# Abstract Nonsense

## Short Exact Sequences Ending in Free Modules Split

Point of Post: We show, in later language, that every free module is projective. Or, in different terms, every short exact sequence sending in a free module splits.

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Motivation

We have seen that every short exact sequence of vector spaces splits and since free modules are “like” vector spaces, we should be able to make a similar statement for them. Indeed, we go one step further (which we could have done for vector spaces, but didn’t). We shall show that whenever we have a short exact sequence of the form

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$0\to L\to M\to F\to 0$

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whre $F$ is free, then the sequence splits. The proof shall be very similar to the case for vector spaces, basically being a consequence of defining an $R$-map on a basis.

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Short Exact Sequence Ending in Free Modules Split

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So, let’s just get right down to the theorem at hand:

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Theorem: Let $R$ be a ring and

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$0\to L\to M\xrightarrow{f}F$

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a short exact sequence of $R$-modules. Then, this chain splits.

Proof: By the splitting lemma it suffices to show that there is a backmap $g:F\to M$ for $f$.  To do this, fis a basis $\{x_\alpha\}_{\alpha\in\mathcal{A}}$ for $F$. Define a map $g:F\to M$ by defining $g(x_\alpha)$ to be any element $L$ which maps to $x_\alpha$ and extending by linearity. Clearly this is an $R$-map $F\to M$ and we note that for any $\displaystyle \sum_{\alpha\in\mathcal{A}}r_\alpha x_\alpha\in F$ one has that

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$\displaystyle f\left(g\left(\sum_{\alpha\in\mathcal{A}}r_\alpha x_\alpha\right)\right)=\sum_{\alpha\in\mathcal{A}}r_\alpha f(g(x_\alpha))=\sum_{\alpha\in\mathcal{A}}r_\alpha x_\alpha$

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and so $f\circ g=\text{id}_F$, and so $g$ is the desired backmap for $f$. The conclusion follows. $\blacksquare$

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References:

[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Blyth, T. S. Module Theory. Clarendon, 1990. Print.

November 20, 2011 -

1. […] since is free, we know this sequence splits and so there exists some backmap and we know that and since is free we […]

Pingback by Submodules of Free Modules Need Not be Free Unless Ring is a PID (pt. II) « Abstract Nonsense | November 21, 2011 | Reply

2. […] In this post we give the definition of  a type of module which, among others, shall occupy a fair amount of our attention in the math (specifically the homological algebra) to come. There are, as there were for tensor products, a multitude of (seemingly) disparate ways to motivate the usefulness/interestingness of projective modules, some of which I understand better than others. A motivation which is somewhat high on the usefulness scale but which is, perhaps, less so on the scale of interestingness is that projective modules are precisely the answer to the following question: “we know that given a module the covariant Hom functor is left-exact, for which modules is the functor actually exact?” Another question for which projective modules answer quite nicely is the question as to which modules have the property that any short exact sequence ending in them splits (for example, we have [perhaps it's slightly opaque from the presentation in that post] seen that this is true for free modules). […]

Pingback by Projective Modules (Pt. I) « Abstract Nonsense | February 7, 2012 | Reply

3. Typo in the end: computation should be “f(g(…”.
Helped with a Hatcher problem! Thanks

Comment by Sicong Zhang | October 10, 2012 | Reply

• No problem man, glad it could help!

Comment by Alex Youcis | October 10, 2012 | Reply