Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Affine Algebraic Sets - D&F Chapter 15, Section 15.1 - Example 2

  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 2 on page 660 reads as follows: (see attachment)

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

    (2) Over any field k, the ideal of functions vanishing at [itex] (a_1, a_2, ... ... ... a_n) \in \mathbb{A}^n [/itex] is a maximal ideal since it is the kernel of the surjective ring homomorphism from [itex] k[x_1, x_2, ... ... x_n] [/itex] to the field k given by evaluation at [itex] (a_1, a_2, ... ... ... a_n) [/itex].

    It follows that [itex] \mathcal{I}((a_1, a_2, ... ... ... a_n)) = (x - a_1, x - a_2, ... ... ... , x - a_n) [/itex]

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

    I can see that [itex] (x - a_1, x - a_2, ... ... ... , x - a_n) [/itex] gives zeros for each polynomial in [itex] k[ \mathbb{A}^n ] [/itex] - indeed, to take a specific example involving [itex] \mathbb{R} [x,y] [/itex] we have for, let us say, a particular polynomial [itex] g \in \mathbb{R} [x,y] [/itex] where g is as follows:

    [itex] g(x,y) = 6(x - a_1)^3 + 11(x - a_1)^2(y - a_2) + 12(y - a_2)^2 [/itex]

    so in this case, clearly [itex] g(a_1, a_2) = 0 [/itex] ... ... ... and, of course, other polynomials in [itex] \mathbb{R} [x,y] [/itex] similarly.

    BUT ... ... I cannot understand D&Fs reference to maximal ideals. Why is it necessary to reason about maximal ideals.

    Since I am obviously missing something, can someone please help by explaining what is going on in this example.

    Another issue I have is why do D&F write [itex] \mathcal{I}((a_1, a_2, ... ... ... a_n)) [/itex] with 'double' parentheses and not just [itex] \mathcal{I}(a_1, a_2, ... ... ... a_n) [/itex]?

    Would appreciate some help.

    Peter

    Note - see attachment for definition of [itex] \mathcal{I}(A) [/itex]
     

    Attached Files:

    Last edited: Oct 29, 2013
  2. jcsd
  3. Oct 30, 2013 #2
    The ideal ##I=\mathcal{I}((a_1,...,a_n))## is the set of all polynomials vanish on the point ##\mathbf{a} = (a_1,....,a_n)##.

    The reasoning is as follows. Note that the polynomials ##X_i-a_i## certainly vanish on this point, as is easily checked. So ##X_i-a_i\in I## certainly. Thus the ideal generated by these polynomials is in ##I## too. So ##(X_1-a_1,...,X_n-a_n)\subseteq I##. But perhaps ##I## contains more! Here is our information about maximality comes in, because if ##I## contains more, it must be the entire ring ##k[X_1,...,X_n]##. But this can not be since the constant polynomial ##1## does not vanish on the set.
     
  4. Oct 30, 2013 #3
    Excellent!!! Now see the link to maximal ideals!

    But will reflect on this further so I am sure I have understood all angles of the theory

    Thank you so much for this critical help ... I can now progress with more confidence!

    Peter
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Affine Algebraic Sets - D&F Chapter 15, Section 15.1 - Example 2
Loading...