TITLE: Which fields have multiplicative group isomorphic to additive group times Z/2Z?
QUESTION [20 upvotes]: Let $K$ be a field, $K_{*}$ its multiplicative group and $K_{+}$ its additive group. As Richard Stanley notes in this answer, the only field for which $K_{+} \simeq K_{*} \times \mathbb{Z}/2\mathbb{Z}$ is $K=\mathbb{Z}/2\mathbb{Z}$. Which fields have $K_{*} \simeq K_{+} \times \mathbb{Z}/2\mathbb{Z}$? For instance, $K=\mathbb{R}$ is an example.

REPLY [5 votes]: Here's another unorderable example.
Consider nonstandard models; let $p$ be an infinite prime such that $(p-1)/2$ is relatively prime to every standard prime, and let $F$ be the internal finite field $\mathbf{F}_p$. The additive group is cyclic of order $p$ and the multiplicative group is the product of $\mathbb{Z} / 2 \mathbb{Z}$ and the cyclic group of order $(p-1)/2$.
By the four square theorem, this field has elements $a,b,c,d$ such that $a^2 + b^2 + c^2 + d^2 \equiv -1 \pmod{p}$.
Viewed externally, the additive group and the multiplicative group of squares are both uniquely divisible abelian groups of the same cardinality, and assuming we can arrange for the cardinality to be non-pathological, we conclude that $F_+ \cong F_\times \times \mathbb{Z} / 2 \mathbb{Z}$.
Furthermore, since $-1$ is a sum of squares, $F$ can't be ordered.