I know that a finite field can be easily encoded into and decoded from a finite Paige loop, which can be defined as a Moufang loop that is simple and isn't a group. However, the finite Paige loop of a field of order $q$ has about $q^7$ elements. Is there any more efficient way to encode a field into a type of structure specified by axioms such that every structure of that type can be uniquely decoded?
2026-03-30 23:11:38.1774912298
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What is the most efficient way to "encode" a finite field into a structure that has only one binary operation?
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Each element of the new set is a pair of a point and a line in the Desarguesian projective plane over the field. Define a binary operation on two points of this set as the element corresponding to the line passing through both of them. Since Desarguesian planes are self-dual, this also corresponds to the operation given by the intersection of two lines. It is simple to define the axioms. There is one step that I'm missing here, but somehow uniquely define which point corresponds to which line. Each field generates a unique plane-structure this way. There should be a simple way to decode a plane-structure into a field. If anyone can complete these two steps, we would get a structure with about $q^2$ elements.
Here you go:
Let $(F,+,\cdot)$ be a field. Define $T=A\times B$ where $A$ is the abelian group $(F,+)$ and $B$ is the commutative monoid $(F,\cdot)$.
The operation in $T$ encodes both operations of $(F,+,\cdot)$ simultaneously, and you can "decode" them separately by projecting onto the first and second coordinates.
Now, perhaps you mean that I'm supposed to give a set of axioms of some sort of algebra and that whatever such object is chosen, it has to give rise to a unique field. I do not have any scheme in mind for that. (One could just define products of pairs of commutative monoids and demand the two coordinates interoperate on the diagonal the way a field has to, but that's hardly pretty.)
What I'm describing above at least accomplishes the goal of encoding and being able to be decoded from the encoded state, and apparently only has $q^2$ elements and one operation, which seems to be very "efficient" as far as how many elements it needs.