Talk:Cancellable Semiring with Unity is Additive Semiring

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? --prime mover (talk) 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 $\struct {S, \circ}$ 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)

Take $R = 2\Z$; it has no identity, and $2$ is not the product of two other elements. --Lord_Farin 03:32, 2 June 2012 (EDT)

I think $\struct {2 \Z, +, \times} \times \struct {D_3, \circ, \mathbf 0}$ furnishes an example of a 'pseudoring' with nontrivial distributor, ánd its distributand is not abelian. --Lord_Farin 03:47, 2 June 2012 (EDT)

So are we saying that it's possible for a structure $\struct {R, *, \circ}$ to satisfy the criteria as given on Axiom:Ring Axioms, but for $\struct {R, *, \circ}$ not to be a ring? That is, $\circ$ to be associative and distributive over $*$ but for $*$ not to be commutative?

If so, this needs to be raised in prominence, because it's something I've never seen in the literature. It would provide a reason for why $*$ is always specified in the ring axioms as being abelian.

BTW I've looked up "pseudo-ring" and it turns out that there's a recent trend (post-1960-ish) to define a ring as a "ring with unity", and therefore by implication that a ring without an identity as "not really being a ring at all" (because some authors got tired of writing "ring with unity" and started using just "ring" for that concept). Having said that, all the books I've studied, and all are post-1960 (except the Nathan Jacobson which I've got most of but haven't got round to studying yet) define the general ring without the identity.

So more work also needed here in an "also defined as".

I will embark on the exercise (later this weekend, perhaps, we lucky Brits get an extra long 4-day one to celebrate some pseudoreligious claptrap or other) to put the ring axioms into the format that everybody knows and loves (i.e. "A" specifying "abelian group", and probably expand them out in the format of Axiom:Field Axioms at the same time. Now we have transcluded pages we have the flexibility to make the subpages more informative without adding too much clutter to the main definition page. --prime mover 04:26, 2 June 2012 (EDT)

I've even seen sources where ring means 'commutative ring with unity'... But indeed, I was trying to say that there are things satisfying the current ring axioms without a commutative distributand. --Lord_Farin 04:42, 2 June 2012 (EDT)

Back to the question of whether this page should exist as stated: currently it seems to be false, although it should hold in any semiring with unity (Dummit and Foote comment in Abstract Algebra that this result does hold in a ring with unity, and their phrasing seems to agree that it doesn't necessarily without unity). I would say we just move it to Cancellable Semiring with Unity is Additive Semiring and amend it slightly as necessary. Any objections? --Alec (talk) 22:13, 4 June 2012 (EDT)

Makes sense to me. --prime mover 01:03, 5 June 2012 (EDT)

Agreed. --Lord_Farin 01:54, 5 June 2012 (EDT)