It is a common exercise in algebra to show that there does not exist a field $F$ such that its additive group $F^+$ and multiplicative group $F^*$ are isomorphic. See e.g. this question.

One of the snappiest proofs I know is that, if we suppose for a contradiction they are, then any isomorphism sends solutions of the equation $2x = 0$ in the additive group to solutions of the equation $y^2 = 1$ in the multiplicative group. Depending on whether the characteristic of $F$ is or isn't 2, the former has either $|F|$ or $1$ solution(s), while the latter has $1$ or $2$ solutions, respectively. There is no field for which these numbers agree, so $F^+ \not\cong F^*$ ever.

One might now ask whether there is a pair of fields, $E$ and $F$, for which $E^+ \cong F^*$ as groups.

Clearly $\def\GF#1{\mathrm{GF}(#1)}\GF2^+ \cong \GF3^*$, $\GF3^+ \cong \GF4^*$, and in general if $p$ is a prime and $p+1$ is a prime power, then $\GF p^+ \cong \GF{p+1}^*$.

You can see that this characterizes the situation in the positive characteristic case, from the same equation trick above: if $\def\c{\operatorname{char}}\c E = 2$ and $\c F \ne 2$, we can make $|E| = 2$ and get a solution. Else we must have $\c E \ne 2 = \c F$. If $\c E = c \ne 0$, then elements of $E$ must get mapped to $c$-th roots of unity in $F$, and there can be at most $c$ of those.

This leaves the case where $\c E = 0$, for which there are no finite fields. In fact, none of the cases above permit any infinite fields, either. This brings me to my question:

Do there exist infinite fields $E$ and $F$ such that $E^+ \cong F^*$?

I believe the answer is no, and it seems unlikely that such an isomorphism would exist, but I can't make heads or tails of it, really. Here's what I have.

As above, I can show that if $E$ and $F$ are infinite, and $\phi: E^+ \to F^*$ is an isomorphism, then we may assume $\c E = 0$—so WLOG it is an extension of $\mathbb Q$—and $\c F = 2$.

Every element of $E$ has infinite additive order, so every element of $F^*$ has infinite multiplicative order, and there are no roots of unity except $1 = \phi(0)$. However, if $a \ne 1$ in $F$, then $a$ has a $k$-th root $\phi(\frac1k \phi^{-1}(a))$ for all $k$, since $E \supseteq \mathbb Q$.

  • $\begingroup$ Slightly tangential: For a positive real number $b\ne1$, the function $\phi(x) = b^x$ is an isomorphism from the additive group of real numbers to the multiplicative group of positive real numbers. Inverse being logarithm to the base $b$. So it is not onto the muliplicative group, but just fails to be surjective; the image has index 2. $\endgroup$ – P Vanchinathan Sep 8 '16 at 5:35
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    $\begingroup$ I think Puiseux series over $\mathbb{F}_2$ might work, but I haven't thought through the details. $\endgroup$ – Eric Wofsey Sep 8 '16 at 7:32

A pair of fields like this exists.

As noted in the question, one may assume that $E$ is a field of characteristic zero. Therefore $E^+$ is an infinite, torsion-free, divisible, abelian group; i.e. an infinite rational vector space. Moreover, any infinite rational vector space is isomorphic to the additive group of some field of characteristic zero. Thus the question reduces to this: is there an infinite field $F$ whose multiplicative group of units is a rational vector space?

Such a field is constructed in

Adler, Allan On the multiplicative semigroups of rings. Comm. Algebra 6 (1978), no. 17, 1751-1753.

Here is the first line of George Bergman's Math Review of this paper:

``Let $\mathbb Q$ denote the additive group of the rational numbers. It is shown that there exists a field whose multiplicative group is isomorphic to $\mathbb Q^{\aleph_0}$, but none whose multiplicative group is isomorphic to $\mathbb Q$.''


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