Paddock - almost the conventional Field

A paddock is a linear context, V = (: (V: a+b ←a :(:V|)) ←b :) = transpose(V), equipped with a symmetric binary operator, construed as mutiplication on V, for which (: (V: u.v ←u :(:V|)) ←v :) is a linear isomorphism ({scalings of V}:|(:V|)), i.e. for each v in (:V|), (V: u.v ←u |(:V|)) is linear and commutes with every linear (V::(:V|)). The symmetry condition says that u.v = v.u for all v, u in (:V|). The fact that all scalings are linear gives us distributivity, i.e. v.(u+w) = (v.u)+(v.w) and (v+u).w = (v.w)+(u.w) for any u, v, w in V. Multiplication is an isomorphism between a linear context and its own collection of scalings.

Now the identity, (:V|) itself, is a scaling of V, so multiplication must map some element of (:V|) to (:V|); this element of V is called 1 and 1.v = v = v.1 for every v in V.

Addition is symmetric and supports cancellation; to what degree can we infer cancellation or completeness for multiplication ?

For any linear context V, composition of linear (V:|(:V|)), restricted to scalings, yields a multiplication on {scalings of V}. I need to check that it's an isomorphism; that it's a symmetric binary operator is part of the definition of scalings. With S = {scalings of V}, we have (:V|) in S so, for any scaling, s, of S we have (:V|) in (:s|) = S and s(:V|) in (|s:) = S, giving us (S: s(:V|) ←s |{scalings of S}) which is linear, since scalings s, t of S have (s+t)(:V|) = s(:V|)+t(:V|). Now, given n in S, s(V)&on;n = n&on;s(V) = (V: n(s((:V|),v)) ←v |(:V|)) and I need to show that n(s(V,v)) = s(n). I know n&on;V = V&on;n, that s commutes with every linear (S:|S) and that s respects addition in S. For any scaling, s, of {scalings of V}, we have V as the identity in {scalings of V} The putative isomorphism, with, is (S: (S: n&on;m ←m |S) ←n |S) and maps S←(:V|) - yes, (:V|) is in S and S is in {scalings of S}. Its inverse is readily enough constructed: (S: s(S) ←s |{scalings of S}).

I expect {scalings of V}, for any linear context V, to form a paddock using composition, within linear (V:|(:V|)), as

the above supercedes the following

Given two symmetric binary operators ({(V:|(:V|))}| :V), one (V itself) construed as addition, the other as multiplication, I'll describe V (with the resulting linear structure) as a paddock precisely if

• a.(b+c) = a.b + a.c for all a, b, c in (:V|) [a.k.a. multiplication distributes over addition]
• addition is cancellable,
• except for any additive identities there may be,
  • multiplication is cancellable,
  • multiplication is complete.

Additive completeness would guarantee an additive identity, thus exercising the exception clause, so two varieties of paddock spring particularly to mind:

those without an additive identity

hence not complete, unless empty. I'll describe these as positive paddocks: lacking an additive identity, these cannot have additive inverses. In a positive paddock, multiplication forms a group: e.g. the positive number line, whether rational, real or surreal.

those which are additively complete

which are known as fields when non-empty. Here addition is a group.

Note that the empty paddock, whose addition and multiplication are both the empty relation, is both additively complete and positive: it is also very boring.

Linearity

Now, a paddock is trivially a linear context (it supports symmetric addition): and distributivity can be expressed as saying that, for each a in V, times(a) = (V| v.a ←v :(:V|)) is linear. The symmetry of multiplication guarantees that outputs of times commute with one another: how about with other linear (V:|(:V|)) ?

Given linear (V:f:(:V|)), consider f(a.(b+c)) = f(a.b) + f(a.c) = a.(f(b)+f(c))