Let $F$ be a free $R$-module with a basis $B$. We know that $B$ satisfies the following property:

For any $R$-module $M$ and any $g:B\rightarrow M$, there exists a unique $R$-map $\varphi:F\rightarrow M$ that extends $g$.

Now suppose that $F$ is any $R$-module and $B$ is any subset of $F$. If $B$ satisfies the above property, is $B$ a basis of $F$?

I think this is true for vector spaces. If $B$ doesn't span $F$ then there is no unique extension of $g$, and if $B$ is linearly dependent then the extension might not exist at all. But I'm having trouble extending my reasoning to modules, due to the lack of division and the presence of torsion elements.

  • 2
    $\begingroup$ Have you heard of projective modules? $\endgroup$ Commented Sep 5, 2012 at 7:37
  • $\begingroup$ @SimonMarkett: No I haven't. Could you please explain further? $\endgroup$
    – wj32
    Commented Sep 5, 2012 at 7:47
  • $\begingroup$ OK, I read a bit about projective modules, but I still can't figure out what it has to do with the situation here :(. $\endgroup$
    – wj32
    Commented Sep 5, 2012 at 7:56
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    $\begingroup$ Ok, I am not necessarily saying that the answer involves projective modules. I just think that in the study of projective modules a lot of questions like yours come up. Also they are defined by inverting the lifting property of vector spaces. So I thought a) that it might interest you and b) that this is the first place to look for ideas. $\endgroup$ Commented Sep 5, 2012 at 7:59
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    $\begingroup$ Isn't this true by some sort of uniqueness of the universal property argument? As in if it satisfies the diagram, then $F$ must be the free module generated by the set $B$? $\endgroup$
    – Juan S
    Commented Sep 5, 2012 at 8:20

2 Answers 2


Juan S's comment is correct. For any set $B$ let me denote by $F(B)$ the free module on that set. In this case there is a canonical map $f : F(B) \to F$, and your hypotheses say that the induced maps

$$f^{\ast} : \text{Hom}_{R\text{-Mod}}(F, M) \to \text{Hom}_{R\text{-Mod}}(F(B), M) \cong \text{Hom}_{\text{Set}}(B, M)$$

are bijections for every $M$. By the Yoneda lemma, $f$ must be an isomorphism. Explicitly, take $M = F(B)$ in the above; then $(f^{\ast})^{-1}(\text{id}_{F(B)})$ is an inverse to $f$.


I'm pretty sure that it also works without Yoneda lemma:

Just consider $F/N$, where $N = (B)$, the R-module generated by B and a map $\psi: B \rightarrow F/N $. $\psi$ sends all $ x \in B$ to $0$, so it's the zero map.

We now extend $\psi$ in two ways:

  • set $\phi_{1}: F \rightarrow F/N$ by mapping everything to $0$. This extends $\psi$
  • let $\phi_{2}: F \rightarrow F/N$ be the canonical epimorphism. This also extends $\psi$

By the above property, the extension has to be unique, so $\phi_{1} = \phi_{2}$. Since the canonical epimorphism is then the zero map, it holds: $F = N$. For R-linear independence let $$\sum_{\text{finite}} r_{i}x_{i} = 0\;,\qquad r_{i}\in R, \:\, x_{i} \in B$$ Now let $r_{j}$ be one of the coefficients and define the map

$$\gamma_{j}:B\rightarrow R:x\mapsto\left\{\begin{array}{ll} 1, & x = x_{j} \\ 0, & x\neq x_{j}\end{array}\right.$$

Then again by above property this extends to: $F_{j}:F\rightarrow R$. And finally it holds $$0 = F_{j}(0) = F_{j}\left(\sum_{\text{finite}} r_{i}x_{i}\right) = \sum_{\text{finite}}r_{i}F_{j}(x_{i}) = r_{j}$$

So B is a basis of F.


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