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I am trying to prove that every module is isomorphic to a free quotient. Is the following a proof for that statement?

Let $M$ be an $R$-module. Choose $S \subset M$ such that $S$ generates $M$ with the generating set $\{s_i \}$. If $s_n$ is a generator that can be obtained from the other, i.e if $s_n = \sum r_is_i$, then remove $s_n$ from $\{s_i \}$. Now consider $\phi: (S) \to M$ given by $s \mapsto s$, then $\phi$ is surjective and $(S)/ \ker \phi \cong M$. Furthermore $(S)/ \ker \phi$ is free since we have killed every linear combination that equals $0$. Is this correct?

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  • $\begingroup$ I tink you misunderstood the notion of "a free quotient". It was presumably intended to mean "a quotient of a free module." You interpreted it to mean "a quotient that is itself free". That's perhaps the most natural interpretation of the wording, but it can't be what you're supposed to prove because it's false in general. $\endgroup$ – Andreas Blass Oct 1 '15 at 8:55
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As an example why this does not work, consider the $\mathbb{Z}$-module $M = \mathbb{Z}/4\mathbb{Z}$. It is generated by $S = \{1 + 4 \mathbb{Z}\}$ and as there is only one element in $S$ we cannot do any elimination. But then $(S) = M$ which is definitely not free.

From a more general point, note that you are trying to take a submodule $N = (S)$ of $M$, restrict the identity map $M \to M$ to a map $N \to M$ and want this to be surjective. Well, this will only work if $N = M$. Also, it will always have a trivial kernel, so using this to get $M$ as the quotient of some free module is bound to fail.

But there is a way to improve your idea to get the result. Instead of taking the submodule generated by $S$ in $M$, you can take $R^S$, the free $R$-module with basis $S$. If $S$ is a generating set (you do not have to do any elimination process here), then the natural map $R^S \to M$ induced by the restriction of the identity $M \to M$ to a map $S \to M$ will give you the result. (Note that you can even take $S = M$ and then you do not have to think about generating sets at all).

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  • $\begingroup$ the construction of $R^{S}$ was the one I wanted to obtain with the elimination process. How do you know that it exists otherwise? How do you obtain it? $\endgroup$ – Ralf Oct 1 '15 at 11:37
  • $\begingroup$ You can define $R^S = \{f : S \to R \:|\: \{s \in S \: | \: f(s) \neq 0\} \text{ is finite }\}$ and show that this $R$-module (with pointwise addition and scalar multiplication) has as a basis $\{f_s \:|\: s \in S\}$ with $f_s(s') = 1$ for $s = s'$ and $f_s(s') = 0$ if $s \neq s'$. $\endgroup$ – Matthias Klupsch Oct 1 '15 at 12:45
  • $\begingroup$ Hmm ok. Just one questions, how does the map $R^{S} \to M$ look like? I takes a function and maps it to what in $M$? $\endgroup$ – Ralf Oct 1 '15 at 12:59
  • $\begingroup$ The map is given by mapping $f \in R^S$ to the element $\sum_{s \in S}f(s) s$. Note that because $f$ is non-zero on only finitely many $s \in S$, this is indeed well-defined. Alternatively you might as well use that $\{f_s \:|\: s \in S\}$ is a basis and map $f_s$ to $s$ which would give you the same map. $\endgroup$ – Matthias Klupsch Oct 1 '15 at 13:01

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