By Foata D., Han G.-N.

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**Additional resources for A basis for the right quantum algebra and the “1 = q” principle**

**Example text**

From Lemma 2 it follows that the quotient algebra A, of the algebra A with respect to the corresponding ideal N , contains a subalgebra isomorphic to A , and therefore also isomorphic to B. Since bak b = 2b · (b · ak ) − ak · b2 it follows that the algebra I is isomorphic to the subalgebra (+) by the two generators a and b that are the images of the elements generated in A a and b. This proves the theorem. I. Malcev [3] that states that any associative algebra over Σ with a ﬁnite or countably inﬁnite number of generators can be embedded into an associative algebra over Σ with two generators.

1 Literally, “J-ring”. We adopt the modern terminology, “Jordan ring”. [Translators] 2 Literally, “Jordan rings with an arbitrary ring Σ of operators”. I. Shirshov Jordan algebras also satisfy the relation (bas )at = (bat )as , (4) which generalizes equation (1). Indeed, suppose that relation (4) holds for exponents s1 and t1 such that s1 + t1 < s + t. Then, from the equation J1 {as+t−3 , a, a, ba} − aJ1 {as+t−3 , a, a, b} = as+t−1 (ba) − (bas+t−1 )a = 0, which is implied by the inductive hypothesis, the validity of equation (4) follows in the case when one of s or t equals 1.

Suppose now that β has height greater than 2, and the height of α is not less than the height of β. Then, β∗ = σi (c∗i ◦ d∗i ) ◦ e∗i + i σj akj ◦ bsj , j where σk are some coeﬃcients. By the bilinearity of all the operations, to prove the lemma it suﬃces to consider in place of β ∗ the elements β1 = (c∗ ◦ d∗ ) ◦ e∗ and β2 = a k ◦ b s . From inductive hypothesis 3) it follows that α∗ ◦ β2∗ = α∗ ◦ ak ◦ bs = α ◦ ak ◦ bs ∗ ∗ = (α ◦ β2 ) . From equation (2) it follows that α∗ ◦ β1∗ = α∗ ◦ [(c∗ ◦ d∗ ) ◦ e∗ ] = J1 {e∗ , c∗ , d∗ , α∗ } − [(α∗ ◦ c∗ ) ◦ e∗ ] ◦ d∗ − [(α∗ ◦ d∗ ) ◦ e∗ ] ◦ c∗ + (α∗ ◦ e∗ ) ◦ (c∗ ◦ d∗ ) + (α∗ ◦ c∗ ) ◦ (d∗ ◦ e∗ ) + (α∗ ◦ d∗ ) ◦ (c∗ ◦ e∗ ), but according to inductive hypothesis 3) and equation (8) we have α∗ ◦ β1∗ = [α ◦ β1 − J1 {e, c, d, a}]∗ = (α ◦ β1 )∗ .