Talk:Cancellable Semiring: Distributand Commutative

Earlier on today I posted up the definition of the Trivial Ring (i.e. one in which the ring product is always zero.

I noticed that the underlying group structure need not be abelian for the ring axioms to be satisfied. This is contrary to the usual ring definition, which insists that it does need to be abelian.

This result, it appears, when I look at it more closely, does not hold for the trivial ring. That is, a structure consisting of a non-abelian group with a zero ring product obeys axioms A, M0, M1 and D of Ring but is not technically a ring as such.

I believe that if it's specified that the ring product is not universally zero, then this result holds, but I think I need to work on it a bit more.

Thoughts, anyone? --Matt Westwood 17:07, 29 November 2009 (UTC)

This is partly in response to your initial question Matt, and partly broader. This proof relies on the fact that any element of a cancellable semiring can be represented as a product of elements. Now this is trivially true in a cancellable semiring with unity, but I see no reason why it has to be true in general. In fact, if every element of the group can be represented as a product, then since $\left({S, \circ}\right)$ forms a semigroup, from Identity Property in Semigroup, the semiring must have a multiplicative identity. So in order for this proof to be valid all semirings (and I think therefore all semigroups) would have to have a multiplicative identity, which clearly isn't right.

In particular, it isn't true for the trivial ring, since in that case only the zero of the ring can be represented as a product.

So I think we have to restrict this to only cancellable semirings with unity, barring a completely different direction to prove this from that I'm not seeing. --Alec (talk) 22:04, 30 May 2011 (CDT)

cancellative semigroup distributand with magma distributor having left or right identity is sufficient for commutative distributand. Expand (x + y) * (r + r) both ways as before where r is a right identity. No need for semiring with unity. But I don't know if necessary nor of any example where the left or right identity is not both or where the distributor is not a semigroup.

... anyone care to take this on board? I'm trying to concentrate on other stuff at the moment, so it will be a while before I make time to address this. --prime mover 02:40, 18 June 2011 (CDT)

I drafted one up in my sandbox, I'll move it over when I've looked at it when I'm awake. That said, I'd still like to resolve the issue of doing this without requiring some form of identity, be it one sided or two sided... --Alec (talk) 23:50, 20 June 2011 (CDT)

Possible that it does only apply when the distributor has an identity. This may be why you usually get as one of the ring axioms that the distributand is commutative. Need to find a non-trivial counterexample, I suppose ... --prime mover 00:31, 21 June 2011 (CDT)