The commutative algebra of T2

October 6, 2006 at 12:01 am (algebra, nim, papers)

I’ve been reading about semigroup rings and other things in commutative algebra. In looking though old notes I found this email from Aaron Siegel from about a year ago, recounting some interesting information from a meeting with David Eisenbud. It corresponds pretty closely to what I was trying to compute myself, this morning

The meeting went well. He showed me a new (to me) way of looking at monoids in terms of commutative rings. Namely, instead of considering the monoid

Q = {a,b : a^2=1,b^3=b }

[ Aside: the monoid Q is the (tame) misere quotient of the game of Nim played using heaps of size 1 and 2 only. In this paper, Q is called T2. ]

we instead look at the ring

R = k[a,b] / (1-a^2,b-b^3)

and now we can apply the techniques of commutative algebra to analyze this ring. One observation is that because Q six elements, R is a 6-dimensional vector space over k. A basis for this vector space is Q itself, but it’s instructive to look at other bases and to study various decompositions of R.

For example, if x is an idempotent of R, then so is 1-x, since (1-x)(1-x) = 1-2x+x^2 = 1-2x+x = 1-x. Further, x(1-x) = 0, so R can be written as the internal direct sum of R/(x) and R/(1-x), with the isomorphism

R -> R/(x) + R/(1-x)

given by y -> ((1-x)y,xy).

Using the above example, we have

R = R/(b^2) + R/(1-b^2),

and it’s not hard to see that R/(1-b^2) is four-dimensional with basis {b^2,ab^2,b,ab} (which you’ll recognize as a maximal subgroup of Q), and R/(b^2) is two-dimensional with basis {1-b^2,a-ab^2} (which of course is equal to {1,a} in R/(b^2)).

Note the parallel with semigroup theory. In the ring R/(1-b^2), we’ve asserted that b^2 = 1, making b^2 into the identity. So the group {b^2,ab^2,b,ab} corresponds exactly with the multiplicative structure of R/(1-b^2). In other words, R/(1-b^2) is isomorphic to the group ring k[Z_2 x Z_2]. Likewise, in R/(b^2) we’ve set b^2 = 0 and the result is isomorphic to the group ring k[Z_2], corresponding to the maximal subgroup {1,a}. So the ring decomposition by idempotents matches the monoid decomposition we’re used to (archimedean components).

All very interesting. Eisenbud also pointed out that the rings can be decomposed further:

1 – b^2 = (1-b)(1+b) so R/(1-b^2) = R/(1-b) + R/(1+b),

and we can further break off the “a” part of each component, since it never interacts with the “b” component. This fully decomposes R as the vector space product of six copies of k and gives a new basis.

Finally, we know that we always have a^2 = 1 in any misere quotient Q. Because of this, the corresponding ring R always decomposes into two isomorphic components

R = R/(1-a) + R/(1+a)

and it’s simpler, therefore, to discard a and study just one of these (R/(1-a) is preferable, since that’s equivalent to just setting a = 1.) Of course, the structure of the P-positions is different on each component. It’s unclear what all of this might mean in game-theoretic terms.

He also recommended a book by Gilmer, “Commutative Semigroup Rings,” which I’ve checked out from Berkeley’s library. The paper by Eisenbud & Sturmfelds, “Binomial Ideals,” might also be useful (it’s on the arxiv).


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