is that right that free module M over R is that M can be generated by a linear independent subset A, and every element of M is a finite sum of elements of A multiplied by coefficients in R(the expression should be unique)?

is a quotient of free module free? and is a direct sum of free module free?

I think the second one is yes, but i do not know how to prove it.


  • $\begingroup$ Yes, direct sums of free modules are free. To show it, take a basis of each free module, the union of these bases will be a basis for the direct sum. However, every module is a quotient of some free module (take a generating set of your module $M$, form the free module $F$ on that set, and consider the map $F\to M$) so the quotient of a free module certainly need not be free. $\endgroup$ – Ben West Apr 8 '17 at 5:07
  • $\begingroup$ As for the first question, $\Bbb Z$ is a free module over $\Bbb Z$. Can you think of quotients that are not free in that case? And without reading too closely, you seem to have a valid definition of a free module there. $\endgroup$ – Arthur Apr 8 '17 at 5:09

Yes, an $R$-module $M$ is free if it has a basis, i.e., a linearly independent generating set.

No, the quotient of free modules need not be free. Consider the $\mathbb Z$-module $\mathbb Z$. This is clearly free and $1$ is a free generator. Similarly, $2\mathbb Z$ is a free $\mathbb Z$-module, and $2$ is a free generator, but the quotient $\mathbb Z/2\mathbb Z$ is not a free $\mathbb Z$-module. In particular, every subset of $\mathbb Z/2\mathbb Z$ is linearly dependent.

On the other hand, direct sums behave much better. If $M_\alpha$ is free for each $\alpha$ in some indexing set $I$, then the direct sum $\bigoplus_{\alpha\in I}M_\alpha$ is free. Take as basis the disjoint union $\bigsqcup_{\alpha\in I} \mathcal B_\alpha$ where $\mathcal B_\alpha$ is a basis for $M_\alpha$.

  • $\begingroup$ I think the key point here is that even though for $R$-modules $M$ and $N$ where $M$ is free and a $R$-module epimorphism $f:M\to N$ it takes a basis set of $M$ to a spanning set of $N$, the set may not be linearly independent (as your example shows). Isn't it? $\endgroup$ – user 170039 May 21 at 4:50
  • $\begingroup$ @user170039: Every $R$-module $N$ can be realized as the quotient of a free $R$-module $M$, or equivalently, for every $R$-module $N$, there is an epimorphism (surjective $R$-module homomorphism) $f\colon M \to N$, with $M$ a free $R$-module. Thus, what you say is true, that a basis is certainly mapped to a generating set, and the question is whether this generating set is linearly independent. $\endgroup$ – Alex Ortiz May 21 at 17:18

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