$\DeclareMathOperator\Ext{Ext}\DeclareMathOperator\Hom{Hom}$I'm self-studying homological algebra and got stuck in my attempt to prove the following:

Let $A$ be a fixed abelian group. Then A is a free abelian group if $\Ext^{1}_{\mathbf Z}(A, F) \cong 0$ for every free abelian group F.

I've attempted to solve the exercise by making long exact sequences from suitable short exact ones (for example we know that there is a short exact sequence $ 0 \to F_1 \to F_0 \to A \to 0$ with $F_1$ and $F_0$ free), and by using that for every abelian group $B$ we know that $\Ext^{2}_{\mathbf Z}(A,B) \cong 0 $ and $\Ext^{0}_{\mathbf Z}(A,B ) \cong \Hom_{\mathbf Z}(A,B)$... but without luck so far.


1 Answer 1


$\DeclareMathOperator\Ker{Ker}$Let $F=\Bbb Z^{\oplus A}$ and $\pi:F\to A$ be the canonical homomorphism. Then we have an exact sequence $$\{0\}\to\Ker\pi\to F\to A\to\{0\}$$ By assumption this splits, hence $A$ is a projective (because it's a direct summand of the free abelian group $F$), hence free, abelian group.

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    $\begingroup$ It's important to notice that this is really specific to abelian groups, that is $\mathbb{Z}$-modules, and relies on the fact that $\mathbb{Z}$ is a principal ring. Over a general commutative ring $R$, you could only say that $A$ is a projective $R$-module. $\endgroup$ Commented Jan 19, 2020 at 10:17
  • $\begingroup$ I don't fully understand @Fabio Lucchinis answer (how does he conclude that $A$ is projetive from the one split exact sequence?), but his suggestion got me on the right track to a proof that I do understand: 1) From the assumption it follows easily that $Ext^{1}_{\mathbf Z}(A, B) \cong 0$ for every abelian group $B$. 2) Now consider a short exact sequence $0 \to X \to Y \to A \to 0$. As $Ext^{1}_{\mathbf Z}(A, X) \cong 0$ we know that every extension of $A$ by $X$ splits. So the SES splits. Hence $A$ is projective pr. definition, hence free (as $\mathbf Z$ is a PID} $\endgroup$
    – Kimarokko
    Commented Jan 19, 2020 at 10:27
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    $\begingroup$ @Kimarokko: I added an explanation of $A$ projective in the answer. $\endgroup$ Commented Jan 19, 2020 at 10:32
  • $\begingroup$ @Captain Lama You can't conclude that $A$ is projective in a similar situation over a general commutative ring $R$, because $\ker \pi$ is not necessarily free (or projective) in this situation. $\endgroup$ Commented Jan 19, 2020 at 16:25

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