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Each semi-ring $R$ comes equipped with a canonical preorder $r\leq r^\prime \Leftrightarrow \exists w: r + w = r^\prime$. If $R$ is a ring this order collapses. However, if $R$ is the positive part of a ring, it should usually be a lattice. More precisely, the only reason it could fail is if, in case of the meet $r\wedge r^\prime$, the supremum of $w$'s as above either wasn't unique because the preorder isn't a partial order, as in the case of rings, or didn't exist somehow.

My question is, can this failure be expressed through more ring-theoretic conditions?

Alexander Praehauser
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  • This preorder is an order for idempotent addition giving a semilattice sttucture. Is your semiring commutative? – Benjamin Steinberg Aug 20 '22 at 15:03
  • Any distributive lattice is a semiring with idempotent addition and it recovers the original order (or maybe it's reversal). – Benjamin Steinberg Aug 20 '22 at 15:04
  • @BenjaminSteinberg Of course! Sorry, I don't know what I was thinking! I guess the question is then, if the addition is not idempotent, when can I be sure that joins and meets exist? – Alexander Praehauser Aug 20 '22 at 15:14
  • To answer your question, I'd be glad about conditions in the commutative case, but conditions in the non-commutative case would be even better. – Alexander Praehauser Aug 20 '22 at 15:22
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    Sorry if I'm being naive, but what does it mean to be "the positive part of a ring"? – Jochen Glueck Aug 20 '22 at 16:00
  • @JochenGlueck Well, thinking naively, the positive real or natural numbers, for instance. More generally you could say the part of a ring generated by its generators before additive inverses are adjoined. Of course, that implies that it would be necessary to actually adjoin those inverses, meaning that they don't just come from another relation, so the additive group of the ring can't have any finite-torsion subgroups. – Alexander Praehauser Aug 20 '22 at 18:52
  • My feeling is that this is rarely a lattice order. – Benjamin Steinberg Aug 20 '22 at 20:33
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    Let $R $ be a commutative partially ordered ring with $1$ as an order unit, and let $R^+$ denote the positive cone, which of course is the corresponding semiring. There are a couple of conditions necessary to be lattice-ordered, and these more or less confirm Ben's belief. (a) Suppose $R$ is (order) prime in the sense that if $a,b$ are nonzero positive elements, then there exists $c \neq 0$ st $0 < c \leq a,b$ (if $R$ is a domain, it satisfies this). Then $R^+$ is lo iff it is totally ordered. (b) Suppose $R^+$ is lo. An extremal positive additive funciton on $R $ or $R^+$ is multiplicative. – David Handelman Aug 21 '22 at 00:38
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    (pt 2). [correction real-valued function] This is quite restrictive. More restrictive is that although for general such $R$, the (normalized) extremal addtive functions can be any compact Hausdorff space (in the weak topology), when $R$ is lattice-ordered, only zero-dimensional Hausdorff spaces (totally disconected) can appear. – David Handelman Aug 21 '22 at 00:41

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