# If $\operatorname{Ext}^{1}_{\mathbf Z}(A, F) \cong 0$ for all free groups F, then A is free?

$$\DeclareMathOperator\Ext{Ext}\DeclareMathOperator\Hom{Hom}$$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.

$$\DeclareMathOperator\Ker{Ker}$$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.
• 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. Commented Jan 19, 2020 at 10:17
• 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} Commented Jan 19, 2020 at 10:27
• @Kimarokko: I added an explanation of $A$ projective in the answer. Commented Jan 19, 2020 at 10:32
• @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. Commented Jan 19, 2020 at 16:25