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Let $R$ be a ring and $M$ a free $R$-module with basis $X$. Is it so that every $m \in M$ can be written uniquely as a linear combination of elements of $X$?

If not, in which cases is that true?

Update: In the book I'm learning from, a basis of $M$ is defined to be a linearly independent, spanning subset of $M$.

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  • $\begingroup$ There are multiple ways to define "basis", which are you using? $\endgroup$ – Jim Mar 26 '14 at 6:35
  • $\begingroup$ The answer is yes by the way, we just need to know which definition you're using in order to explain why the answer is yes. $\endgroup$ – Jim Mar 26 '14 at 6:39
  • $\begingroup$ @user138093: the definition you use is equivalent to the definition I used below.see for example: 6.51& 6.52 of the book Steps in Commutative Algebra (Second edition) $[R. Y. Sharp]$. $\endgroup$ – user 1 Mar 26 '14 at 7:07
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let $m = ∑r_i e_i = ∑s_i e_i$ then $∑(r_i-s_i) e_i=0$ so by independence we have $r_i=s_i$

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  • $\begingroup$ Don't we also need the choice of basis elements to be unique? i.e. to show that $\sum r_ie_i=\sum s_if_i \Rightarrow r_i=s_i$ and $e_i=f_i$, or am I misunderstanding something? $\endgroup$ – Christopher.L Mar 6 at 21:58
  • $\begingroup$ @Christopher.L : Well, I suppose since all $e_i, f_i$:s are in $X$, we could just rearrange and "fill out" with $r_i, s_i =0$ when needed to get both representations on the form $\sum v_ie_i$. $\endgroup$ – Christopher.L Mar 7 at 10:32

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