Results 1 to 1 of 1

Thread: Algebriac Geometry - Morphisms of Algebraic Sets

  1. #1
    Super Member Bernhard's Avatar
    Joined
    Jan 2010
    From
    Hobart, Tasmania, Australia
    Posts
    594
    Thanks
    2

    Algebriac Geometry - Morphisms of Algebraic Sets

    I am reading Dummit and Foote (D&F) Section 15.1 on Affine Algebraic Sets.

    On page 662 (see attached) D&F define a morphism or polynomial map of algebraic sets as follows:

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

    Definition. A map $\displaystyle \phi \ : V \rightarrow W $ is called a morphism (or polynomial map or regular map) of algebraic sets if

    there are polynomials $\displaystyle {\phi}_1, {\phi}_2, .......... , {\phi}_m \in k[x_1, x_2, ... ... x_n] $ such that

    $\displaystyle \phi(( a_1, a_2, ... a_n)) = ( {\phi}_1 ( a_1, a_2, ... a_n) , {\phi}_2 ( a_1, a_2, ... a_n), ... ... ... {\phi}_m ( a_1, a_2, ... a_n)) $

    for all $\displaystyle ( a_1, a_2, ... a_n) \in V $

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


    D&F then go on to define a map between the quotient rings k[W] and k[V] as follows: (see attachment page 662)


    -------------------------------------------------------------------------------------------------------
    Suppose F is a polynomial in $\displaystyle k[x_1, x_2, ... ... x_n] $.

    Then $\displaystyle F \circ \phi = F({\phi}_1, {\phi}_2, .......... , {\phi}_m) $ is a polynomial in $\displaystyle k[x_1, x_2, ... ... x_n] $

    since $\displaystyle {\phi}_1, {\phi}_2, .......... , {\phi}_m $ are polynomials in $\displaystyle x_1, x_2, ... ... x_n $.

    If $\displaystyle F \in \mathcal{I}(W)$, then $\displaystyle F \circ \phi (( a_1, a_2, ... a_n)) = 0 $ for every $\displaystyle ( a_1, a_2, ... a_n) \in V $

    since $\displaystyle \phi (( a_1, a_2, ... a_n)) \in W $.

    Thus $\displaystyle F \circ \phi \in \mathcal{I}(V) $

    It follows that $\displaystyle \phi $ induces a well defined map from the quotient ring $\displaystyle k[x_1, x_2, ... ... x_n]/\mathcal{I}(W) $

    to the quotient ring $\displaystyle k[x_1, x_2, ... ... x_n]/\mathcal{I}(V) $ :

    $\displaystyle \widetilde{\phi} \ : \ k[W] \rightarrow k[V] $

    $\displaystyle f \rightarrow f \circ \phi $

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

    My problem is, how exactly does it follow (and why?) that $\displaystyle \phi $ induces a well defined map from the quotient ring $\displaystyle k[x_1, x_2, ... ... x_n]/\mathcal{I}(W) $ to the quotient ring $\displaystyle k[x_1, x_2, ... ... x_n]/\mathcal{I}(V) $ ?

    Can someone (explicitly) show me the logic of this - why exactly does it follow?

    Peter
    Attached Files Attached Files
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. 'Curly' Z and I - Affine algebraic sets
    Posted in the LaTeX Help Forum
    Replies: 2
    Last Post: Dec 20th 2013, 08:10 PM
  2. Ideals of Algebraic Sets
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Oct 9th 2011, 11:46 AM
  3. algebraic laws of sets
    Posted in the Algebra Forum
    Replies: 1
    Last Post: May 29th 2010, 07:36 AM
  4. algebraic laws of sets
    Posted in the Algebra Forum
    Replies: 11
    Last Post: May 27th 2010, 01:31 PM
  5. Bijection between sections and vector bundle morphisms
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Apr 1st 2010, 12:02 PM

Search Tags


/mathhelpforum @mathhelpforum