Whitehead problem

I was reading Hilton and Stammbach’s A Course in Homological Algebra, when I spotted this rather forlorn passage:

Of course, if A is free, Ext(A,Z) = 0, but it is still an open question whether, for all abelian groups A, Ext(A,Z) = 0 implies A free.

It is forlorn because we now know that we’ll never know: this is the Whitehead problem, and in 1973 Saharon Shelah proved that it is independent of the axioms of set theory.

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3 thoughts on “Whitehead problem

  1. Alas, for we shall never know whether the number two has a square root. This is the “irrationality of sqrt(2)” problem, and some smartass greek proved it to be independent of the axioms of field theory.

    More interestingly, this is one of the few problems which is resolved in the opposite way than you’d expect. There are essentiall two main directions these problems fall in – there’s the constructible heierarchy and largeish fragments of set theory with proper forcing. Typically both of these will resolve independent problems (at least those you get in ‘normal’ mathematics) and do so in opposite ways. Normally the constructible heirarchy says pathological examples exist and proper forcing says that they don’t.

    In this particular case (if my memory serves me correctly) you can use the diamond principle (which holds in the constructible heierarchy) to prove that any such group is free, and there’s a group (which is perfectly well defined in normal set theory) which you can use Martin’s axiom for aleph_1 to show that it has Ext(G, Z) = 0.

    It’s been a while since I looked at this, so I might be misremembering, but I remember having been caught out by this before.

  2. I’m not committed to the idea that we’ll truly never know, but it was my initial reaction.

    Does the group that’s perfectly well defined in normal set theory turn out to be free assuming CH?

  3. Well, my point is that saying “We’ll never really know” is silly. We know the answer. The answer is that it depends on the underlying set theory. Better yet, we’ve got two natural cases in which we know *how* it depends on the underlying set theory. What more do you want? :-)

    Looking up the result in Fremlin’s “Consequences of Martin’s axiom” I find the following statement (paraphrased):

    There is a nonfree abelian group G of cardinality aleph_1 such that if Martin’s axiom for aleph_1 holds (actually something ever so slightly weaker, but I can’t be bothered to make the distinction) such that if H is an abelian group then every surjective homomorphism f : H -> G with countable kernel splits. I forget exactly how Ext works, but this seems to imply that Ext(G, Z) = 0.

    The exact construction of G is a bit unpleasant. It’s not hard, but it would require me to remember various things in set theory which I don’t at the moment. It involves creating a chain of countable abelian groups G_t such that for t

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