I am wondering if this two definitions of free module are equivalent:

Definition 1: A module X over a commutative ring R is called a free module if there is a subset $E\subset X$ such that every element of $x\in X$ can be written as

\begin{equation} x = \sum_{b\in E} \alpha_b.b \end{equation}

in which $\alpha_b\in R$ and $\alpha_b = 0$ but for finitely many $b\in E$. And in the equation above we have $x=0$ if and only if $\alpha_b=0$ for all $b\in E$.


Definition 2: A module X over a commutative ring R is called a free module if there is a subset $E\subset X$ such that for every $R$-module $Y$ we have that any function $f:E\rightarrow Y$ can be extended in a unique way to a homomorphism $\tilde{f}:X\rightarrow Y$.

I was able to prove that (1)$\Rightarrow$(2), but I'm having problems in showing that the opposite direction holds, this is one way I tried to start:

We want to show that for every $x\in E$ we can write it as in Definition 1, so define

$$ V = \{x\in X\ |\ \text{$x$ can't be written as a combination of elements of $E$} \} $$

And by using the property of definition 2 I try to show that $V$ has to be an empty set, but I can't find the right function to prove this, any hints?

  • $\begingroup$ Your definition of $\;V\;$ is lacking. It needs to include that the combination is unique ... $\endgroup$
    – DonAntonio
    Mar 19, 2018 at 15:18
  • $\begingroup$ @DonAntonio I was thinking of first proving that they can all be written as a combination, and then prove the uniqueness, you think it's easier to do it all together? $\endgroup$
    – D18938394
    Mar 19, 2018 at 15:20
  • 2
    $\begingroup$ In definition 2 you must include uniqueness of the extending linear map $\bar f$. $\endgroup$ Mar 19, 2018 at 15:21
  • $\begingroup$ Yes, thanks, I forgot to include that, already edited it. $\endgroup$
    – D18938394
    Mar 19, 2018 at 15:23

1 Answer 1


In order to prove $2\Rightarrow 1$, first note that $$X=\sum_{e\in E}Re$$ for $N=\sum_eRe$ is an $R$-submodule of $X$, and if $\varphi:X\to X/N$ is the projection onto the factor module, then $\varphi(e)=0$ for all $e\in E$ hence, by uniqueness requirement in Definition 2, we have $\varphi=0$, that's $N=X$.

Consequently, each $x\in X$ can be written in the form $$x=\sum_{e\in E}x_ee$$ where $x_e\in R$ and $x_e=0$ for almost every $e\in E$.

To prove uniqueness, note that by Definition 2, for all $e\in E$ there exists one and only one $R$-linear map $\pi_e:X\to R$ such that $\pi_e(e)=1$ and $\pi_e(e')=0$ for all $e'\in E$ with $e'\neq e$.

Consequently, if $x_e\in R$, $x_e=0$ for almost every $e\in E$ and $\sum x_ee=0$, then \begin{align} 0=\pi_e\Bigl(\sum x_ee\Bigr)=x_e \end{align} for all $e\in E$, and this proves Definition 1.


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