This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 5126 Monday, April 14 Tenth Homework Solutions 1. Let I = ( x , x 2 y , x 3 y 2 ,... ) and suppose I is finitely generated. Then there is a positive integer p such that I = ( x , x 2 y , x 3 y 2 ,..., x p + 1 y p ) . We now show that x p + 2 y p + 1 / I , which will yield a contradiction. The general element of R is a klinear sum of prod ucts (including the empty product, which we interpret as 1) of the form x i + 1 y j . If x p + 2 y p + 1 I , then we may write x p + 2 y p + 1 = x f 1 + x 2 yf 2 + + x p + 1 y p f p + 1 with f i R for all i . We note that when x i y i 1 f i is written out as a klinear sum of elements in the form x r y s , only those with either r = s + 1 with s p (corresponding to the constant term of f s ) or r s + 2. Thus x p + 2 y p + 1 cannot appear in the sum and we have the required contradiction. For the final sentence, if R was a finitely generated kalgebra, then R would be Noethe rian (by Hilberts basis theorem), which would mean that all ideals, and in particular...
View
Full
Document
 Fall '07
 PALinnell
 Math, Algebra

Click to edit the document details