# ahw10 - Math 5126 Monday April 14 Tenth Homework Solutions...

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
This is the end of the preview. Sign up to access the rest of the 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 k-linear 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 k-linear 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 k-algebra, then R would be Noethe- rian (by Hilbert’s basis theorem), which would mean that all ideals, and in particular...
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

ahw10 - Math 5126 Monday April 14 Tenth Homework Solutions...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online