# Compatibility with multiplication of a cyclic order on a ring

Considering a linear order on the additive group of a ring is compatible with multiplication if:

• $$a < b \implies ax < bx$$ and $$xa < xb$$ for any positive $$x$$,

we could define compatibility for a cyclically ordered additive group in a similar way:

A cyclic order on the additive group of a ring is compatible with multiplication if:

• $$[a, b, c] \implies [ax, bx, cx]$$ and $$[xa, xb, xc]$$ for any positive $$x$$.

One can notice the induced cyclic order C of any linearly ordered ring L is compatible with multiplication:

• Definition of C: $$[a, b, c] \iff a < b < c \lor b < c < a \lor c < a < b$$;
• Compatibility with multiplication of L: $$a < b < c \lor b < c < a \lor c < a < b \implies ax < bx < cx \lor bx < cx < ax \lor cx < ax < bx$$ for a positive $$x$$;
• From the properties of the Natural cut of a cyclically ordered group:
$$x$$ is positive in L $$\iff x$$ is positive in C;
• Definition of C: $$ax < bx < cx \lor bx < cx < ax \lor cx < ax < bx \iff [ax, bx, cx]$$.

It looks like the only rings with non-linearly cyclically ordered additive groups (with three or more elements) compatible with multiplication in this way are $$\mathbb Z_3$$ and $$\mathbb Z_4$$. Is this correct?

My attempt to prove it:

Lemma. In a ring with cyclically ordered additive group compatible with multiplication:
if both $$a$$ and $$b$$ are positive or negative, then $$ab$$ is positive;
if one of $$a$$ or $$b$$ is positive, and another one is negative, then $$ab$$ is negative.

Proof:

1. Case $$a,b$$ are positive $$\implies ab$$ is positive:

• Definition: $$a$$ is positive $$\iff [0, a, -a]$$;
• Compatibility with multiplication on $$b$$: $$[0, a, -a] \implies [0, ab, -ab]$$;
• Definition: $$[0, ab, -ab] \iff ab$$ is positive.
2. Case $$a$$ is negative, $$b$$ is positive $$\implies ab$$ is negative:

• Definition: $$a$$ is negative $$\iff [0, -a, a]$$;
• Compatibility with multiplication on $$b$$: $$[0, -a, a] \implies [0, -ab, ab]$$;
• Definition: $$[0, ab, -ab] \iff ab$$ is positive.
3. Case $$a$$ is positive, $$b$$ is negative $$\implies ab$$ is negative:

• Same as Case 2.
4. Case $$a$$ is negative, $$b$$ is negative $$\implies ab$$ is positive:

• Definition: $$a$$ is negative $$\iff [0, -a, a]$$;
• $$-b$$ is positive;
• Compatibility with $$-b$$: $$[0, -a, a] \implies [0, ab, -ab]$$;
• Definition: $$[0, ab, -ab] \iff ab$$ is positive.

Now, back to the main statement:

• From the property of Quadrants of a cyclically ordered group:
if there are two different positive elements, then every quadrant of the group is not empty;
• Taking $$a$$ from the first quadrant, and $$b$$ from the second quadrant:
• $$(a + a)b = a(b + b)$$;
• $$a + a$$ is positive, $$b + b$$ is negative;
• $$(a + a)b$$ is positive, but $$a(b + b)$$ is negative, contradiction.

Thus, there cannot be more than one positive element in order for non-linearly cyclically ordered additive group of a ring to be compatible with multiplication.

Are there better approaches for compatibility with multiplication for rings?