Let $M$ be an $R$-module and consider the following statement.

$M$ is projective whenever the obvious group map $\tau: \text{Hom}_R(M, R) \to \text{Hom}_R(M, {R}/{I})$ is surjective for any ideal $I\subset R$.

I have read that when $R = \mathbb{Z}$, determining the truth of this statement is equivalent to deciding the Whitehead problem, which asks whether or not the abelian group $M$ must be free so long as every group extension of $M$ by $\mathbb{Z}$ is trivial.

I know that being free is equivalent to being projective for abelian groups. But why is the surjectivity of $\tau$ equivalent to the condition that $\text{Ext}(M, \mathbb{Z}) =0$? I don't see how the fact that $\tau$ is surjective implies that any such extension splits.

I may be missing something obvious, but could someone please explain?

Here is one reference for what I am talking about.

Google book page.


This is incorrect: surjectivity of $\tau$ is strictly weaker than $\operatorname{Ext}(M,\mathbb{Z})=0$. Specifically, surjectivity of $\tau$ is instead equivalent to $\operatorname{Ext}(M,\mathbb{Z})$ being torsion-free. The statement that you quoted is not equivalent to Whitehead's problem for $R=\mathbb{Z}$ and in fact is provably false for $R=\mathbb{Z}$. For instance, if $M=\mathbb{Q}$, $\operatorname{Hom}(\mathbb{Q},\mathbb{Z}/I)=0$ for any ideal $I\subseteq\mathbb{Z}$, so $\tau$ is trivially always surjective, but $\mathbb{Q}$ is not projective.

Here's how you prove that surjectivity of $\tau$ is equivalent to $\operatorname{Ext}(M,\mathbb{Z})$ being torsion free. Note that for any $I\subseteq \mathbb{Z}$, we have a long exact sequence $$0\to \operatorname{Hom}(M,I)\to\operatorname{Hom}(M,\mathbb{Z})\stackrel{\tau}\to\operatorname{Hom}(M,\mathbb{Z}/I)\stackrel{\delta}\to\operatorname{Ext}(M,I)\stackrel{f}\to\operatorname{Ext}(M,\mathbb{Z})\to\operatorname{Ext}(M,\mathbb{Z}/I)\to 0.$$

In particular, we see that $\tau$ is surjective iff $\delta=0$ iff $f$ is injective. Note moreover that if $I$ is nontrivial, then $I=n\mathbb{Z}$ for some $n>0$ and so we can identify $f$ with the multiplication by $n$ map $\operatorname{Ext}(M,\mathbb{Z})\to\operatorname{Ext}(M,\mathbb{Z})$. So, $\tau$ is surjective for $I=n\mathbb{Z}$ iff $\operatorname{Ext}(M,\mathbb{Z})$ has no $n$-torsion. We conclude that $\tau$ is surjective for all $I$ iff $\operatorname{Ext}(M,\mathbb{Z})$ is torsion-free.

  • $\begingroup$ Thank you, but I keep reading that my quoted statement is the dual of Baer's criterion for injective modules and that this reduces to the Whitehead problem for $R= \mathbb{Z}$. Do you have any idea what the correct version of this claim is supposed to look like? $\endgroup$ – CuriousKid7 Jan 20 at 19:05
  • $\begingroup$ Where are you reading that? I don't think there's any sense in which the Whitehead problem is literally dual to Baer's criterion, though they are certainly related. $\endgroup$ – Eric Wofsey Jan 20 at 19:35
  • $\begingroup$ I've given the link of one reference in my OP. But you are certainly right. $\endgroup$ – CuriousKid7 Jan 20 at 19:45
  • 2
    $\begingroup$ I think it's simply an error in that text where the authors were a bit over-eager to fit a remark on Whitehead's problem into their narrative. $\endgroup$ – Eric Wofsey Jan 20 at 19:50
  • $\begingroup$ Yes, I agree. In fact, the correct formulation is much more delicate. It seems to be known as "Faith's problem on $R$-projectivity" if you're interested. $\endgroup$ – CuriousKid7 Jan 20 at 19:52

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