9
$\begingroup$

In $\mathbb Z$-Mod, is $\mathrm{Ext}^1(\prod\mathbb Z,\mathbb Z)=0$?

I think it is not equal to 0, but I do not know how to prove it.

  • First I select a short exact sequence: $0\rightarrow\mathbb Z \rightarrow \mathbb Q\rightarrow \mathbb Q/\mathbb Z \rightarrow 0$. Then we get $\mathrm{Hom}(\prod \mathbb Z,\mathbb Q)\stackrel{f}\rightarrow\mathrm{Hom}(\prod \mathbb Z,\mathbb Q/\mathbb Z) \rightarrow \mathrm{Ext}^1(\prod \mathbb Z,\mathbb Z) \rightarrow 0$. Is $f$ is epic? I can not see the Hom between infinite product to $\mathbb Q$ or $\mathbb Q/\mathbb Z$. Thanks for you help!
$\endgroup$
2
  • $\begingroup$ Your short exact sequence is an injective resolution. $\endgroup$ Commented Aug 9, 2017 at 12:12
  • $\begingroup$ @MarianoSuárez-Álvarez yeah $\endgroup$
    – Jian
    Commented Aug 9, 2017 at 12:12

1 Answer 1

6
$\begingroup$

I presume that by $\prod\mathbb{Z}$ you mean an infinite product. Since every infinite product contains a countable product, and $\text{Ext}^1(-,\mathbb{Z})$ is right exact, to prove that $\text{Ext}^1(\prod\mathbb{Z},\mathbb{Z})\neq0$ we may as well assume that it's a countable product.

I'll just write $\Pi$ for the product of countably many copies of $\mathbb{Z}$.

By a theorem of Baer, $\Pi$ is not a free abelian group, but if it were, then $\text{Ext}^1(\Pi,\mathbb{Z})$ would be zero. So you can expect any proof to be at least as hard as Baer's theorem.

One proof of Baer's theorem uses a famous result of Specker that describes $\text{Hom}(\Pi,\mathbb{Z})$, and in particular shows that it is countable. This can be used to prove your statement.

Although $\text{Hom}(\Pi,\mathbb{Z})$ is countable, $\text{Hom}(\Pi,\mathbb{Z}/2\mathbb{Z})$ is uncountable (in fact, of cardinality $2^{2^{\aleph_0}}$), since $\Pi/2\Pi$ is a vector space of infinite dimension over the field $\mathbb{F}_2$ of two elements, and so has uncountably many homomorphisms to $\mathbb{Z}/2\mathbb{Z}$.

Now apply the functor $\text{Hom}(\Pi,-)$ to the short exact sequence $0\to\mathbb{Z}\stackrel{\times2}{\to}\mathbb{Z}\to\mathbb{Z}/2\mathbb{Z}\to0$, producing an exact sequence $$\text{Hom}(\Pi,\mathbb{Z})\to\text{Hom}(\Pi,\mathbb{Z}/2\mathbb{Z})\to\text{Ext}^1(\Pi,\mathbb{Z}).$$ Since the first term is countable, but the second term is uncountable, the third term is non-zero.

Probably there are other proofs based on different approaches to Baer's theorem.

$\endgroup$
0

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .