commutative rigs and the Grothendieck Group - MathOverflow

commutative rigs and the Grothendieck Group

2012-08-30T19:51:38Z

If I start with a commutative rig, and apply the Grothendieck Group construction to it, twice, once to the additive structure and once to the multiplicative structure, is the result well-known? Does the order of application of the construction matter?

In particular, starting from the semiring $\mathbb{N}$ of the naturals, what do I get for the multiplicative structure? It can't be quite isomorphic to the field $\mathbb{Q}$, but could it be $\mathbb{Q}$ endowed with a meadow structure [link is to a pdf]?

Answer by Johannes Hahn for commutative rigs and the Grothendieck Group

2012-08-30T20:36:02Z

The Grothendieck group of a commutative group is group itself, hence applying the construction twice doesn't change the result. The semiring case is a special case of this.
What do you mean by "order of application". There is no order if you apply the same thing twice. G(G(M)) is the same as G(G(M)) ...
The Grothendieck group of the naturals is the ring of integers. This is the standard construction of the integers.

This sounds somewhat to trivial ... May be you mean something different here. Maybe you want to apply the construction to the multiplicative monoid instead of the additive structure in one of the two steps? The result in this case is (independent of the order) the zero ring because there is an absorbing element in the multiplicative monoid (the zero element of the semiring) so that all elements of the monoid get identified in the Grothendieck group.

Answer by Will Sawin for commutative rigs and the Grothendieck Group

2012-08-30T20:52:56Z

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 ring. If you only take the Grothendieck group of the multiplicative semigroup of non-zero-divisors, you get the unit group of the total quotient ring. In particular, for an integral domain you get the multiplicative group of the field of fractions. This construction is clearly the same as localization by whatever subset you pick.

If the original semiring is a subsemiring of a field, then it does not matter what order you pick. This is because you can do all the arithmetic inside the field, so the additive Grothendieck group is just the subset generated by a semiring under addition, multiplication and subtraction, and the multiplicative Grothendieck group is the subset generated under addition, multiplication, and division. Applying both gets the field generated by that semiring, regardless of order.