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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 [itex] V = \mathcal{Z}(x^3 - y^2) [/itex] in [itex] \ \ \mathbb{A}^2 [/itex].
If [itex] (a, b) \in \mathbb{A}^2 [/itex] is an element of V, then [itex] a^3 = b^2 [/itex].
If [itex] a \ne 0 [/itex], then also [itex] b \ne 0 [/itex] and we can write[itex] a = (b/a)^2, \ b = (b/a)^3 [/itex].
It follows that V is the set [itex] \{ (a^2, a^3) \ | \ a \in k \} [/itex].
For any polynomial [itex] f(x,y) \in k[x,y] [/itex]. we can write [itex] f(x,y) = f_0(x) + f_1(x)y + (x^3 - y^2)g(x,y) [/itex]
... ... ... etc etc
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I cannot follow the line of reasoning:
"For any polynomial [itex] f(x,y) \in k[x,y] [/itex]. we can write [itex] f(x,y) = f_0(x) + f_1(x)y + (x^3 - y^2)g(x,y) [/itex]"
Can anyone clarify why this is true and why D&F are taking this step?
Peter
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Let [itex] V = \mathcal{Z}(x^3 - y^2) [/itex] in [itex] \ \ \mathbb{A}^2 [/itex].
If [itex] (a, b) \in \mathbb{A}^2 [/itex] is an element of V, then [itex] a^3 = b^2 [/itex].
If [itex] a \ne 0 [/itex], then also [itex] b \ne 0 [/itex] and we can write[itex] a = (b/a)^2, \ b = (b/a)^3 [/itex].
It follows that V is the set [itex] \{ (a^2, a^3) \ | \ a \in k \} [/itex].
For any polynomial [itex] f(x,y) \in k[x,y] [/itex]. we can write [itex] f(x,y) = f_0(x) + f_1(x)y + (x^3 - y^2)g(x,y) [/itex]
... ... ... etc etc
----------------------------------------------------------------------------------------------
I cannot follow the line of reasoning:
"For any polynomial [itex] f(x,y) \in k[x,y] [/itex]. we can write [itex] f(x,y) = f_0(x) + f_1(x)y + (x^3 - y^2)g(x,y) [/itex]"
Can anyone clarify why this is true and why D&F are taking this step?
Peter