If possible, how could one prove that every short exact sequence $0 \to A \xrightarrow f B \xrightarrow g C \to 0$ of vector spaces (here $A$, $B$ and $C$) splits without using any basis of $A$, $B$ or $C$. I was not able to exhibit a morphism $h \colon B \to A$ such that $h \circ f=Id_A$ without considering a basis.
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6 Answers
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You can prove this using Zorn's lemma on the pairs (D,h) where D is a subspace of B and h is a partial section.
And yes, you do need Zorn's lemma: without it, there may exist vector spaces none of whose nontrivial subspaces has a complement (Herrlich, Axiom of Choice, LNM 1876, Disaster 4.43).
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So, for example, $0 \to \mathbb{Q} \to \mathbb{R} \to \mathbb{R}/\mathbb{Q} \to 0$ ... the inclusion of the rationals into the reals, these are all vector spaces over the rationals. Can you write down your split? Not constructively!
answered Dec 1, 2009 at 13:15
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$\begingroup$ this is really a great example! $\endgroup$ Commented Mar 15, 2010 at 2:11
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One can show the short exact sequence by proving that a field is a semisimple ring—alhough one could argue that somewhere hidden in that proof a basis is considered...
I cannot imagine a negative answer to your question in any other form than a model of ZF without the axiom of choice, so that there are vector spaces without bases, and in which maybe there are short exact sequences of vector spaces which do not split.
answered Dec 1, 2009 at 5:33
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Without using any basis of A, B or C specifically, but rather using the existence of bases for all vector spaces, we may observe that all vector spaces are free as modules; and all free modules are projective.
Finally, the projectivity property gives us a splitting by lifting the identity map on C along the surjection from B by projectivity of C. This gives us a splitting, thus splitting the entire sequence.
ETA With the arguments in the comments to Ben Websters answer, it is pointed out that without AC, and without at least the existence of bases, things can fail badly. Obviously, if we disallow AC, this answer fails as badly.
answered Dec 1, 2009 at 17:42
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Apply $\mathrm{Hom}(C,-)$ to your short exact sequence. It remains exact, so the identity map from $C$ to $C$ has at least one preimage. In fact, the splittings are exactly its preimages.
I would guess the problem with the axiom of choice is that you need to actually choose one of them. Or it hidden somewhere else?
answered Dec 1, 2009 at 17:50
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$\begingroup$ How do you know that it remains exact? $\endgroup$ Commented Dec 1, 2009 at 18:03
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2$\begingroup$ A more explicit form of that comment is: Under AC, C is free so Hom(C,-) is exact. But if AC fails sufficiently badly, C may not have a basis, and now suddenly it's not so clear that Hom(C,-) is exact I guess. $\endgroup$ Commented Dec 1, 2009 at 19:15
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$\begingroup$ Fair enough. I'm blissfully unconcerned about axiom of choice issues. $\endgroup$– Ben Webster ♦Commented Dec 2, 2009 at 4:01
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1$\begingroup$ The obvious rule of thumb is: any construction where the standard argument involves Zorn or AC will almost certainly not work without AC (i.e. there will be models of ZF where it fails), becuase if it worked without Zorn then everyone would say "use Zorn, although there is a non-Zorn argument". Every vector space has a basis because of Zorn, and you've never heard anyone say "...and it can be done without Zorn", so you can be sure there will be some crazy model of ZF containing a vector space with no basis. $\endgroup$ Commented Dec 2, 2009 at 21:54
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$\begingroup$ I have a vague idea, but still where does the
Extcomputation fail without AC? $\endgroup$ Commented Dec 2, 2009 at 23:31
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It should feel natural that splittings and bases go together; in the opposite direction, suppose you only consider short exact sequences $0\rightarrow 0 \rightarrow V\rightarrow k^\beta\rightarrow 0$ where by $k^\beta$ I mean of course the free vector space on the set $\beta$ taken with its standard basis $\{x\in\beta\}$. In this case a splitting is (essentially the same thing) as a basis for $V$. Why would you want to avoid that sort of thing?
answered Dec 3, 2009 at 23:27
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1$\begingroup$ But you can only prove that such a short exact sequence exists for all $V$ iff you can pove that all vector spaces are free (obviously), and you can only do that if they are. That depends on your assuming AC or not. $\endgroup$ Commented Dec 3, 2009 at 23:41
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$\begingroup$ true! I think this emphasizes something about the original question, but I've too much of a head-cold to think just what right now... $\endgroup$ Commented Dec 4, 2009 at 1:57