Math 2.0

Field theory and polynomials Definition

Quasifield

Builds on GroupDivision ring

Definition

Let QQ be a set with two binary operations ++ and ⋅\cdot, and let Q∗=Q∖{0}Q^* = Q \setminus \set{0}. The triple (Q,+,⋅)(Q, +, \cdot) is a (left) quasifield if the following hold.

  1. (Q,+)(Q, +) is a group with identity element 00.
  2. a⋅0=0⋅a=0a \cdot 0 = 0 \cdot a = 0 for all a∈Qa \in Q.
  3. (Q∗,⋅)(Q^*, \cdot) is a loop: there is 1∈Q∗1 \in Q^* with 1⋅a=a⋅1=a1 \cdot a = a \cdot 1 = a for all a∈Q∗a \in Q^*, and for a,b∈Q∗a, b \in Q^* the equations a⋅x=ba \cdot x = b and y⋅a=by \cdot a = b have unique solutions in Q∗Q^*.
  4. a⋅(b+c)=a⋅b+a⋅ca \cdot (b + c) = a \cdot b + a \cdot c for all a,b,c∈Qa, b, c \in Q.
  5. For all a,b,c∈Qa, b, c \in Q with a≠ba \ne b, the equation a⋅x=b⋅x+ca \cdot x = b \cdot x + c has exactly one solution x∈Qx \in Q.

A right quasifield is defined in the same way with (b+c)⋅a=b⋅a+c⋅a(b + c) \cdot a = b \cdot a + c \cdot a in clause 4 and x⋅a=x⋅b+cx \cdot a = x \cdot b + c in clause 5.

Remarks

Example: every division ring DD is a quasifield. Clauses 1 to 4 are axioms of a division ring. For clause 5, by distributivity ax=bx+ca x = b x + c is equivalent to (a−b)x=c(a - b) x = c, which has the unique solution x=(a−b)−1cx = (a - b)^{-1} c since a≠ba \ne b.

Non-example: the integers ℤ\Z with the usual operations. Clause 3 fails, since 2⋅x=12 \cdot x = 1 has no solution in ℤ∖{0}\Z \setminus \set{0}.

One can prove that the additive group (Q,+)(Q, +) of a quasifield is always abelian. A quasifield in which both distributive laws hold is a semifield, and a quasifield whose multiplication is associative is a near-field (on the matching side).

Reading path

To understand this page from the ground up, read these 3 pages in order. The foundations come first.

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  1. Division ring concept
  2. Group concept
  3. Quasifield definition

Further reading

  1. Web https://en.wikipedia.org/wiki/Quasifield
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DefinesQuasifield
UsesGroup, Division ring

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