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 $K^\*$ is a finite direct sum of these types, then it also works; e.g. $\mathbb{R}^\* = \mathbb{Z}/2 \times \mathbb{R}^+$.

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 characterizations^{1} 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.

^{1}R.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