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An ordered ring is a (not necessarily commutative) ring with a total order $≤$ respecting the operations:

  • if $a ≤ b$ then $a + c ≤ b + c$,

  • if $0 ≤ a$ and $0 ≤ b$ then $0 ≤ ab$ (equivalently: if $a ≤ b$ and $0 ≤ c$, then $ac ≤ bc$).

An ordered ring is discretely ordered if there is no element between $0$ and $1$:

  • if $0 ≤ a ≤ 1$ then $a = 0$ or $a = 1$.

This answer

https://math.stackexchange.com/a/621121/77349

gives an example of an ordered ring that is non-commutative. I do not have the competence to quickly tell whether that example is a discretely ordered ring (I would guess it is not).

Is there a non-commutative discretely ordered ring?

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    $\begingroup$ If you replace the base ring $\mathbb{R}$ with $\mathbb{Z}$ and otherwise use the same construction you may obtain a non-commutative discretely ordered ring, but I am not familiar enough with Weyl algebras to be sure of my answer $\endgroup$
    – Louis Hainaut
    Aug 27, 2020 at 14:57
  • $\begingroup$ Neither am I! But if no-one else chimes in I may take a deeper look. $\endgroup$
    – Anders Lundstedt
    Aug 27, 2020 at 14:59
  • $\begingroup$ i agree with the first comment: I think the subset $\mathbb R$ is going to be ordered in the normal way as a subset of the Weyl algebra. $\endgroup$
    – rschwieb
    Aug 27, 2020 at 15:01
  • $\begingroup$ and that replacing $\mathbb R$ with $\mathbb Z$ would eliminate the problem. $\endgroup$
    – rschwieb
    Aug 27, 2020 at 15:12

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