I am trying to understand kind answer (somewhat generalized) by David Speyer on my previous question What is growth of ass. algebra with 3 generators and relation a1a2a3 + a2a3a1 +a3a1a2 - a1a3a2 - a2a1a3 -a3a2a1 ?
Consider free associative algebra $A$, denote $a_1, ... a_N$ to be its generators; take any homogeneous non-commutative polynomial $P(a_i)$. Consider $A/P$.
Let us choose any monomial $a_1^{k_1}a_2^{k_2}...a_N^{k_N}$ in $P$.
Question what are the conditions on $P$ such that monomials non-containing subsequence $a_1^{k_1}a_2^{k_2}...a_N^{k_N}$ will form a basis in $A/P$ ?
Remark: obviously such monomials will always span $A/P$, because we can rewrite relation $P=0$ in the form $a_1^{k_1}a_2^{k_2}...a_N^{k_N} = P - a_1^{k_1}a_2^{k_2}...a_N^{k_N}$ so using this relation we can always avoid $a_1^{k_1}a_2^{k_2}...a_N^{k_N}$.
**Example Positive ** Consider relation $ab=ba$, so factor is commutative, then obviously monomials not-containing $ab$ i.e. monomials $b^pa^q$ form basis.
(Moreover it is true for $ab = q ba$, even for $q=0$).
**Example Negative ** Consider $a^2 = ab$, the monomials not-containing $a^2$ are linear dependent: for example $aba= a b^2$.
**Example non-clear ** It not clear for whether in the case of my original question i.e. $P = a_1a_2a_3 + a_2a_3a_1 +a_3a_1a_2 - a_1a_3a_2 - a_2a_1a_3 -a_3a_2a_1$, the construction works or not ?
I do not fully understand David's argument: " Let $u_1$ be the lexicographically first $u_i$ and, if that monomial occurs as $u_i$ more than once, choose the one so that $v_i$ is lexicographically earliest. Then the monomial $u_1(a_1a_2a_3)v_1$ only occurs in the term $u1Δv1$ and no others, so it can't cancel out. So the right hand of the sum contains the nonstandard term $u1(a_1a_2a_3)v_1$, a contradiction."
Because: Consider $u_1 = a_3$ - "lexigraphically biggest", take $v_2=a_3$ then $u_1(a_1a_2a_3 )= a_3a_1a_2a_3 = (a_3a_1a_2)a_3 = (a_3a_1a_2)v_2$ . So the $u_1(a_1a_2a_3 )$ is contained in $Δv1$ . This contradicts the claim "monomial only occurs in the term and no others". Am I wrong ?