Let $K$ be a field, $K^\times$ its multiplicative group and $I$ an infinite set. Is then $(K^\times)^{(I)} \subseteq (K^\times)^I$ a direct summand? If not, is it possible to characterize the fields for which this is true?

In any case, it's a pure subgroup. If $K$ is finite, the answer is yes. If $K$ has arbitrary roots, that is $K^\times$ is divisible, then it's also true. If $K^\times$ is the additive group of a vector space (i.e. it's elementary abelian for some prime or uniquely divisible), you can use linear algebra. If $K^\times$ is a finite direct sum of these types, then it also works; e.g. $\mathbb{R}^\times = \mathbb{Z}/2 \times \mathbb{R}^+$.

Now what about $K = \mathbb{Q}$. Here $K^\times = \mathbb{Z}/2 \oplus \mathbb{Z}^{(\mathbb{P})}$. If $I=\mathbb{N}$ and $\hom((\mathbb{Z}^{(I)})^I,\mathbb{Z})$ is countable, then it's false. But I don't know if this is true, the argument of Specker computing $\hom(\mathbb{Z}^\mathbb{N},\mathbb{Z})$ does not seem to take over. Another case would be that $K^\times$ is torsion, i.e. $K$ is an algebraic extension of $\mathbb{F}_p$ for some prime $p$, e.g. $K = \mathrm{colim}\_s \mathbb{F}\_{p^{q^s}}$ for some prime $q$ and $K^\times = \mathrm{colim}\_s \mathbb{Z}/(p^{q^s}-1)$. This is a subgroup of $\mathbb{Q}/\mathbb{Z}$, which does not have to be divisible.

I don't know an example of an abelian group $G$ such that $G^{(I)}$ is not a direct summand of $G^I$, but I'm pretty sure that there is one. But does this $G$ also arise as $K^\times$? (EDIT: I know that $G=\mathbb{Z}, I = \mathbb{N}$ does it, but $\mathbb{Z}$ is no $K^x$.) There are several characterizations1 when $G$ has the form $K^\times$ for some field $K$. Perhaps this is useful here. The whole question is motivated by the study of $K \otimes_K \otimes_K ...$ as defined here.

1R.M. Dicker, A set of independent axioms for a field and a condition for a group to be the multiplicative group of a field, Proc. London Math. Soc., 18, 1968, p.114 - 124

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    $\begingroup$ by the way, $Tor(K^*)$, the subgroup of the roots of unity, does not have to be a direct summand of $K^*$: P. M. Cohn, Eine Bemerkung über die multiplikative Gruppe eines Körpers, Arch. Math 13, p. 344 - 348 $\endgroup$ – Martin Brandenburg Jan 27 '10 at 16:30

Re: "I don't know an example of an abelian group $G$ such that $G^{(I)}$ is not a direct summand of $G^I$, but I'm pretty sure that there is one."

Let $G$ be the the integers, and $I$ a countable indexing set. If $G^{(I)}$ were a direct summand, let $P$ be a complement summand.

We arrive at a contradiction as follows: First, $P$ is isomorphic to $G^{I}/G^{(I)}$ which contains the element $(2,4,8,16,32,\ldots)$ modulo $G^{(I)}$, which (as we can peal off any of the initial terms) is a non-zero element which is infinitely divisible by 2. Second, $P$ is a subgroup of $G^{I}$, which has no infinitely divisible elements (other than zero).

I think this argument might be modified to show that the algebraic closure of a finite field will give you the counter-example you need (changing "divisibililty" to some sort of degree consideration), but I don't have a lot of time to think about it right now. I'll come back later if someone else doesn't answer your question fully.

Back now. Try the following. Let $K$ be the field obtained by adjoining to $\mathbb{Q}$ the $2^{p}$th root of each prime prime $p$ (in $\mathbb{Z}$). Let $G$ be the multiplicative group of $K$. Suppose by way of contradiction that $G^{(I)}$ is a direct summand of $G^{I}$, and let $C$ be a complement. As before, $C$ is isomorphic to $G^{I}/G^{(I)}$, and the element $(2,3,5,\ldots )$ modulo $G^{I}$ is infinitely divisible by $2$ (thinking of ``divisibility'' multiplicatively in this case--in other words, after chopping off the front, we can take square roots as many times as we want). However, I believe it is the case that there is no element of $G^{I}$ which is infinitely divisible in this sense. (I'll leave it to the experts to prove this, but I think some form of Kummer theory would suffice. But it may be difficult to prove it.) [One last edit: I think it may even be easier to look at $\mathbb{Q}(x_1,x_2,\ldots)$ and adjoin a $2^{n}$th root of $x_{n}$, and modify the example accordingly.]

  • $\begingroup$ Sorry, I already knew the example $\mathbb{Z}$ (cf. the edit in the question). But your proof is better than mine, which uses the result of Specker. The multiplicative group of an algebraically closed field is divisible, thus we get a direct summand. I'll check out your other example. $\endgroup$ – Martin Brandenburg Jan 25 '10 at 23:46
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    $\begingroup$ Gotcha. I thought that the example with $\mathbb{Z}$ is well known. I've figured out a simplification to my counter-example to your original question. Just let $K=\mathbb{Q}(x_{1},x_{2},\ldots)$. Notice that no element has $2^{n}$ roots for all $n$ (except $1$) by a simple degree argument. Then consider $(x_{1}^{2},x_{2}^{4},x_{3}^{8},\ldots)$ (modulo the appropriate subgroup). etc... $\endgroup$ – Pace Nielsen Jan 26 '10 at 0:00
  • $\begingroup$ fine, thank you! so the remaining question is: can we classify the fields $K$ for which we get a direct summand? $\endgroup$ – Martin Brandenburg Jan 26 '10 at 0:02
  • $\begingroup$ I should note that your counterexample gives many insights which fields are excluded! ... $\endgroup$ – Martin Brandenburg Jan 26 '10 at 0:07
  • $\begingroup$ Yes. I just noticed that if we let $x_{i}=2$ for all $i$, we see that $\mathbb{Q}$ also doesn't work. :-) $\endgroup$ – Pace Nielsen Jan 26 '10 at 0:10

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