# Read e-book online An Introduction to Noncommutative Noetherian Rings PDF

By K. R. Goodearl, R. B. Warfield Jr

ISBN-10: 0511217293

ISBN-13: 9780511217296

ISBN-10: 0521836875

ISBN-13: 9780521836876

This advent to noncommutative noetherian earrings, available to someone with a uncomplicated historical past in summary algebra, can be utilized as a second-year graduate textual content, or as a self-contained reference. broad explanatory fabric is given, and routines are built-in all through. New fabric comprises the elemental different types of quantum teams.

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J j bj x in S. 12. Let R be a ring and α an automorphism of R. Suppose that S = R[x; α] and S = R[x ; α]. Then there is a unique ring isomorphism ψ : S → S such that ψ(x) = x and ψ|R is the identity on R. Proof. 11 with φ : R → S being the inclusion map; we obtain a unique ring homomorphism ψ : S → S such that ψ(x) = x and ψ|R = φ. We may rephrase the last property by saying that ψ|R is the identity on R. 11 also provides a ring homomorphism ψ : S → S such that ψ (x ) = x and ψ |R is the identity on R.

14 follows immediately from the right noetherian case. 15. Let α be an automorphism of a ring R and T = R[x±1 ; α]. If R is right (left) noetherian, then so is T . Proof. Set S = R[x; α] and remember that S is a subring of T . We proceed by relating the right ideals of T to those of S, as follows. Claim: If I is a right ideal of T , then I ∩ S is a right ideal of S and I = (I ∩ S)T . It is clear that I ∩ S is a right ideal of S and that (I ∩ S)T ⊆ I. If p ∈ I, then p = am xm + am+1 xm+1 + · · · + an xn for some integers m ≤ n and coeﬃcients ai ∈ R.

Since φ(y i xj ) = yˆi x ˆj i j for all i, j, conclude that the monomials y x are linearly independent and thus that φ is an isomorphism. (b) Since R = k[ˆ y ] is a polynomial ring over k, there is a unique k-algebra homomorphism η : R → Oq (k 2 ) such that η(ˆ y ) = y. 11 that η extends uniquely x) = x. Finally, show to a ring homomorphism ψ : S → Oq (k 2 ) such that ψ(ˆ that φ and ψ are inverses of each other. Now that we have Oq (k 2 ) ∼ = S, we can say that Oq (k 2 ) is a skew polynomial ring.

### An Introduction to Noncommutative Noetherian Rings by K. R. Goodearl, R. B. Warfield Jr

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