# Does there exist a field $(F,+,*)$ so that $(F,+) \cong (F^*,*)$?

This question occurred to me earlier today.

I can see that if the field has a unit, then there is an element of multiplicative order $2$, namely $-1$. Thus if there was an isomorphism $(F,+) \cong (F^*,*)$ then the characteristic of the field would have to be divisible by two, so that it must be of characteristic two. It would also have to be an infinite field, otherwise $N= N-1$. So I am looking in particular for some infinite field of characteristic 2, or for some other group invariants that would help.

Another possible lead is that $(F^*, *) \subseteq Aut(F, +)$. Does the exponent of a group impact the exponent of its automorphism group? Or something along those lines.

Other than that I am stuck, however. Does anyone know if this is an accessible problem to me? I only know the basics of field theory, extensions and Galois theory. Does anyone have an answer (or hint) that they would like to share?

Edit: Daniel Fischer answered my question.

What about the case when $R$ is a commutative ring and we are trying to compare $(R,+)$ with the group of units of $R$?

• Of course we can have $(F,+)\cong (G^\times,\cdot)$ if we allow different fields: If $p=2^n-1$ is a Mersenne prime, let $F=\mathbb F_p$ and $G=\mathbb F_{2^n}$. This uses that the multiplicative group of a finite field is cyclic. – Hagen von Eitzen Jan 10 '14 at 22:05
• @AreaMan: By the way, this is problem 10B, 3) in Halmos Problems for Mathematicians, Young and Old. – Watson Jul 24 '16 at 14:10

Since you already ruled out all characteristics $\neq 2$, it remains to rule out characteristic $2$.

In $(F,+)$, every element $\neq 0$ has order $2$, thus, if we had $(F,+) \cong (F^\ast, *)$, we would have $x^2 = 1$ for all $x \neq 0$. But in a field of characteristic $2$, $x^2 -1 = (x-1)^2$ has only one root, $1$.

Thus $(F,+) \not\cong (F^\ast,*)$ for all fields $F$.

Regarding the addendum, a commutative ring whose group of units is isomorphic to its additive group, that is possible. One example uses the three finite fields $\mathbb{F}_2,\, \mathbb{F}_3,\,\mathbb{F}_4$. The multiplicative group of $\mathbb{F}_k$ is isomorphic to the cyclic group $C_{k-1}$ for $k = 2,3,4$. The additive group of $\mathbb{F}_4$ is isomorphic to $C_2^2$, and hence to the multiplicative group of $\mathbb{F}_3^2$, so

$$R = \mathbb{F}_2 \times \prod_{k=0}^\infty \left(\mathbb{F}_3^{2^k}\times \mathbb{F}_4^{2^k}\right)$$

has isomorphic additive group and group of units:

$$(R^\ast,*) \cong \{1\} \times \prod_{k=0}^\infty \left(C_2^{2^k}\times C_3^{2^k}\right) \cong C_2\times C_3 \times C_2^2 \times C_3^2 \times C_2^4 \times C_3^4\times \dotsc,$$

and

$$(R,+) \cong C_2 \times \prod_{k=0}^\infty \left(C_3^{2^k}\times C_2^{2^{k+1}}\right) \cong C_2 \times C_3 \times C_2^2\times C_3^2\times C_2^4 \times C_3^4\times\dotsc\,.$$

• Thanks! It seems that sorts of weird objects can be constructed by absorbing blemishes into infinite products... – Lorenzo Najt Jan 11 '14 at 16:18
• The example given here for rings is easier. – Watson Feb 3 '17 at 10:31