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I had a very basic question about free modules that I wanted to use as an ingredient in a proof but I am not sure if it is true in general.

Let $R$ be a commutative ring with multiplicative identity. Suppose $ \{ F_i \}_{i \in \mathbb{N} }$ are a countable family of free $R$-modules.

Is the countable direct sum $F_1 \oplus F_2 \oplus F_3 \oplus F_4 \oplus \ldots$ free?

If so is there an easy way to prove this?

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You need some kind of hypothesis. Take, for example, $R=k\times k$ with $k$ a field. Then the module $R/R(1,0)$ is projective, and countably many copies of it do not sum to a free one. –  Mariano Suárez-Alvarez Sep 13 '11 at 3:55
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@Mariano: I'm confused: it says that you have a family of free modules over $R$, not of projective ones. –  Arturo Magidin Sep 13 '11 at 3:57
    
Oh. I misread. Maybe there is a typo in the question? :/ –  Mariano Suárez-Alvarez Sep 13 '11 at 4:01

2 Answers 2

up vote 4 down vote accepted

Yes, the direct sum of free modules is free. One way to show it is to show it has a basis. Let $\beta_i$ be a basis for $F_i$; then $\cup \beta_i$ is a basis for the direct sum (interpret an element of $\beta_i$ as tuple that has $0$ $j$th entry for every $j\neq i$).

Another is note that a module $M$ is free if and only if there exists a set $X$ such that $M\cong R^{(X)}$, where $R^{(X)}$ is the set of functions $f\colon X\to R$ with finite support, with coordinatewise addition and scalar multiplication; then show that $$\bigoplus_{i\in I} R^{(X_i)} \cong R^{(\mathop{\amalg}\limits_{i\in I}X_i)}$$ where $\amalg$ is the disjoint union.

Yet a third way is to remember the "free module" is the left adjoint of the underlying set functor, and therefore respects coproducts; so the direct sum (coproduct) of free modules is the free module on the coproduct of their bases (this is just the abstract nonsense way of justifying the second proof mentioned above).

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Arturo: Hmm, it seems your answer is 50% better than mine. :) –  Pete L. Clark Sep 13 '11 at 3:59
    
@Pete: Not really: as I note in the third paragraph, the "third way" is really just the hand-wavy/abstract-nonsense way of justifying the isomorphism in the second one. So this is really just $2+\epsilon$ ways. (-: –  Arturo Magidin Sep 13 '11 at 4:01

Yes. In fact arbitrary direct sums of free modules are free.

This is straightforward enough to prove that I'm honestly not sure what to tell you. First you need to pick one specific definition of free module and work with that. For instance, if a free module is one which has a basis, then you can prove this by showing that the union of bases for the submodules is a basis for the direct sum. Or if your definition of a free module is one which is isomorphic to $\bigoplus_{i \in I} R$, then taking the disjoint union of the index sets $I_n$ corresponding to each free module $F_n$ will give you an index set for $\bigoplus_{n=1}^{\infty} F_n$.

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