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I was recently reading this question about free modules. In this post, the definition of a free module is given as

An $R$-module $F$ is said to be free on the set $A \subset F$ if for every nonzero $x \in F$, there exist unique $r_1, r_2, \cdots, r_n \in R - \{0\}$ and unique $a_1, a_2, \cdots, a_n \in A$ such that $x = r_1a_1 + r_2a_2 + \cdots + r_na_n$.

This definition is taken from the text by Dummit and Foote. In this post, Alex M. made the following comment:

Notice that, according to your definition, if $a \in A$, then $Ra$ is a free module on A over R, which is clearly not what you want. Something is missing from it, maybe a maximality condition?

  1. How does the $a \in A \implies [Ra$ is free on A$]$ follow? It is obvious that any $x \in Ra$ can be represented as $x = ra$, but how do you know this is unique?

  2. Why is this "clearly not what you want"?

  3. Is Alex M. correct that there is something missing from this definition?

EDIT

As Eric Wofsey pointed out to me, there are additional comments in the original post (which I failed to read) that clears up the confusion about the definition given in Dummit and Foote. Hence questions (2) and (3) are unnecessary, because the (additional) comments show Alex M. is not contending that this definition is lacking . But this still leaves question (1)...

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  • $\begingroup$ Read the rest of the comments in the same thread: Alex M was misreading the definition. $\endgroup$
    – Eric Wofsey
    Commented Mar 14, 2017 at 17:34
  • $\begingroup$ @EricWofsey Should have seen that earlier. So that takes care of (2) and (3), but leaves (1). I will edit my question. $\endgroup$
    – Sam Y.
    Commented Mar 14, 2017 at 18:04

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If you omit the requirement that $A\subseteq F$ in the definition, then if $F$ is free on $A$, so is any submodule of $F$. Indeed, the definition of "$F$ is free on $A$" just says that for any $x\in F$ a certain condition holds (without any further reference to the set $F$), so if you take a submodule you just have fewer elements $x$ for which the condition needs to hold. In particular, this applies to a submodule of the form $Ra$.

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  • $\begingroup$ It seems like everything you wrote would be true even if you keep the requirement that $A \subset F$? $\endgroup$
    – Sam Y.
    Commented Mar 14, 2017 at 23:32
  • $\begingroup$ Yes, but the only submodule of $F$ containing $A$ is $F$ itself (since every element of $F$ is a linear combination of elements of $A$). $\endgroup$
    – Eric Wofsey
    Commented Mar 14, 2017 at 23:34
  • $\begingroup$ True. But suppose we require $A \subset F$. Then everything you said about the implication $[F$ is free on $A] \implies [$ any submodule is free on $A]$ still holds, right? $\endgroup$
    – Sam Y.
    Commented Mar 14, 2017 at 23:43
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    $\begingroup$ I think I see your point now...we can't have "any submodule free on A" because D/F require the module to contain $A$, and the only such module is $F$. Thanks for explaining things in multiple ways! $\endgroup$
    – Sam Y.
    Commented Mar 14, 2017 at 23:47

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