Which fields have multiplicative group isomorphic to additive group times Z/2Z?

Question

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.

Answer 1

EDIT: The answer below is incorrect: in addition to every element having odd roots and exactly one of $a$ and $-a$ having a square root, there must be no roots of unity in $K$ besides $\pm 1$. This still holds for any ordered field with all odd roots and positive square roots, but my argument for an unorderable example does not work. I do not know whether there is any example that cannot be ordered.

---

It is easy to see that any such field must have characteristic 0. In characteristic 0, this holds iff the multiplicative group has the form $\mathbb{Z}/2\oplus V$, where $V$ is a $\mathbb{Q}$-vector space. Having such a splitting is equivalent to the following condition: for all $a\in K_*$, $a$ has a $p$th root for all odd $p$ and exactly one of $a$ and $-a$ has a square root. It's easy to construct fields that satisfy this. For instance, an ordered field satisfies this iff it has odd roots of all elements and square roots of all positive elements.

However, there are also examples that cannot be ordered. In fact, any field of characteristic 0 that does not contain a square root of $-1$ is contained in such a field. Indeed, you can just take a maximal algebraic extension $K$ that does not contain a square root of $-1$. First, I claim every odd degree polynomial has a root over $K$. Let $f\in K[x]$ be a polynomial of minimal odd degree that does not have a root. Then by maximality of $K$, $K[x]/(f)$ must have a square root of $-1$. That is, there exist polynomials $g$ and $h$ such that $h^2+1=fg$, and $h$ can be chosen to have degree strictly less than $f$. But then $g$ must have odd degree less than $f$ and $K[x]/(g)$ also contains a square root of $-1$. Thus $h$ cannot have a root in $K$, contradicting minimality of the degree of $f$.

Second, for any $a\in K$, either $a$ or $-a$ is a square. Suppose $-a$ is not a square; by maximality it suffices to show $K[\sqrt{a}]$ does not contain a square root of $-1$. But if $(b+c\sqrt{a})^2=-1$, then either $b^2=-1$ or $c^2=-1/a$, contradicting that $-a$ and $-1$ are not squares in $K$.

Answer 2

A subfield of $\mathbb R$ has this property if it is closed under $\exp$ and $\log$.

Given any finite set of numbers,the set they generate under the operations of $+,\times, -, /,\exp,\log$ is always countable. Starting with $0$ and $1$, we get a countable subfield of $\mathbb R$ with this property. We can add an element of $\mathbb R$ not in this subfield, getting a larger countable field, then repeat, getting uncountably many fields with this property

We can also make fields larger than $\mathbb R$ with this property - the nonstandard reals for any ultrafilter will provide an example.

The most obvious question this kind of mucking about won't answer is whether there is an unordered field with this property. Clearly all exponentials are perfect squares, and vice versa. So the question is essentially whether the sum of two exponentials must be an exponential.

Answer 3

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.

Answer 4

I think one can construct an unorderable example as follows.

Let $E$ be the field of real constructible numbers. On the one hand, the only roots of unity in $E$ are $\pm1$. On the other hand, if $\zeta_n$ denotes a primitive root of $1$, the extension $E(i,\zeta_n)/E(i)$ is abelian and has no subextension of degree 2, hence $[E(i,\zeta_n):E(i)]$ is odd.

Let $F$ be an unorderable extension of $E$ with no roots of unity besides $\pm1$ (e.g., the fraction field of $E[x,y]/(x^2+y^2+1)$), and let $K$ be a maximal algebraic extension of $F$ with respect to this property. We need to show that every $a\in K$ has a $p$th root in $K$ for every odd prime $p$, and $a$ or $-a$ has a square root.

Recall that for prime $p$, a polynomial $x^p-a$, $a\in K$, is either irreducible or has a root in $K$. [Its roots are $\alpha,\alpha\zeta_p,\dots,\alpha\zeta_p^{p-1}$, hence the constant coefficient of its proper factor would be of the form $\alpha^d\zeta_p^i\in K$ for some $0<d<p$ and $i$; as $(p,d)=1$ and $\alpha^p\in K$, this implies $\alpha\zeta_p^j\in K$ for some $j$.]

I claim that if $p$ is odd, $x^p-a$ has a root in $K$. If not, then the extension $K(\alpha)$, $\alpha^p=a$, contains a root of unity $\zeta_n\ne\pm1$. By the above-mentioned property, $[K(\alpha):K]=p$ is prime, hence $K(\alpha)=K(\zeta_n)$ is a cyclotomic extension, and therefore normal. But then it contains the conjugates of $\alpha$, and in particular, $\zeta_p$; however, $[K(\zeta_p):K]<p$, hence $\zeta_p\in K$, a contradiction.

Second, assume that $\alpha=\sqrt a\notin K$. Then $K(\alpha)=K(\zeta_n)$ for some $n>2$. Since $[K(\zeta_n):K]=2$, $[K(\zeta_n,i):K(\zeta_n)]\le2$, $[K(i):K]=2$, and $[K(\zeta_n,i):K(i)]$ is odd, we must have $i\in K(\zeta_n)$, i.e., $K(\alpha)=K(i)$. Writing $\alpha=u+iv$ with $u,v\in K$, we have $u^2-v^2=a$ and $2uv=0$. As $v\ne0$, we get $u=0$ and $v^2=-a$, i.e., $\sqrt{-a}\in K$.