being a torsionfree divisible abelian group of cardinality $2^{\aleph_0}$.)

This result is the subject of this short article by James Wiegold. A second, shorter proof is sketched by J. Rotman in his MathSciNet reiew of Wiegold's article (he says the result is "well known", which indeed seems plausible). A third (still shorter?) proof is sketched at the end of $\S$ 6.4 of these rough course notes of mine.


Note also that the same section in my notes contains a proof of a mild generalization of your main question: if $R$ is a Dedekind domain, for every torsionfree module $M$ and every module $N$, $\Ext^1_R(M,N)$ is divisible. (Note that over a Dedekind domain it is true that torsionfree = flat and divisible = injective. Nevertheless I stand by these two replacements, since the argument actually uses "only" the torsionfree condition and yields "only" the conclusion about divisibility.)

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.