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.

247 anal sex amateur anal sex anal asian gay sex anal asian hot sex anal asian picture sex anal black sex anal demo sex video anal free sex anal gay sex anal hardcore sex anal movie sex anal penetration sex video anal porn free movie sex xxx anal porn sex anal sex anal sex clip anal sex free anal sex free pic anal sex free trailer anal sex gallery anal sex guide anal sex movie anal sex photo anal sex pic anal sex picture anal sex porn anal sex position anal sex pregnant anal sex site anal sex story anal sex technique anal sex teen anal sex tip anal sex toy anal sex trailer anal sex trailer video anal sex video anal sex video clip anal sex video virgin anal sex with young girl asian anal sex best lube for anal sex black anal sex black white anal sex blonde chick thirsty for anal sex catalog anal sex anal sex china doll rough sex anal pee christine young free anal sex video deep anal sex ebony anal sex extreme anal sex first anal sex first time anal sex first time anal sex video foxy lady 4 queen of anal sex free anal porn sex movie download free anal sex free anal sex clip free anal sex gallery free anal sex movie free anal sex movie gallery free anal sex pic free anal sex picture free anal sex video free granny anal sex free pic of anal sex gay anal sex gay anal sex video guide to anal sex hard anal sex hardcore anal sex hardcore gay anal sex her first anal sex her first anal sex devon lee homemade anal sex toy hot anal sex how to have anal sex interracial anal sex kinky anal sex latina anal sex lesbian anal sex male anal sex mature anal sex painful anal sex rough anal sex sex anal teen anal sex xxl sex anal young anal sex

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

Comments are closed.