Is it true that the structure of a commutative ordered semiring with identity is unique on a commutative ordered monoid (i.e., the structure of the monoid and the order are consistent)? I am not asking about existence.
The question is caused by the fact that an ordered monoid is the minimum of the structure cod-a of the metric $d \colon X \times X \to M$. Therefore, I wonder if there is a natural multiplication in this case. Since multiplication in ordinary R-metric spaces is very useful
The answer is no.
Consider the set $\mathbb{N}^{<\omega}$, consisting of $\omega$-tuples with only finitely many nonzero entries. This set is totally ordered under the relation $\prec$, where $(a_0,a_1,\ldots)\prec (b_0,b_1,\ldots)$ exactly when $a_i<b_i$ (in the usual ordering $<$ on $\mathbb{N}$) for the least value of $i$ such that $a_i\neq b_i$. In other words, $\prec$ is the ``compare the first differences'' relation.
Under component-wise addition, $(\mathbb{N}^{<\omega},+,\prec)$ is a commutative ordered monoid.
We can make this into a commutative ordered semiring by thinking of elements of $\mathbb{N}^{<\omega}$ as the coefficients of polynomials (starting at the constant term, and going up in degree). This polynomial multiplication respects the nonstrict order $\preceq$.
We can also identify elements of $\mathbb{N}^{<\omega}$ with coefficients in $\mathbb{N}[x,y]$ via $$(a_0,a_1,a_2,\ldots)\mapsto a_0 + a_1x + a_2y + a_3x^2+a_4xy+a_5y^2+\cdots.$$ Multiplication here also respects $\preceq$.
