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7

It is at least sometimes called a "monoid semiring" by analogy with "group ring". As such it would be notated $S = \mathbb{N_0}[M]$ (or $\mathbb{N}[M]$ depending how you define things). By the way, the ring $\mathbb{Z}[M]$ you define in #1 is a commutative ring, but not a field because the element $2$ (the identity of M plus itself in $S$) has no inverse. ...


4

The Grothendieck group of a semiring is a ring. So you are just asking what will happen if you take the multiplicative Grothendieck group of a ring. Well, it depends on whether you include $0$ in the multiplicative structure! If you include $0$, then $a/b=0a/0b=0/0$ so you get the trivial group. Similarly, inverting zero-divisors will collapse part of the ...


4

Tarski, cardinal algebras, pag. 10 writes: Formulas of the type "$a+b=b$" [...] can be read "$a$ is absorbed by $b$" or "$b$ absorbs $a$". The relation of absorption plays an important role in the arithmetic of C.[ardinal]A.[lgebra]'s. Wehrung, Injective positively ordered monoids I [ http://www.sciencedirect.com/science/article/pii/002240499290104N ] pag. ...


3

Some observations: Consider the monoid $P = \{ (N, p)\in\mathbb{N}\times\mathbb{Z} \,:\, N = 0 \Rightarrow p = 0\}$ with the semiring structure $(N, p)(M, q) = (NM, Mp+Nq)$. I claim that the map $(N, p)\mapsto N-1+H^p$ is an isomorphism between $P$ and your semiring. In particular, the additive monoid underlying your semiring is not finitely generated. ...


3

You are describing free constructions between finitary varieties. A finitary variety is an equationally defined class of algebras for (i) an arbitrary set $\Sigma$ of operation symbols each $\sigma \in \Sigma$ having finite arity, (ii) an arbitrary set of equations $E$ consisting of pairs $(\phi_1,\phi_2)$ where each $\phi_i$ is a term built from the ...


1

I don't think you can ask for a characterisation of $\mathcal{A}^{op}$ that is any more concrete than the definition. However, $\mathcal{A}^{op}$ can be described in alternative terms via scheme theory. This won't describe it in any simpler terms, and in fact it introduces a good deal of extra complication, but it can perhaps be useful sometimes because ...


1

I think the answer is no. Let $S=\{0\}\cup[1,\infty)$ be the subsemiring of the (usual) reals. A non-zero ideal of $S$ is of the form $\{0\}\cup[a,\infty)$ where $a\geq 1$. Clearly, the only prime ideal of $S$ (according to your definition) is $\{0\}$ and it is subtractive. But no proper non-zero ideal of $S$ is subtractive. Correction: My argument is not ...



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