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

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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 Hilberts basis theorem), which would mean that all ideals, and in particular...
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ahw10 - Math 5126 Monday, April 14 Tenth Homework Solutions...

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