1
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.