Download Ideals of Identities of Associative Algebras by Aleksandr Robertovich Kemer PDF

By Aleksandr Robertovich Kemer

This booklet matters the examine of the constitution of identities of PI-algebras over a box of attribute 0. within the first bankruptcy, the writer brings out the relationship among sorts of algebras and finitely-generated superalgebras. the second one bankruptcy examines graded identities of finitely-generated PI-superalgebras. one of many effects proved matters the decomposition of T-ideals, that is very priceless for the learn of particular forms. within the 5th element of bankruptcy , the writer solves Specht's challenge, which asks even if each associative algebra over a box of attribute 0 has a finite foundation of identities. The e-book closes with an software of tools and effects confirmed prior: the writer reveals asymptotic bases of identities of algebras with cohesion enjoyable all the identities of the total algebra of matrices of order .

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Ideals of Identities of Associative Algebras

This e-book issues the learn of the constitution of identities of PI-algebras over a box of attribute 0. within the first bankruptcy, the writer brings out the relationship among different types of algebras and finitely-generated superalgebras. the second one bankruptcy examines graded identities of finitely-generated PI-superalgebras.

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Algebras C(1) , ... , C(S) exist satisfying the conditions com(CU)) < com(I', TZ[A]), (20) S n r, i c(i) Indeed, it is obvious that t(T', TZ[B(`)]) < t(B(`)). If this inequality is strict, then algebras C(l) , ... 7. But if tar t(r, TZ[A]) 5 then d(I', T2[A]) < c(BM) and the existence of C(l) , ... 7. From (20) it follows that s II (j=1 Therefore in (19) TZ[B(`)] can be replaced by fli TZ[C(')]. The theorem is proved. 1. s. s. algebras A(') ... , A(m) such that m nT2[A(')] = T2[A]. i=I §3. Trace identities Let R be a commutative algebra with unity over the field F , and let A be an R-superalgebra (RA C AO , RA1 C A,).

I=1 (36) §3. SEMIPRIME VARIETIES 25 We put IF = {g}T, where n is an arbitrary positive integer. Then IF C T[Mk(F)] but IT Z T[Mk+l(F)](F)], sF(X)/T[Mk+l(F)] has no zero divisors [1]. 2 implies that the T-ideal n`1" T[Bt ] is nilpotent modulo IF. , g is an H-polynomial. Now let f be an H-polynomial with f E T[Mk(F)] but f T[Mk+l (F)] . We have already proved that (36) is valid for some s. 2 and (36) that the largest nilpoHence, since tent T-ideal modulo IF coincides with the T-ideal n'1" IF = { f is an H-polynomial, m fEn (37) It is obvious that Var(MZk k) is the largest prime variety containing Mk(F) but not containing Mk+l (F) .

Ak and bj , ... , bs . Since G C E0 U El , for any g E G we have g2 E E0. Since EoR is an algebraic algebra over R , this implies that all the elements of G are algebraic over R. 48 II. s. algebra; therefore F(X)/F satisfies a nontrivial (ordinary) identity. Hence E is a PI-algebra. Then, by the theorem of Shirshov on height [75], elements gl , ... , gp E G exist such that E is generated as an F-space by elements of the form gl 1 gl q , where q < h = h (E) and sl > 1 . Since gl , ... , gp are algebraic over R, there are a positive integer n and elements 1 q rl j E R, 1< i< p, 1< j < n such that n n+1 j gi ri j gi j=1 We denote by R' the subalgebra of R generated by the finite number of elements rl j and the unit of R.

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