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Affine Algebraic Sets - D&F Chapter 15, Section 15.1 - Example 3 - pag

  1. Oct 29, 2013 #1
    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)

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

    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
     
  2. jcsd
  3. Oct 30, 2013 #2
    It suffices to prove this for ##f(x,y) = x^n y^m##.

    We prove it by induction on ##m##:
    For ##m=0##, take ##f_0(x) = x^n## and the rest ##0##.
    For ##m=1##, take ##f_1(x) = x^n## and the rest ##0##.

    If it holds for ##m<m^\prime##, then write ##x^n y^m = - x^ny^{m-2} (x^3 - y^2 ) - x^{n+3}y^{m-2}##.
    By induction, express ##x^{n+3}y^{m-2}## in the required form, then you can also express ##x^n y^m## in such form.
     
  4. Oct 30, 2013 #3
    Thank you so much for your help with this problem

    I had nearly given up on it, and with it my progress into algebraic geometry!

    So thanks again!

    Peter
     
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