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- Jun 22, 2012

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I am reading Dummit and Foote Ch 15, Commutative Rings and Algebraic Geometry. In Section 15.1 Noetherian Rings and Affine Algebraic Sets, Example 3 on page 660 reads as follows: (see attachment)

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Let [TEX] V = \mathcal{Z}(x^3 - y^2) [/TEX] in [TEX] \ \ \mathbb{A}^2 [/TEX].

If [TEX] (a, b) \in \mathbb{A}^2 [/TEX] is an element of V, then [TEX] a^3 = b^2 [/TEX].

If [TEX] a \ne 0 [/TEX], then also [TEX] b \ne 0 [/TEX] and we can write[TEX] a = (b/a)^2, \ b = (b/a)^3 [/TEX].

It follows that V is the set [TEX] \{ (a^2, a^3) \ | \ a \in k \} [/TEX].

For any polynomial [TEX] f(x,y) \in k[x,y] [/TEX]. we can write [TEX] f(x,y) = f_0(x) + f_1(x)y + (x^3 - y^2)g(x,y) [/TEX]

... ... ... etc etc

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I cannot follow the line of reasoning:

"For any polynomial [TEX] f(x,y) \in k[x,y] [/TEX]. we can write [TEX] f(x,y) = f_0(x) + f_1(x)y + (x^3 - y^2)g(x,y) [/TEX]"

Can anyone clarify why this is true and why D&F are taking this step?

Peter

[This has also been posted on MHF]

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Let [TEX] V = \mathcal{Z}(x^3 - y^2) [/TEX] in [TEX] \ \ \mathbb{A}^2 [/TEX].

If [TEX] (a, b) \in \mathbb{A}^2 [/TEX] is an element of V, then [TEX] a^3 = b^2 [/TEX].

If [TEX] a \ne 0 [/TEX], then also [TEX] b \ne 0 [/TEX] and we can write[TEX] a = (b/a)^2, \ b = (b/a)^3 [/TEX].

It follows that V is the set [TEX] \{ (a^2, a^3) \ | \ a \in k \} [/TEX].

For any polynomial [TEX] f(x,y) \in k[x,y] [/TEX]. we can write [TEX] f(x,y) = f_0(x) + f_1(x)y + (x^3 - y^2)g(x,y) [/TEX]

... ... ... etc etc

-------------------------------------------------------------------------------------------------------------------------------------

I cannot follow the line of reasoning:

"For any polynomial [TEX] f(x,y) \in k[x,y] [/TEX]. we can write [TEX] f(x,y) = f_0(x) + f_1(x)y + (x^3 - y^2)g(x,y) [/TEX]"

Can anyone clarify why this is true and why D&F are taking this step?

Peter

[This has also been posted on MHF]

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