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
  • 2
    $\begingroup$ 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 $\mathbb Q$.''

EDIT: 8/21/20

In response to the August 18 question of mr_e_man below, I asked my library to get a scan of Adler's paper. Adler's construction of a field whose multiplicative group is isomorphic to the direct power $\mathbb Q^{\aleph_0}$ goes like this:

Let $\mathbb F_{2^n}$ be Galois field of order $2^n$. Let $P$ be the set of primes. The desired field can be taken to be any field $\mathbb K$ that is an ultraproduct of the form $\prod_{\mathcal U} \mathbb F_{2^p}$ over a nonprincipal ultrafilter $\mathcal U$ defined on $P$. Let's show that the multiplicative group of $\mathbb K$ is isomorphic to $\mathbb Q^{\aleph_0}$.

Up to isomorphism, $\mathbb Q^{\aleph_0}$ is the unique torsion-free divisible abelian group of cardinality continuum. The multiplicative group $\mathbb K^{\times}$ is isomorphic to the abelian group $\prod_{\mathcal U} \mathbb F_{2^p}^{\times}$. The factors in this ultrproduct have sizes $2^2-1< 2^3-1 < 2^5-1 < \cdots$, so the sizes are a strictly increasing countable sequence. This is enough to prove that the ultraproduct has size continuum. (I am asserting that any nonprincipal ultraproduct of countably many finite sets of strictly increasing size has cardinality continuum.)

If $p\neq q$, then $\gcd(|\mathbb F_{2^p}^{\times}|,|\mathbb F_{2^q}^{\times}|)=\gcd(2^p-1,2^q-1)=1$. Thus, for any $m\leq \textrm{min}(2^p-1, 2^q-1)$, at most one of the groups $\mathbb F_{2^p}^{\times}$ has $m$-torsion. In fact, the map $x\mapsto x^m$ must be a bijection on all $\mathbb F_{2^p}^{\times}$ except possibly finitely many of them. In the ultraproduct $\prod_{\mathcal U} \mathbb F_{2^p}^{\times}=\mathbb K^{\times}$ the maps $x\mapsto x^m$, $m\in\mathbb Z^+$, must all be bijections. This is enough to show that $\mathbb K^{\times}$ is torsion-free and divisible.

  • $\begingroup$ Is the construction too complicated to describe here? Also, does $\mathbb Q^{\aleph_0}$ indicate the direct sum (sequences with only finitely many terms) or the direct product (arbitrary sequences)? $\endgroup$ – mr_e_man Aug 18 '20 at 23:16
  • 1
    $\begingroup$ @mr_e_man: I can access the Math Review, but not the paper, so I can't say more than what I wrote in my answer above. But, regarding the second question, it is shown that any torsion-free divisible abelian group of infinite rank is isomorphic to the multiplicative group of a field in: Contessa, Mott, Nichols, Multiplicative groups of fields. Advances in Commutative Ring Theory, Lect. Notes Pure Appl. Math., vol. 205, Fez 1997, Dekker, New York (1999), pp. 197-216. Hence you can take $\mathbb Q^{\aleph_0}$ to mean either product or sum. $\endgroup$ – Keith Kearnes Aug 19 '20 at 2:01
  • $\begingroup$ @mr_e_man: see August 21 edit. $\endgroup$ – Keith Kearnes Aug 22 '20 at 5:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.