# Extensions of an infinite product of copies of Z by Z

The question is simple:

Let $P$ be an infinite direct product of copies of $\mathbb Z$. Do there exist any nontrivial extensions $$0 \to \mathbb Z \to E \to P \to 0$$ in the category of commutative groups?

In other words, I am asking whether the group $\mathrm{Ext}^1(P,\mathbb Z)$ is trivial. The problem here is of course that the group $P$ is not a free group.

Already a funny thing happens with $\mathrm{Hom}(P,\mathbb Z)$. For any finite or infinite index set $I$, the canonical evaluation map $$\bigoplus_{i\in I}\mathbb Z \to \mathrm{Hom}\Big(\mathrm{Hom}\Big(\bigoplus_{i\in I}\mathbb Z,\:\mathbb Z \Big),\:\mathbb Z \Big) \cong \mathrm{Hom}\Big(\prod_{i\in I}\mathbb Z,\:\mathbb Z \Big)$$ is an isomorphism! That is a nontrivial statement (due to??), whose proof is not a formality at all. Replacing $\mathbb Z$ by, say, $\mathbb Z/p\mathbb Z$, the corresponding statement is wrong for infinite $I$.

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The statement is due to Ernst Specker ("Additive Gruppen von Folgen ganzer Zahlen", Port. Math. 9, 1950). –  Martin Brandenburg Sep 29 '10 at 17:32
The asserted canonical isomorphism works only if there are no measurable cardinals $\leq |I|$. In this generality, it's due to Los (in TeX that should be {\L}o\'{s}, but I have no idea how to produce that here). Specker did the case of countable $I$. –  Andreas Blass Sep 29 '10 at 19:59
I think John Irwin and I once computed `$\text{Ext}^1(P,{\mathbb Z})$ and found that it's very large. I'll look for my notes on this when I get home tonight. Meanwhile, have you looked in Fuchs's book "Infinite Abelian Groups"? The answer might be there. –  Andreas Blass Sep 29 '10 at 20:02
If P is the countable direct product of $\mathbb{Z}$ and S the countable direct sum, then P/S looks like (countable product) Ext($\mathbb{Q}/\mathbb{Z}$,$\mathbb{Z}$) x (uncountable sum) $\mathbb{Q}$. So Ext(P,Z) = Ext(X,Z) x (uncountable product) $\mathbb{R}$ where X = (countable product) (product over primes p) $\mathbb{Z}_p$. –  Steve D Sep 29 '10 at 20:51
@Andreas: A simple way is to cut & paste from Wikipedia: "Łoś". :-) –  Hans Lundmark Sep 29 '10 at 21:31

Here is a complete answer; I think it is more or less what Steve wrote in his comment, except I don't understand the appearance of $\mathbb{R}$ there. If $I$ is the infinite index set, let $L=\mathbb{Z}^{(I)}\subset P$ be the obvious free submodule. Then $\mathrm{Ext}^1(P,\mathbb{Z})=\mathrm{Ext}^1(P/L,\mathbb{Z})$.
Now $P/L$ has a big divisible subgroup $D$, whose inverse image in $P$ consists of maps $I\to\mathbb{Z}$ converging to zero in $\widehat{\mathbb{Z}}$ (the profinite completion of $\mathbb{Z}$). (For instance, if $I=\mathbb{N}$ take the sequence $n\mapsto n!$). Since $P/L$ is torsion-free (imediate), $D$ is a nonzero $\mathbb{Q}$-vector space. Since $D$ is divisible it is a direct summand of $P/L$; hence, $P/L$ admits $\mathbb{Q}$ as a direct summand. But it is well known (and easy to see) that $\mathrm{Ext}^1(\mathbb{Q}/\mathbb{Z},\mathbb{Z})\cong\widehat{\mathbb{Z}}$, hence $\mathrm{Ext}^1(\mathbb{Q},\mathbb{Z})=\widehat{\mathbb{Z}}/{\mathbb{Z}}\neq0$.
The long exact sequence together with the result of Specker shows that $Ext^1(P/L,Z) \to Ext^1(P,L)$ is surjective with kernel isomorphic to $P/L$. Thus I don't understand $Ext^1(P,Z) = Ext^1(P/l,Z)$. –  Martin Brandenburg Sep 30 '10 at 12:29
Oops, I'm afraid you're right. So it remains to look at the image of $P$ into $\mathrm{Ext}^1(P,\mathbb{Z})$ (easy, I think) and then see if we can exhibit something else (not so easy?). –  Laurent Moret-Bailly Sep 30 '10 at 15:50
$\mathbb{R}$ is Ext($\mathbb{Q}$,$\mathbb{Z}$). –  Steve D Oct 1 '10 at 5:13
If we're just showing Ext(P,$\mathbb{Z}$) is non-zero, we're done, since Ext(P/S, $\mathbb{Z}$) is strictly bigger than $|\mathbb{R}|$, while |P/S| = $|\mathbb{R}|$. –  Steve D Oct 1 '10 at 5:35
@Steve: Thanks, I hadn't realized that $\widehat{\mathbb{Z}}/\mathbb{Z}\cong\mathbb{R}$. –  Laurent Moret-Bailly Oct 5 '10 at 13:10