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?