I've been trying to find a rig-like structure (a set R with a monoid structure (R,\cdot,1) and

Sallie Banks

Sallie Banks

Answered question

2022-01-13

I've been trying to find a rig-like structure (a set R with a monoid structure (R,,1) and a (commutative) monoid structure (R,+,0) such that the multiplication distributes over addition) in which 0 is not an absorber with respect to multiplication, i.e. 0r0 for some rR.
It should be possible to construct such a structure since you have to require absorption as a separate axiom in a rig.
So far I haven't been successful as I don't have any experience with rigs but maybe someone else has a neat idea?

Answer & Explanation

yotaniwc

yotaniwc

Beginner2022-01-14Added 34 answers

Step 1
Let (R,,1) be a commutative monoïd such that
r2=r
for all
rR
(for example, you can start with any Boolean ring (R,+,,0,1). Concrete example are given, by (Z2Z)I for any nonempty set I, for example)
Then set += and 0=1.
Hence (R,+,0)=(R,,1) is a commutative monoïd. Moreover, for all r,s,tR, we have
r(s+t)=r(st)=rst and
rs+rt=(rs)(rt)=r2st=rst,
so you have the distributivity property. Now
r0=r1=r
for all rR.
hysgubwyri3

hysgubwyri3

Beginner2022-01-15Added 43 answers

Step 1
Let E be the set of intervals of the real line of the form [a,b] where a0 and b0.
Define
[a,b][c,d]=[M(a,c),Max(b,d)]
and
[a,b][c,d]=[a+c,b+d].
One can check that (E,) and (E,) are abelian monoids, both having identity [0,0], and furthermore distributes over .
If it had an absorbing zero element, it would be a full-fledged semiring... but it obviously does not, since the identity is the same as the identity, it is obviously not absorbing.

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