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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 $|X|$?

2 Answers 2

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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.2011-01-06
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    Ah! I had completely missed your "most imporant" observation. That pretty much does it. Thanks!2011-01-07
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    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.2011-01-07
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    @Jack. When $|X|=0$, why is $B=Z/6Z$?2011-01-07
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    @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.2011-01-07
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    @Jack. But $Z/3Z$ also satisfies the conditions. Why is it not $Z/3Z$ also?2011-01-07
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    @TCL: Oh, sorry. By "free" Mariano means all the possible examples are quotient rings of the "free" example. So Z/6Z is the free one with no extra generators. The other examples with no generators, like Z/3Z and Z/2Z are quotient rings of Z/6Z. The examples with one generator are all quotients of that ring with 108 elements, etc.2011-01-08
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    If the "free" examples are not too big, then they can be extremely helpful in understanding all examples. The similar problem in group theory is called the Burnside problem, but the free examples are actually too big to be of much practical use. I think the ones for rings might be pretty good, at least for "3".2011-01-08
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    What is a free ring?2014-05-19
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    @TheSubstitute: A "free" X with generators Y is an X so that for any other X with generators Y' there is a homomorphism of Xs taking Y to Y'. The free ring on no generators is the integers, the free ring on one generator is the ring of integer polynomials in a single variable, the free ring on two generators is the ring of polynomials in two non-commuting indeterminates with integer coefficients, etc. This answer does not discuss free rings. It discusses free "rings in which the cube of every element is itself".2014-05-19
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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 $B$.