I am reading this notes.

Definition 1:

Let $R$ be a commutative ring with $1$. Let $S$ be a set. A free $R$-module $M$ on generators $S$ is an $R$-module $M$ and a set map $i:S\rightarrow M$ such that, for any $R$-module $N$ and any set map $f : S \rightarrow N$, there is a unique $R$-module homomorphism $\bar{f} : M \rightarrow N$ such that $\bar{f}\circ i = f : S \rightarrow N$. The elements of $i(S)$ in $M$ are an $R$-basis for $M.$

Definition 2: $M$ is a free $R$-module if $M$ has a basis.

We know that Definition 1 is equivalent to Definition 2. So we can choose anyone of them to define free module. If we choose the Definition 2 then it is very easy to give examples of free module, for example, the $R$-módule $_{R}R$ is free, but we have to work a lot more in order to show that free modules really exist by using the Definition 1.

My question is: Is there a practical or theoretical advantage to choose the Definition 1 instead of the Definition 2?

PS I am self-studying Module Theory, and I find more natural the Definition 1.

  • 1
    $\begingroup$ This is the "category theory" definition of free module. It is sometimes preferred because it generalizes to other notions of freeness. (Specifically, it is defining "free module" as an adjoint of the "forgetful' function from the category of $R$-modules to the category of sets.) $\endgroup$ – Thomas Andrews Jan 27 '15 at 16:57
  • $\begingroup$ For some details of @ThomasAndrews' answer, see for instance en.wikipedia.org/wiki/Free_object $\endgroup$ – Andreas Caranti Jan 27 '15 at 16:58
  • 2
    $\begingroup$ From the notes directly above the defintion: "The following definition is an example of defining things by mapping properties, that is, by the way the object relates to other objects, rather than by internal structure." Please clarify you question in view of this. $\endgroup$ – quid Jan 27 '15 at 16:59
  • $\begingroup$ And @quid's quote is exactly what category theory is - trying to understand the universe of objects we are studying in terms of maps only, as well as we can. $\endgroup$ – Thomas Andrews Jan 27 '15 at 17:02
  • $\begingroup$ @quid: I tried to improve my question. $\endgroup$ – spohreis Jan 27 '15 at 20:52

In practice one often views the map $i$ (as in definition $1$) as an inclusion; then definition $1$ defines a free $R$-module $M$ in terms of how you can define $R$-linear maps from $M$ to another $R$-module: just assign the values on $S$ as you wish, and you can linearly extend in a unique way. This is used all the time. Definition $1$ also makes the connection between the terms "free $R$-module", "free commutative $R$-algebra" (aka polynomial ring over $R$), "free group", etc. very clear: it basically the same universal property. This leads to the categorical point of view, e.g., that the functor $F:\mathbf{Set}\rightarrow\mathbf{R-Mod}$ (assigning to a set $S$ the free $R$-module $F(S)$ generated by $S$) is left adjoint to the forgetful functor $U:\mathbf{R-Mod}\rightarrow\mathbf{Set}$.

Also, the first definition is often used to define important objects in various branches of mathematics, for instance the singular chain groups. Let $X$ be a topological space and $S$ be the set of all continuous maps from the standard $n$-simplex $\Delta_n$ to $X$. Then the free $\mathbf{Z}$-module generated by $S$ is by definition the abelian group of singular $n$-chains of $X$ (denoted $S_n(X)$).

  • $\begingroup$ This is all true but one could also define a basis for an algebraic structure $A$ as a set $X$ such that $F(X) \cong A$. To really see the point of the universal property you'd probably want to look at other examples of adjoints. $\endgroup$ – Ibrahim Tencer Feb 6 '15 at 21:22

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