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I Affine Algebraic Sets - Dummit and Foote, page 660, Ex. 3

  1. Jun 17, 2016 #1
    I am trying to gain an understanding of the basics of elementary algebraic geometry and am reading Dummit and Foote Chapter 15: Commutative Rings and Algebraic Geometry ...

    At present I am focused on Section 15.1 Noetherian Rings and Affine Algebraic Sets ... ...

    I need someone to help me to fully understand the reasoning/analysis behind one of the statements in Example (3) on Page 660 of D&F ...

    On page 660 (in Section 15.1) of D&F we find the following text and examples (I am specifically focused on Example (3)):


    ?temp_hash=90edf888d4a48ef79bbbb2b712fc9e1e.png



    In the above text, in Example (3), we find the following:

    "... ... For any polynomial ##f(x,y) \in k[x,y]## we can write

    ##f(x,y) = f_0(x) + f_1(x)y + (x^3 - y^2) g(x,y).##"


    Can someone explain ( frown.png slowly and carefully) exactly how/why this is true ... ...

    Peter


    ====================================================

    In order for readers of the above post to understand the context of the question and the notation employed I am providing the introductory pages on affine algebraic sets in the D&F text ... ... as follows:



    ?temp_hash=90edf888d4a48ef79bbbb2b712fc9e1e.png
    ?temp_hash=90edf888d4a48ef79bbbb2b712fc9e1e.png
    ?temp_hash=90edf888d4a48ef79bbbb2b712fc9e1e.png
     

    Attached Files:

  2. jcsd
  3. Jun 17, 2016 #2

    fresh_42

    Staff: Mentor

    Hi Peter,

    let us start and consider ##f(x,y)## as polynomial in ##y##. Then it can be written:
    ##f(x,y) = c_0 \cdot f_0(x) + c_1 \cdot y \cdot f_1(x) + c_2 \cdot y^2 \cdot f_2(x) + \dots##
    What we are doing next is to gather everything, that "disturbs" us and putting it into the terms ##f_0## and ## f_1##.
    There are no restrictions on those, beside that they aren't allowed to contain ##y##'s. so our first substitution will be ##f_0^{new} = c_0 f_0## and ##f_1^{new} = c_1 f_1##. For the next steps I will drop the ##^{new}## because it is not necessary to keep a record on them.

    Now ##c_2 \cdot y^2 \cdot f_2(x) = c_2 \cdot y^2 \cdot (d_0 + d_1 \cdot x + d_2 \cdot x^2 + \dots) = c_2d_0 y^2 + c_2d_1xy^2+c_2d_2x^2y^2 + \dots## and every term is of the form ##e_n \cdot x^n \cdot y^2.##
    This can be written as ##e_n \cdot x^n \cdot y^2 = (x^3-y^2) \cdot (-e_n x^n) + e_n \cdot x^{n+3}.##
    The first factor at ##(x^3-y^2)## will be used to build up ##g(x,y)## and the second will be added to ##f_0.##

    You can proceed this way with every following term ##c_n y^n f_n(x) \; (n>2)##, i.e. perform the long division on polynomials.
     
  4. Jun 17, 2016 #3

    mathwonk

    User Avatar
    Science Advisor
    Homework Helper

    To summarize fresh34's nice answer, for any domain R, we can divide by any polynomial in the ring R[Y] whose lead coefficient is a unit. E.g. we can always divide by Y^2 + aY + b. Moreover, as is usual in long division, the remainder has lower degree than the divisor.

    Thus if we take R = k[X], then in the ring R[Y] = k[X][Y] ≈ k[X,Y], we can divide any polynomial by Y^2 - X^3, and get a remainder of degree lower than 2 (in Y). I.e. the remainder will have form f1Y + f0 where f0 and f1 lie in the coefficient ring k[X].

    The point of this somewhat tedious example is to illustrate that over a finite field, one does not keep much information about a polynomial from looking only where it vanishes, since over Z/2Z, any multiple of the polynomial x(x-1) vanishes everywhere. to insure that the zero locus has as much information as possible, one usually assumes the field algebraically closed. see e.g. hilbert's nullstellensatz in DF, p.675.
     
    Last edited: Jun 17, 2016
  5. Jun 17, 2016 #4

    Thanks fresh_42 ... that post was most helpful ...

    You really helped me when you wrote:

    "This can be written as ##e_n \cdot x^n \cdot y^2 = (x^3-y^2) \cdot (-e_n x^n) + e_n \cdot x^{n+3}.##
    The first factor at ##(x^3-y^2)## will be used to build up ##g(x,y)## and the second will be added to ##f_0.##"

    I was really perplexed at how to deal with this situation ...

    Thanks again for the help,

    Peter
     
  6. Jun 17, 2016 #5

    Thanks mathwonk ... really helpful to get a general perspective on this matter ...

    Thank you for moving my understanding forward ... really appreciate it ...

    Peter
     
  7. Jun 17, 2016 #6

    fresh_42

    Staff: Mentor

    You're welcome. It's always a pleasure to read your carefully prepared and precise questions.
     
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