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It is decades since I've done math, so please forgive the lack of correct terminology and lack of latex etc. I'm endeavoring to write a simple CAS calculator that can handle structures that undergraduates could run into.

Thanks to wikipedia, I've found that the way of adding a multiplicative inverse to a ring to create a field is called 'localization' - where R is the ring, S = R - zero divisors of R,

the new field is basically an ordered pair (S, R), with the meaning S^-1 * R I guess there is an implied cancellation law added so that (x,x) => (1,1), along with standard rules for rational addition, multiplication, and inverse.

Have I got this right so far?

Anyway, I figure that creating an additive inverse for a semiring is R[n], where n^2 = 1, and x+xn=0, n is the symbol for negative. Is this the quotient group R[n]/(n^2 - 1, 1 + n)?

Is there any proper name for this process, and do I have it correct?

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  • $\begingroup$ It depends on your semiring. In some cases, semirings do not obey the distributive law, and adding elements will not fix it, and often quotients also will not. Gerhard "Ask Me About System Design" Paseman, 2012.03.07 $\endgroup$ Mar 8, 2012 at 1:23
  • $\begingroup$ Thanks Gerhard, The version of semiring that I have been using is an additive commutative monoid and a multiplicative monoid related by the destributive law and 0 annihilates over multiplication. I would specify that n is a new element, commutative over R like a standard polynomial extension, though R not necessarily commutative. With distributivity specified, does the above extension and quotient of the semi-ring always work, or are there other problems? Also it seems like creating additive inverses should have a proper name ... $\endgroup$
    – Richard
    Mar 8, 2012 at 4:00
  • $\begingroup$ Even if it is distributive, you are right that it can't always work. I can see how some semirings (like Max-plus) only can have inverses when quotiented with itself making a trivial ring. $\endgroup$
    – Richard
    Mar 8, 2012 at 13:38

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I think it is often called forming the Grothendieck ring. The construction is really defined for commutative monoids, where one calls it the Grothendieck group and then you carry the multiplication along for the ride in the semiring case. Peter May uses this terminology on page 199 of his concise course on algebraic topology.

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  • $\begingroup$ That is exactly what I was looking for. The standard construction for Grothendieck rings is very similar to the above method of localization - using a Cartesian product and defining new operations - and I had in fact considered something like it. I'm not sure if my proposed construction is isomorphic or not, I suspect that my equivalence relation is suboptimal (the newly constucted group will be larger than necessary), though it still should work. $\endgroup$
    – Richard
    Mar 8, 2012 at 8:33
  • $\begingroup$ I can't really understand much in the book you referenced, and I can't see where it defines multiplication over the Cartesian product, but I have defined it distributively (p1,n1)(p2,n2)=(p1 p2 + n1 n2, p1 n2 + n1 p2) I've realized that an optimal solution defined by the Grothendieck group is necessary, otherwise inverses are not unique = no ring. Max-plus semi-rings under this construction collapse to a single element (as idempotent operations can't have inverses). I guess a quotient creating a trivial ring is always guaranteed to create a ring!! Though not what I was hoping for. $\endgroup$
    – Richard
    Mar 8, 2012 at 13:32
  • $\begingroup$ You can construct the Grothendieck group of the additive monoid via a fraction style construction (except the operation is addition). Then the multiplication can be extended in a natural way. Of course if addition is idempotent then you will collapse the whole semiring. $\endgroup$ Mar 8, 2012 at 19:59
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I've found a useful reference about when semi-groups can be turned into groups

http://en.wikipedia.org/wiki/Cancellative_semigroup#Embeddability_in_groups

Which can be extended to turning semi-rings into rings as described in the above answer.

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