I am looking for a couple of examples of a $\mathbb Z$-module $M \neq 0$ with $$ \mathrm{Hom}_{\mathbb Z}(M, \mathbb Z) = 0 \\ \mathrm{Ext}^1_{\mathbb Z}(M, \mathbb Z) = 0 $$

If $M$ was finitely generated this would not be possible since $M$ would be the direct sum of a free part and a torsion part.

To put this into context: I would like to realize $M$ as the first homology group of a CW complex $X$. My ultimate goal would be to construct a complex $X$ such that its first homology group is nonzero but all cohomology groups (except in degree 0) vanish. The above constraints arise by looking at the exact sequence of the universal coefficient theorem in cohomology.

  • $\begingroup$ The question can be reformulated to asking whether there exist an abelian group $M$ such that $\mathbb Q \to \mathbb Q/\mathbb Z$ is an $M$- colocal isomorphism, i.e., $\hom (M, \mathbb Q)\to \hom (M, \mathbb Q/\mathbb Z)$ is an isomorphism. As the answer indicates, this problem seems to be very sensitive to the underlying set theory. $\endgroup$ – Justin Young May 23 '13 at 13:31

One may ask if $\mathrm{Ext}^1(M,\mathbb{Z})=0$ implies that $M$ is free, this is the famous Whitehead problem. Shelah has shown that it is independent from ZFC. It is false in certain models of ZFC, but true for example in Gödel's constructible universe L. But if $M$ is free with basis $B \neq \emptyset$, then of course $\hom(M,\mathbb{Z}) \cong \mathbb{Z}^B \neq 0$. This means that for example in L there is no example.


I dare say there does not exist such $M$. Indeed, $\mbox{Ext}^1_\mathbb{Z}(M,\mathbb{Z})$ arises from the short exact sequence $$ 0 \to F_1 \stackrel{r}{\to} F_0 \to M \to 0 $$ (where $F_0$ and $F_1$ are free) as the defect of exactness of the $\mbox{Hom}(-,\mathbb{Z})$ functor : $$ 0 \to \mbox{Hom}(M,\mathbb{Z}) \to \mbox{Hom}(F_0,\mathbb{Z}) \to \mbox{Hom}(F_1,\mathbb{Z}) \to \mbox{Ext}^1_\mathbb{Z}(M,\mathbb{Z}) \to 0 $$ Now if $\mbox{Ext}^1_\mathbb{Z}(M,\mathbb{Z})=0$ and $\mbox{Hom}(M,\mathbb{Z})=0$, then this reduces to $$ 0\to \mbox{Hom}(F_0,\mathbb{Z}) \to \mbox{Hom}(F_1,\mathbb{Z}) \to 0 $$ which means that $\mbox{Hom}(r,\mathbb{Z}):\mbox{Hom}(F_0,\mathbb{Z}) \to \mbox{Hom}(F_1,\mathbb{Z})$ is an isomorphism, in particular it's injective, which means that $r$ is surjective, and hence $M=0$.

  • 1
    $\begingroup$ While $Hom(-,X)$ does map epimorphisms to monomorphisms, it can do that to things that aren't epimorphisms as well, so your observation isn't enough to infer the conclusion. $\endgroup$ – Hurkyl May 23 '13 at 18:00
  • $\begingroup$ But here $F_0,F_1$ are free. Doesn't this help somehow? $\endgroup$ – Martin Brandenburg May 23 '13 at 21:24
  • $\begingroup$ Also note that in the end you do not use that the Ext term vanishes, you only use that Hom is zero. The conclusion that $M=0$ cannot be right. $\endgroup$ – Jim May 24 '13 at 10:49

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