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Let $\cal A$ be the (noncommutative) unitary $\mathbb Z$-algebra defined by three generators $a,b,c$ and four relations $a^2=a,b^2=b,c^2=c,(a+b+c)^2=a+b+c$. Is it true that $ab\neq 0$ in $A$ ?

This question is natural in the context of an older question here on MSE.

My thoughts : The following two relations follow easily from the axioms :

$$ \begin{array}{lcl} cb &=& -(ab+ac+ba+bc+ca) \\ cab &=& ca+ab+2(ba+ac+bc)+aba+abc+aca+bac+bca \\ \end{array}\tag{1} $$

Denote by $W$ the set of words on $a,b,c$ (they are called monomials in the algebra $\cal A$). We order $W$ with the shortlex $a<b<c$ ordering (which we denote by $\prec$). The two relations above express $cb$ or $cab$ in terms of $\prec$-smaller monomials. Iterating those two relations and using induction on $\prec$ ,any monomial can be transformed in $\cal A$ into a term whose monomials do not contain any of $aa,bb,cc,cb,cab$. Denote by $W'$ the set of all monomials satisfying this condition. We therefore have a surjection $s : {\cal A}' \to {\cal A}$ where ${\cal A}'=\oplus_{w\in W'} {\mathbb Z}w$.

Conjecture 1. The mapping $s$ is bijective, in other words $W'$ is a $\mathbb Z$-basis for $\cal A$.

Note that the action of $a$ or $b$ on $W'$ is trivial to describe : for any monomial $w\in W'$, if $w$ does not start with an $a$ then $aw$ stays in $W'$, and $aw=w$ otherwise. Similarly for $b$.

The action of $c$ is more complicated. Using the two relations in (1) and induction on $\prec$ again, we see that there is a unique $\mathbb Z$-linear map $C:{\cal A}' \to {\cal A}'$ such that $C(1)=c,C(a)=ca$ and

$$ \left\lbrace\begin{array}{lcl} C(abw) &=& (Ca+ab+2(ba+aC+bC)+aba+abC+aCa+baC+bCa)w \ ( \ \text{if} \ bw\in W') \\ C(acw) &=& cacw \ ( \ \text{if} \ acw\in W') \\ C(bw) &=& -(ab+aC+ba+bC+Ca)w \ ( \ \text{if} \ bw\in W') \\ C(cw) &=& cw \ ( \ \text{if} \ cw\in W') \end{array}\right.\tag{2} $$

Indeed, any monomial distinct from $1$ or $a$ starts with exactly one of $ab$, $ac$, $b$ or $c$. It is not clear however (at least to me) how to show that

Conjecture 1 (equivalent form). This $C$ satisfies $C^2=C$ and $(a+b+C)^2=a+b+C$.

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Ewan Delanoy
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    To show that $ab \neq 0$ in $\mathcal{A}$, just observe that there is a ring homomorphism $\mathcal{A} \to \mathbb{F}_2$ sending each of $a$, $b$ and $c$ to $1$ (by the universal property). Of course, this completely sidesteps your more interesting conjectures :) – darij grinberg May 05 '16 at 19:11
  • (You are aware of Bergman's "diamond lemma" paper http://www.sciencedirect.com/science/article/pii/0001870878900105 and the updates at https://math.berkeley.edu/~gbergman/papers/updates/diamond.html , right?) – darij grinberg May 05 '16 at 19:15
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    (Oh, wait, your Conjecture 1 appears on page 184 of Bergman's paper...) – darij grinberg May 05 '16 at 19:17
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    @darijgrinberg Thanks for the reference! I was very much aware of diamond lemma techniques in groups and semigroups, but I didn't realize they could be generalized this easily to algebras ... "All the main results in this paper are trivial" is the best first sentence ever in a mathematical paper! – Ewan Delanoy May 06 '16 at 11:08

1 Answers1

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Your Conjecture 1 is indeed true. It can be proved using diamond lemma techniques and the proof is sketched in Section 2.1 of the paper

Bergman, George M., The diamond lemma for ring theory, Adv. Math. 29, 178-218 (1978). ZBL0326.16019.

Essentially, using the diamond lemma, it turns out to suffice to check that the two ways to reduce $ccab$ and $cabb$ using (either by first reducing $cc$ or $bb$, or by first using your identity for $cab$) are equal, and this is a straightforward computation.

(This answer was adapted from darij grinberg's comments on the question.)

Eric Wofsey
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