Reports of the Academy of Sciences of the USSR

1. Vol. 133, No. 6

MATHEMATICS

A. T. GAINOV

COMMUTATIVE FREE AND ANTICOMMUTATIVE FREE PRODUCTS OF ALGEBRAS

(Presented by Academician A. I. Maltsev on 14 IV 1960)

1. Let

[
p_i(x_1,\ldots,x_{n_i}) \qquad (i=1,2,\ldots)
]

be some collection of elements of the free nonassociative algebra over a field (\Omega) of characteristic zero. All polynomials (p_i) are assumed to be multilinear and homogeneous of degree greater than 1.

If (A_\alpha,\ \alpha\in S), are arbitrary algebras over the same field (\Omega), then by

[
\bar G=\prod_{\alpha\in S}^{*} A_\alpha
]

we denote the nonassociative free product of the algebras (A_\alpha), constructed for some choice of bases (L_\alpha) of the algebras (A_\alpha). Let (I) be the set of all elements of (\bar G) of the form

[
\sum ab_1b_2\ldots b_k p_i(\mathfrak A_1,\ldots,\mathfrak A_{n_i})c_1c_2\ldots c_l,
\tag{1}
]

where (\alpha\in\Omega), (b,\mathfrak A,c) with indices are regular words in the sense of Kurosh (see (1), p. 245) of the algebra (\bar G), and, moreover, if the words (\mathfrak A_1,\mathfrak A_2,\ldots,\mathfrak A_{n_i}) all have length 1, then at least two of them lie in different algebras (A_\alpha). In the product (b_1b_2\ldots b_k p_i(\mathfrak A_1,\ldots,\mathfrak A_{n_i})c_1c_2\ldots c_l), parentheses are placed in some manner. It is not difficult to verify that (I) is a two-sided ideal in the algebra (\bar G).

Definition 1. The factor algebra (G=\bar G/I) will be called the reduced free product of the algebras (A_\alpha,\ \alpha\in S), defined by the identical relations

[
p_i(x_1,\ldots,x_{n_i})=0 \qquad (i=1,2,\ldots),
\tag{P}
]

or, more briefly, the (P)-free product of the algebras (A_\alpha), and will be denoted by

[
G=\prod_{\alpha}^{\nabla} A_\alpha=A_1\nabla A_2\nabla\cdots .
]

It turns out that the algebra (G) contains subalgebras (\mathfrak A_\alpha,\ \alpha\in S), isomorphic respectively to the algebras (A_\alpha), and is generated by them. It is not difficult to show that the definition of the algebra (G) does not depend on the choice of the bases (L_\alpha) of the algebras (A_\alpha).

We give a second definition of the (P)-free product of algebras.

Definition 2. Let each algebra (A_\alpha,\ \alpha\in S), be given by a system of generators (M_\alpha) and a system of defining relations (\Phi_\alpha), with the sets (M_\alpha,\ \alpha\in S), pairwise disjoint. Denote by (M) the union of all the sets (M_\alpha): (M=\bigcup_{\alpha\in S} M_\alpha), by (\Phi) the union of all the systems (\Phi_\alpha): (\Phi=\bigcup_{\alpha\in S}\Phi_\alpha), and by (\Psi) the system of all possible defining

relations of the form

[
\sum \alpha x_1x_2\ldots x_k p_i(y_1,\ldots,y_{n_i}) z_1z_2\ldots z_l=0,
]

where (\alpha\in\Omega), and (x,y,z) with indices are arbitrary (M)-words (i.e., products of generators from (M)); moreover, in the notation of each polynomial (p_i(y_1,\ldots,y_{n_i})), generators from at least two different sets (M_\alpha) occur. Then the algebra (G), with system of generators (M) and system of defining relations (\Phi\cup\Psi), is called the (P)-free product of the algebras (A_\alpha,\ \alpha\in S), defined by the identity relations (P).

Definitions 1 and 2 are equivalent, which can be shown analogously to how this is done in group theory ((2), pp. 209–210).

Let us note the simplest properties of (P)-free products of algebras.

I. (A\nabla B=B\nabla A).

II. If (G=\prod_{\alpha\in S}^{\nabla} A_\alpha), if the set (S) is the union of pairwise disjoint subsets (S_\beta): (S=\bigcup_\beta S_\beta), and if (B_\beta=\prod_{\alpha\in S_\beta}^{\nabla} A_\alpha), then

[
G=\prod_\beta^{\nabla} B_\beta .
]

III. If (G=\prod_\alpha^{\nabla} A_\alpha) is the (P)-free product of the algebras (A_\alpha), and if in each algebra (A_\alpha) a subalgebra (B_\alpha) has been chosen, then the subalgebra (B) generated in (G) by all (B_\alpha) will be their (P)-free product:

[
B=\prod_\alpha^{\nabla} B_\alpha .
]

IV. If (G=\prod_\alpha^{\nabla} A_\alpha) and if some factors (A_\alpha) themselves are decomposed into a (P)-free product:

[
A_\alpha=\prod_\beta^{\nabla} A_{\alpha\beta},
]

then

[
G=\prod_{\alpha,\beta}^{\nabla} A_{\alpha\beta}.
]

This latter decomposition of the algebra (G) we shall call a continuation of the original decomposition.

V. If (G=A\nabla B) and (\bar A) is the two-sided ideal generated in (G) by the subalgebra (A), then (G/\bar A\cong B).

Any algebra in which, for any of its elements (x_1,x_2,\ldots,x_{n_i}), all equalities (P) hold is called a (P)-algebra. According to A. I. Mal'cev (3), an algebra with a system of generators (M) and a system of identity defining relations (P) is called a free (P)-algebra with system (M) of free generators.

VI. The (P)-free product of algebras (A_\alpha) is a (P)-algebra if and only if each algebra (A_\alpha) is a (P)-algebra.

VII. The (P)-free product of free (P)-algebras (A_\alpha) with systems (M_\alpha) of free generators is a free (P)-algebra with the system

[
M=\bigcup_\alpha M_\alpha
]

of free generators.

VIII. A free (P)-algebra with system ({a_\alpha}) of free generators can be decomposed into the (P)-free product of free (P)-algebras (A_\alpha) on one generator (a_\alpha).

IX. Every (P)-free product (G) of algebras (A_\alpha) contains a two-sided ideal (J) which is a free (P)-algebra and such that (G/J) is the direct product of the same algebras (A_\alpha).

Let us give several examples of (P)-free products of algebras.

Example 1. The system (P)

[
xy+yx=0,
]

[
(xy)z+(yz)x+(zx)y=0
]

defines the left-free product of algebras.

Example 2. The reduced free products defined by the systems

[
xy-yx=0,
\tag{K}
]

[
xy+yx=0,
\tag{AK}
]

will be called, respectively, commutative free ((K)-free) and anticommutative free ((AK)-free) products of algebras.

Example 3. The system ((P))

[
xy=0
]

defines the direct product of algebras.

The nonassociative free product of algebras may be regarded as a (P)-free product, assuming that the system ((P)) is empty in this case.

1. In what follows we shall speak only about (K)-free and (AK)-free products of algebras. Therefore, in the expressions “(P)-free products,” “(P)-algebras,” the letter (P) is everywhere assigned one and the same meaning, (K) or (AK).

We shall show how to find a basis (B) of an algebra (G) which is a (P)-free product of algebras (A_\alpha), if bases (L_\alpha) of the algebras (A_\alpha), (\alpha\in S), are given. Let (\mathfrak G) be the free (P)-algebra with the set (L=\bigcup_{\alpha\in S}L_\alpha) of free generators. According to (4), as a basis (\mathfrak B) of the algebra (\mathfrak G) one may take the ordered aggregate of all so-called (P)-regular words. The set of all those (P)-regular words which do not contain, in any pair of brackets, two elements of one and the same algebra (A_\alpha), forms precisely a basis (B) of the algebra (G). An arbitrary element (x\in G) can be expressed through the elements of the basis (B) in the following way. First we write (x) in regular form, regarding it as an element of the nonassociative product of the algebras (A_\alpha) ((1), p. 245). Then, regarding each regular word occurring in this expression as an element of the above-mentioned free (P)-algebra (\mathfrak G), we express them in (P)-regular form in the same way as A. I. Shirshov does this ((4), p. 82). As a result we obtain the desired expression of the element (x) through the basis (B).

By modifying in the appropriate way the methods of proof of the theorem on subalgebras and Zhukov’s theorem for nonassociative free products of algebras, proposed by A. G. Kurosh ((5)), one can prove the following theorems.

Theorem 1. Let

[
G=\prod_{\alpha\in S}^{\triangledown} A_\alpha \triangledown F
]

be the (P)-free product of arbitrary algebras (A_\alpha) and a free (P)-algebra (F), and let (H) be a subalgebra of the algebra (G) which is a (P)-algebra. Then

[
H=\prod_{\alpha\in S}^{\triangledown} B_\alpha \triangledown V,
]

where (\prod^{\triangledown}) is the same (P)-free product, (B_\alpha=H\cap A_\alpha), and (V) is some free (P)-algebra.

A special case of this theorem, when all (A_\alpha=0), is the theorem of A. I. Shirshov that every subalgebra of a free (P)-algebra is itself free (4). (The condition that (H) must be a (P)-algebra is essential.)

Let

[
g_1,\ g_2,\ldots,\ g_n
\tag{2}
]

be some finite system of generators of the algebra (G). According to A. G. Kurosh ((5), p. 257), elementary transformations of the system (2) are transformations of one of the following two types:
a) addition to any (g_i) of an arbitrary polynomial in all the remaining elements of the system (2);
b) multiplication of an element (g_i) by some distin-

a nonzero number of elements of the ground field (\Omega). An elementary transformation of the system (2) again leads to a system of generators of the algebra (G).

Theorem 2. Let (G) be a (P)-algebra with a finite number of generators (2), and

[
G = A_1 \nabla A_2 \nabla \cdots \nabla A_k
]

a (P)-free product. Then by elementary transformations the system (2) can be brought to such a form that all generators (2) will lie in the factors (A_1,\ldots,A_k); moreover, if (g_1,g_2,\ldots,g_l) are all the generators from the system (2) lying, for example, in (A_1), then they generate (A_1).

Corollary 1. Every (P)-algebra with a finite number of generators can be decomposed into the corresponding (P)-free product of a finite number of algebras indecomposable into such a product.

Corollary 2. (P)-free decompositions of a (P)-algebra with (n) generators consist of no more than (n) free factors.

Corollary 3. The minimal number of generators of a (P)-algebra with a finite number of generators is equal to the sum of the corresponding numbers for all factors of any decomposition of it into a (P)-free product.

In conclusion let us consider the question of the isomorphism of decompositions into a (P)-free product of algebras ((P=K) or (P=AK)).

Definition 3. Two (P)-free decompositions of an algebra shall be called isomorphic if every free factor of one of these decompositions that is not a free (P)-algebra is a free factor in the other decomposition as well, and conversely, i.e., if

[
G=\prod_{\alpha}^{\nabla} A_{\alpha} \nabla F
=\prod_{\beta}^{\nabla} B_{\beta} \nabla F',
]

where (F,F') are free (P)-algebras, and (A_{\alpha}, B_{\beta}) are nonfree (P)-algebras, then (A_{\alpha}=B_{\alpha}) under a corresponding numbering.

By property V of (P)-free products, here (F \cong F').

Theorem 3. Any two (P)-free decompositions of a (P)-algebra (G) possess isomorphic refinements. In particular, if a (P)-algebra (G) admits a (P)-free decomposition with indecomposable factors, then all such decompositions are isomorphic to one another.

The proof of this theorem is carried out on the basis of Theorem 2 and properties I–VIII of (P)-free products literally in the same way as for nonassociative free decompositions ((^{1}), pp. 258–259).

I express my gratitude to A. I. Mal’cev and A. I. Shirshov for their attention to the present work and for valuable comments.

Institute of Mathematics
Siberian Branch of the Academy of Sciences of the USSR

Received
6 IV 1960

CITED LITERATURE

(^{1}) A. G. Kurosh, Matem. sborn., 20, no. 2, 239 (1947).
(^{2}) A. G. Kurosh, Theory of Groups, 1953.
(^{3}) A. I. Mal’cev, Matem. sborn., 26, no. 1, 19 (1950).
(^{4}) A. I. Shirshov, Matem. sborn., 34, no. 1, 81 (1954).
(^{5}) A. G. Kurosh, Matem. sborn., 37, no. 2, 251 (1955).