If I start with a commutative rig, and apply the [Grothendieck Group](http://en.wikipedia.org/wiki/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](http://staff.science.uva.nl/~janb/FAM/meadowsVNR.pdf)  structure [link is to a pdf]?



commutative rigs and the Grothendieck Group
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.


