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

This advent to noncommutative noetherian jewelry, 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. vast explanatory fabric is given, and workouts are built-in all through. New fabric comprises the fundamental different types of quantum teams.

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Rk is a finite list of nonzero generators for J. There exist p1 , . . , pk ∈ I such that pi has leading coefficient ri and some degree ni . Set n = max{n1 , . . , nk }, and note that pi xn−ni is an element of I with leading coefficient ri but with degree n. Thus, there is no loss of generality in assuming that all the pi have the same degree n, that is, pi = ri xn + [lower terms]. Step 3. Set N = R + Rx + · · · + Rxn−1 , the set of elements of S with degree less than n. Observe that N = R + xR + · · · + xn−1 R, since b0 + b1 x + · · · + bn−1 xn−1 = b0 + xα−1 (b1 ) + · · · + xn−1 α1−n (bn−1 ) c0 + xc1 + · · · + xn−1 cn−1 = c0 + α(c1 )x + · · · + αn−1 (cn−1 )xn−1 for all bj , cj ∈ R.

Set N = R + Rx + · · · + Rxn−1 , the set of elements of S with degree less than n. Observe that N = R + xR + · · · + xn−1 R, since b0 + b1 x + · · · + bn−1 xn−1 = b0 + xα−1 (b1 ) + · · · + xn−1 α1−n (bn−1 ) c0 + xc1 + · · · + xn−1 cn−1 = c0 + α(c1 )x + · · · + αn−1 (cn−1 )xn−1 for all bj , cj ∈ R. Consequently, N is a right (as well as left) R-submodule of S. 4. Hence, its submodule I ∩ N is a finitely generated right R-module; say q1 , . . , qt generate I ∩ N . 18 CHAPTER 1 Step 4. Let I0 be the right ideal of S generated by p1 , .

1Y. Show that R[x±1 ; α]op = Rop [x±1 ; α−1 ] if α is an automorphism of a ring R. (Cf. ) 1Z. 17 if the skew-Laurent ring T = R[x±1 ; α] is replaced by a skew polynomial ring S = R[x; α], where we continue to assume that α is an automorphism of R. We have already noted that S always has some nontrivial ideals, namely Sx, Sx2 , . . (a) If I is any ideal of R, show that Ixn + Sxn+1 is an ideal of S for each n ∈ N. Thus, if R is not simple, S has nonzero ideals not generated by powers of x. (b) If R is simple and no positive power of α is inner, show that the only nonzero ideals of S are S, Sx, Sx2 , .

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