×

Subalgebras of free commutative and free anticommutative algebras. (Die Unteralgebren der freien kommutativen und der freien antikommutativen Algebren.) (Russian) Zbl 0055.02703

For brevity refer to commutative and anticommutative algebras as \(K\)-algebras and \(AK\)-algebras, respectively. Then the author proves the proposition: “Every subalgebra of a free \(\varepsilon\)-algebra is again a free \(\varepsilon\)-algebra” for \(\varepsilon = K, AK\). The method of proof follows closely that of the author’s paper [Mat. Sb., N. Ser. 33(75), 441–452 (1953)] reviewed in Zbl 0052.03004. If \(R\) is any set, the \(\varepsilon\)-regular words in \(R\) are defined analogously to the regular words in a free Lie algebra (loc. cit.): The \(\varepsilon\)-regular words of length 1 are the nonassociative words of length 1 (i. e. the elements of \(R\) itself) ordered in any way.
If the \(\varepsilon\)-regular words of length \(< n\) have been defined and ordered (so that all words of length \(< m\) precede all words of length \(m\)), then a word \(w\) of length \(n\) is called \(\varepsilon\)-regular, if 1. \(w = u v\), where \(u\) and \(v\) are \(\varepsilon\)-regular, and 2. \(u \geq v\) for \(\varepsilon = K\); \(u > v\) for \(\varepsilon = AK\).
The \(\varepsilon\)-regular words of length \(n\) are then ordered in any way to follow the \(\varepsilon\)-regular words of length \(< n\).
Let \(\mathfrak A\) be the free \(\varepsilon\)-algebra on the set \(R\) over a field \(F\) (of characteristic \(\neq 2\) if \(\varepsilon= AK\)). Then the \(\varepsilon\)-regular words form a basis of \(\mathfrak A\).
For any subalgebra \(\mathfrak B\) of \(\mathfrak A\) ( a set \(\mathfrak M\) can be defined in exactly the same way as for a Lie algebra (loc. cit.) and this is shown to be a set of free generators of \(\mathfrak B\); hence \(\mathfrak B\) is a free \(\varepsilon\)-algebra.

MSC:

17A50 Free nonassociative algebras

Citations:

Zbl 0052.03004
PDFBibTeX XMLCite
Full Text: EuDML