# Show from the axioms: Addition in a quasifield is abelian

According to wikipedia a quasifield is an algebraic structure $(Q,+,\cdot)$ such that

1. $(Q,+)$ is a group. (As usual, we denote its identity element by $0$.)
2. $(Q\setminus\{0\},\cdot)$ is a loop. (Its identity element will be denoted by $1$.)
3. Left distributive law: $a\cdot (b+c) = a\cdot b + a\cdot c$ for all $a,b,c\in Q$.
4. Each equation $a\cdot x = b\cdot x + c$ with $a,b,c\in Q$ and $a\neq b$ has a unique solution $x\in Q$.

Now the Wikipedia article claims that

One can prove that the axioms imply that the additive group (Q,+) is abelian.

My question is: How?

For ordinary rings, we are in a similar situation. It is possible to drop the commutativity of the addition from the axioms, since it follows from the other axioms: $a + b = b + a$ is implied by

$$a + a + b + b = (a\cdot 1 + a\cdot 1) + (b\cdot 1 + b\cdot 1) \overset{\text{L}}{=} a\cdot (1 + 1) + b\cdot (1+1)\\\overset{\text{R}}{=} (a+b)\cdot (1+1) \overset{\text{L}}{=} (a+b)\cdot 1 + (a + b)\cdot 1 = a + b + a + b.$$

This is not too bad in our situation, since the proof doesn't rely on the associativity of the multiplication (which we don't have in a quasifield). However, the problem is that both distributive laws have been used (tagged with L and R). More precisely, step R is not covered by the above axioms. I wonder if we can find an alternative justification for that equality based on property 4, which looks a bit related to the missing right distributive law.

## 1 Answer

This answer follows M. Bommireddy et al, Arch. Math. 42, 573-576 (1984).

Consider $a,b\in Q$. Even in a non-abelian group, $0+b=b+0$, so the only non-trivial case is $a\neq 0$. Since $(Q\setminus \{0\},\cdot)$ is a loop, there exists $s\in Q\setminus \{0\}$ such that $sa=b+a-b$. We want to prove $s=1$. If $s\ne 1$, by axiom 4, there is an $x_1$ with $$1\cdot x_1 = sx_1+b\qquad (1)\\ x_2 := x_1+a \qquad(2) \\ \Rightarrow sx_2 = sx_1+sa \qquad(3)\\$$

Now: $$sx_1+b+a-b=_{(1)}x_1+a-b=_{(2)}x_2-b \qquad (4)$$

but by the definition of $s$ also: $$sx_1+b+a-b=sx_1+sa=s (x_1+a)=_{(2)}sx_2 \qquad (5)$$

From (4) and (5), we have $$1\cdot x_2=sx_2+b$$

Comparing with (1) and using axiom 4, either $s=1$ or $x_1=x_2$. The former proves the theorem; the latter applied to (2) yields $a=0$, a contradiction.

• Thanks for that great answer, and for digging up that paper! The combination "Bommireddy", "Reddy" and "Reddy" of the authors is quite funny. – azimut Mar 28 '14 at 14:35
• To make your answer more readable, you may want to add a justification why you assume $a\neq 0$, and to replace "let $sa := b + a - b$" by something like "Since $(Q,\cdot)$ is a loop, there is a $s\in X\setminus\{0\}$ with $sa = b + a - b$." – azimut Mar 28 '14 at 14:44
• @azimut: Thanks, I've added the answer accordingly, fixing an incorrect eq reference along the way. – Ansgar Esztermann Mar 28 '14 at 21:02