Extensions of an infinite product of copies of Z by Z - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T15:15:05Z http://mathoverflow.net/feeds/question/40499 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/40499/extensions-of-an-infinite-product-of-copies-of-z-by-z Extensions of an infinite product of copies of Z by Z Xandi Tuni 2010-09-29T17:13:02Z 2010-10-05T13:08:58Z <p>The question is simple:</p> <p><em>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?</em></p> <p>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.</p> <p>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 $I$.</p> http://mathoverflow.net/questions/40499/extensions-of-an-infinite-product-of-copies-of-z-by-z/40586#40586 Answer by Laurent Moret-Bailly for Extensions of an infinite product of copies of Z by Z Laurent Moret-Bailly 2010-09-30T08:29:10Z 2010-10-05T13:08:58Z <p>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 $\mathrm{Ext}^1(P,\mathbb{Z})=\mathrm{Ext}^1(P/L,\mathbb{Z})$. </p> <p>EDIT: the last formula is wrong, see Martin's and Steve's comments below.</p> <p>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 $\mathrm{Ext}^1(\mathbb{Q},\mathbb{Z})=\widehat{\mathbb{Z}}/{\mathbb{Z}}\neq0$.</p>