I have the following definition of free module:

A free module $F$ is an $R$-module together with a set map $i:X \to F$ such that given any other $R$-module $M$ and a set map $f:X \to M$ there exists a unique $R$-module homomorphism $\overline{f}:F \to M$ such that $\overline{f} \circ i = f$

However I see that very often $i$ is taken as an inclusion and $X$ as a subset of $F$.

Please can you clarify me what kind of identification is done here?

Note: this definition was taken from here


This definition does not give new free modules when $i$ is not an inclusion. Say $i(a)=i(b)$ where $a,b\in X$. Let $M$ be a nonzero module and $f:X\rightarrow M$ be a map with $f(c)$ for $c\in X$ with $c\neq b$ and say $f(b)\neq 0$. Then $f$ does not extend to a map $g:F\rightarrow M$ since $f(a)=0$ but $f(b)\neq 0$.

In other words, there is no free module $(F,X,i)$ with $i$ is not an inclusion.

  • $\begingroup$ what you mean is that necessarily $i$ is an inclusion and $X$ a subset of $F$ to have a good definition? $\endgroup$ – Javier Mar 8 '18 at 13:49
  • $\begingroup$ Your definition is also a definition of a free module. But as you say, it just hides the fact that there is no free module with $i$ is not an inclusion. $\endgroup$ – Levent Mar 8 '18 at 13:53
  • 1
    $\begingroup$ To be a bit more clear : your argument shows that for every free module $i$ must be injective. Once you know that, you can prove that WLOG one can assume that it actually is an inclusion. $\endgroup$ – Arnaud D. Mar 8 '18 at 13:59
  • $\begingroup$ @ArnaudD. Yes, of course. Thanks for pointing out. I used the terms inclusion and injective in a wrong way. $\endgroup$ – Levent Mar 8 '18 at 14:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.