# $A$ PID, $M$ flat (i.e., torsion-free). Then $\operatorname{Ext}_A^1(M,N)$ is injective, for all $N$.

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?

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$\text{Ext}_A^\bullet(-,-)$ 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. – Andrea Gagna Jan 28 '13 at 21:38

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.

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Usefull! I will try to think this way. – Andrea Gagna Jan 28 '13 at 21:46
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 $P_0 \to P_0\otimes Q_0$). – Andrea Gagna May 31 '13 at 23:58
@AndreaGagna books.google.ro/… – user26857 Jun 1 '13 at 0:26
Sure, balacing Ext! Great, thank you. – Andrea Gagna Jun 1 '13 at 0:37


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.)

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