## Hot answers tagged semirings

8

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

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 ...

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. ...

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

There is no bijection satisfying the addition and multiplication
identities. In fact, even
without the assumption that $f$ is a bijection
there
are only three functions satisfying both identities: for all $x$ and $y$
either
$f_1(x,y) = x$, $f_2(x,y) = y$ or $f_3(x,y) = 0$.
To see this, use the addition identity repeatedly to deduce that
(i) $f(a,b)$ $=$ ...

3

In the non-commutative case you can't say much. The monoid of all maps on n letters is generated by the symmetric group and an idempotent. In fact semigroups generated by idempotents can be quite wild. The singular nxn matrices over a field are generated by idempotents.
In the commutative case much more can be said. In the finite case your semigroup will ...

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.
...

2

not a solution
Anything that satisfies 1,2,3; then define $x^\dagger = 1$ for all $x$.
added December 27
Try this example...
$U := \mathbb Z \times \mathbb Z$.
multiplication is performed by componentwise addition, $(a,b)(c,d) = (a+c,b+d)$
the lattice operations are also performed componentwise, $(a,b) \vee (c,d) = (a\vee c,b\vee d)$, $(a,b) \wedge ...

1

An easy answer along traditional lines is available iff the measure has "bounded variation" in a suitable sense.
To undestand this, first note that integration with values in Banach algebras (which are rings, but not compact) contains the integration of scalar valued functions with respect to vector (Banach) valued measures (embed the Banach space in a ...

1

It is very easy to see that this is not feasible. It suffices to try $q=0$, so we'd look for a function $f(\cdot,\cdot)$ such that $x+y+z-1=f(x,y)+f(x,z)$. As we can switch variables, this would yield $f\equiv const$, which is impossible.

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 ...

Only top voted, non community-wiki answers of a minimum length are eligible