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I preferred to call here a semiring $(R, +, .)$ to be an idempotent if $x+x=x$ and $x.x=x~\forall~x\in R$. It is apparent that we can define certain partial order relations on an idempotent semiring. Can we define a partial order relation on a none idempotent semiring? Or, is every partially ordered semiring an idempotent ?

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$\mathbb R$ is a totally ordered semiring that isn't idempotent, for example.

I guess what made you forget about this is that the natural order on idempotent semirings is defined by $a\leq b$ when $a\oplus b=b$, which obviously doesn't occur in the example I gave.

If the semiring isn't additively idempotent, then you do not have reflexivity of this candidate for order. So in this sense, yes, for that relation to be reflexive, you would need the semiring to be idempotent.

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  • $\begingroup$ can the natural order on idempotent semiring defined by $a\leq b$ when $a\oplus b=b$ be equivalently written as $a\oplus x=b$ for some $x\in \Bbb R$? Also, can the multiplicative monotonicity be written as $ax=b $ for some $x\in~\Bbb R$? $\endgroup$
    – gete
    Nov 27, 2018 at 16:33
  • $\begingroup$ @gete I would think it would be impossible to prove anti-symmetry, then. $\endgroup$
    – rschwieb
    Nov 27, 2018 at 16:52
  • $\begingroup$ You are right, as i am also stucked in anti-symmetricity but i found such additive monotonicity in a book"Algebraic Theory and application in computer science " by U. Hebisch and H.J Weinert, Vol-5 , chapter 5 under the heading-partially ordered semirings (page-144). $\endgroup$
    – gete
    Nov 27, 2018 at 17:02
  • $\begingroup$ @gete For example, using your proposed definition and $\mathbb Z$, all elements are related to one another. $\endgroup$
    – rschwieb
    Nov 27, 2018 at 17:21
  • $\begingroup$ Sorry! would you please clearify the last comment again? $\endgroup$
    – gete
    Nov 27, 2018 at 17:26

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