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](https://mathoverflow.net/questions/12503/z-ni-is-a-direct-summand-of-z-ni). 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](https://mathoverflow.net/questions/11767/infinite-tensor-products).


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



Split powers of the multiplicative group of a field
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).




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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.




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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.]


