TITLE: Is it possible to produce a field with three operations?
QUESTION [13 upvotes]: A Field $F$ is defined such that:

There is a first operation, say $+$ (addition), that is closed, associative and commutative over $F$, and has an identity element $0 \in F$ and an inverse element $-x \in F$ for each $x \in F$ (that is, $(F, +)$ is a commutative group);  
There is a second operation, say $\cdot$ (multiplication), that is closed, associative and commutative over $F \setminus \{0\}$, and has an identity element $1 \in F \setminus \{0\}$ and an inverse $1/x \in F \setminus \{0\}$ for each $x \in F \setminus \{0\}$ (that is, $(F\setminus \{0\}, \cdot)$ is a commutative group);  
The second operation distributes over the first operation.

My question is this: 
Is it possible to define a third operation, say $*$, for $F$ such that $(F\setminus \{0,1\}, *)$, is a commutative group and $*$ distributes over multiplication?
In other words, is it possible to have $(F, +, \cdot, *)$ where both $(F,+, \cdot)$ and $(F\setminus \{0\},\cdot, *)$ are fields?
My gut feeling is no.
In all fields I am familiar with $(\mathbb{Q,R,C,Z}_p)$, the multiplication is provably equivalent to that field's analytic continuation of iterated addition, but the analytic continuation of iterated multiplication (that is, exponentiation) is provably not a commutative group operation on $(F \setminus \{0,1\})$ (and catastrophically so, it fails on all counts! It's not even well-defined!)
However, the coincidence that multiplication is equivalent to iterated addition feels like just that, a coincidence, as the definition of field makes no claims about the relation between the two operations other than the distributive property.
Is my gut feeling correct, or am I right to be skeptical?

REPLY [4 votes]: Yes, it is possible, at least in the finite case.
Suppose that $K_1$ and $K_2$ are two fields with the additive group of $K_1$ isomorphic to the multiplicative group of $K_2$.
If $K_1$ and $K_2$ are finite, then, since the multiplicative group of a finite field is cyclic, the field $K_1$ must be a prime field. Let the orders of $K_1$ and $K_2$ be $p$ and $q^n$ respectively, so that $p=q^n-1$. If $n=1$, then $p$ and $q$ must be two and three respectively. If $n>1$, then $q$ must be two and $p$ must then be a Mersenne prime.
Hence, $K_1$ and $K_2$ could be one of the following:

$GF(2)$ and $GF(3)$
$GF(2^n-1)$ and $GF(2^n)$, where $2^n-1$ is a Mersenne prime (of course, $n$ must also be prime)

In the infinite case, all the elements of $K_2$ other than zero and one must have the same multiplicative order. If this order were finite (lets say its $n$), then it would contradict the fact that a nonzero polynomial over a field cannot have more roots than its degree (in this case, the polynomial of interest is $x^n-1$). Hence, $K_2$ must have a torsion-free multiplicative group, which implies that $K_1$ and $K_2$ must have characteristic zero and two respectively.
But I do not know whether any infinite examples exist.