By José Bueso, José Gómez-Torrecillas, Alain Verschoren (auth.)

ISBN-10: 9048163285

ISBN-13: 9789048163281

ISBN-10: 9401702853

ISBN-13: 9789401702850

The already extensive variety of purposes of ring conception has been more desirable within the eighties by means of the expanding curiosity in algebraic constructions of substantial complexity, the so-called classification of quantum teams. one of many primary houses of quantum teams is they are modelled through associative coordinate earrings owning a canonical foundation, which permits for using algorithmic buildings in keeping with Groebner bases to check them. This ebook develops those equipment in a self-contained method, focusing on an in-depth learn of the inspiration of an enormous category of non-commutative jewelry (encompassing such a lot quantum groups), the so-called Poincaré-Birkhoff-Witt jewelry. We comprise algorithms which deal with crucial features like beliefs and (bi)modules, the calculation of homological measurement and of the Gelfand-Kirillov measurement, the Hilbert-Samuel polynomial, primality exams for high beliefs, etc.

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**Extra info for Algorithmic Methods in Non-Commutative Algebra: Applications to Quantum Groups**

**Example text**

Then r may be written as a (inite product PI . . Pn of irreducible elements of R. PROPOSITION PROOF. Let us start by showing that r has at least one irreducible factor. If r is irreducible itself, nothing has to be proved. Otherwise, we may find elements rI,SI fi=. U(R) such that r = rISI. Obviously Rr c RrI, and this inclusion is strict. Indeed, if Rr = RrI, then r and rl are right associated, which would imply 51 to be a unit a contradiction. Iterating, this would yield an infinite chain of left ideals Rr c RrI c Rr2 c ...

4. 1. Ifr and sare elements ofR, then we say thatr divides s on the right or that r is a right divisor of s, and we will denote this by ris, ifthere exists some q E R with the property that s = qr. We will then also say that s is a left multiple ofr. Left divisibility is defined in a similar way. 2. , with yx = xy + 1. It is then obvious that xlxy + 1. On the other hand, x does not divide xy + 1 on the left. i,j aijXiyj E Al (1<) with ExAMPLE i,j which is impossible, as the right hand side has no constant term.

PROOF. Define a I<-algebra homomorphism u: I«x, y) by sending x,y E -+ I<[x][y; a] I«x,y) to x,y E I<[x][y; a]. It is dear that u(yx - qxy) = u(y)u(x) - qu(x)u(y) = yx - qxy = a(x)y - qxy = O hence u vanishes on Iq. So u induces a I<-algebra homomorphism = I«x,y)/Iq - + I<[x][y; a]. which we stiU denote by u. It is obvious that u is surjective. To show that u is actually an isomorphism, let us exhibit an inverse v for it. u: I

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