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Take $X$ a set and $B_X$ the quotient of $\mathbb Z[X]$ by the ideal generated by all elements $u^3-u$ with $u\in \mathbb Z[X]$, which came up in this answer to this question.

Can you describe $B_X$ explicitly for small values of $|X|$?

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up vote 6 down vote accepted

Just checking exhaustively:

When |X|=0, B=Z/6Z of size 6.

When |X|=1, B=Z[x]/(6,xxx-x,3xx+3x)={a+bx+c(xx+x):a,b in Z/6Z, c in Z/3Z} of size 108.

When |X|=2, B=Z[x,y]/(6,xxx-x,yyy-y,3xx+3x,3yy+3y) = {a+bx+cy+dxy+e(xx+x)+f(xx+x)y+g(xx+x)(yy+y)+hx(yy+y)+i(yy+y) : a,b,c,d in Z/6Z, e,f,g,h,i in Z/3Z} of size 314928.

Presumably the pattern holds, so that if |X|=n, the size is 6^(2^n)*3^(3^n-2^n) with similar additive invariants, but I didn't check n=3 due to the size. At any rate, its size is a divisor.

For larger exponents (varying the "3"), the characteristic of the ring is related to the denominators of the Bernoulli numbers as in this OEIS entry.

Since 3 + 4 = 1, 3*3 = 3, 4*4 = 4, 3*4 = 0, 3*x = x*3, and 4*x = x*4 in any ring B satisfying the law uuu=u, one has that 3 and 4 are orthogonal central idempotents and B decomposes as (3B) × (4B). 3B has characteristic 2 and 4B has characteristic 3, so it suffices to consider algebras over the fields Z/2Z and Z/3Z (satisfying the law uuu=u) and take their direct product (X finite or not, B free or not).

In characteristic 2, (u+1)^3 - (u+1) = u^2-u, so the algebra 3B is always a Boolean algebra. For B free, 3B is a free Boolean algebra since the relevant homomorphisms all have 4B in their kernel.

In characteristic 3, (u+v)^3 - (u+v) = u^3-u + v^3-v, so it suffices to check the law on a basis. Since the law is clearly multiplicative, it then suffices to check it on an algebra generating set, or more importantly, to impose it there. This gives the simpler description of 4B = Z[x,y]/(3,xxx-x,yyy-y), etc.

This gives a reasonably explicit description of the free B as a direct product of a free boolean algebra and the "obvious" characteristic 3 algebra.

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Yes, this is 2^{2^n} * 3^{3^n}, which is the size of A x B in my answer when X is finite. – Qiaochu Yuan Jan 6 '11 at 23:03
Ah! I had completely missed your "most imporant" observation. That pretty much does it. Thanks! – Mariano Suárez-Alvarez Jan 7 '11 at 1:13
No problem. I had wondered about these rings for a while myself. I think this describes the "3" replaced by prime power case reasonably well (modulo any funny business with bernoulli numbers). I wonder if u^6-u is interesting or just strange. – Jack Schmidt Jan 7 '11 at 1:31
@Jack. When $|X|=0$, why is $B=Z/6Z$? – TCL Jan 7 '11 at 13:57
@TCL: When X has nothing in it, the ring is just generated by its one, 1. You still have to impose x^3=x, so for instance 2^3-2 = 6 must in fact be 0. Then you check that in Z/6Z every element really does cube to itself: 0^3=0, 1^3=1, 2^3=8≡2, 3^3=27≡3, 4^3≡4, 5^3≡5, good. – Jack Schmidt Jan 7 '11 at 16:02

The arguments I gave in the other question show the following:

  • Every prime ideal is a maximal ideal, and the Jacobson radical is zero.
  • The residue field of every prime ideal is either $\mathbb{F}_2$ or $\mathbb{F}_3$.

It follows that $B_X$ has characteristic $6$. A morphism $B_X \to \mathbb{F}_2$ is freely determined by its values on $X$, so the subspace of $\text{Spec } B_X$ with residue fields $\mathbb{F}_2$ is homeomorphic to $\{ 0, 1 \}^X$. Similarly the subspace of $\text{Spec } B_X$ with residue fields $\mathbb{F}_3$ is homeomorphic to $\{ 0, 1, -1 \}^X$.

So $B_X$ embeds into $A \times B$ where $A$ is the ring of continuous functions $\{ 0, 1 \}^X \to \mathbb{F}_2$ and $B$ is the ring of continuous functions $\{ 0, 1, -1 \}^X \to \mathbb{F}_3$. It is precisely the subring of this ring generated by the images of $X$, where $x \in X$ corresponds to the product of the evaluation $e_x : \{ 0, 1 \}^X \to \mathbb{F}_2$ and the evaluation $e_x': \{ 0, 1, -1 \}^X \to \mathbb{F}_2$. So I guess it is a certain fiber product of $A$ and $B$.

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