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Algebriac Geometry - Morphisms of Algebraic Sets

  1. Nov 1, 2013 #1
    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 [itex] \phi \ : V \rightarrow W [/itex] is called a morphism (or polynomial map or regular map) of algebraic sets if

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

    [itex] \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)) [/itex]

    for all [itex] ( a_1, a_2, ... a_n) \in V [/itex]

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


    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 [itex] k[x_1, x_2, ... ... x_n] [/itex].

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

    since [itex] {\phi}_1, {\phi}_2, .......... , {\phi}_m [/itex] are polynomials in [itex] x_1, x_2, ... ... x_n [/itex].

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

    since [itex] \phi (( a_1, a_2, ... a_n)) \in W [/itex].

    Thus [itex] F \circ \phi \in \mathcal{I}(V) [/itex]

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

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

    [itex] \widetilde{\phi} \ : \ k[W] \rightarrow k[V] [/itex]

    [itex] f \rightarrow f \circ \phi [/itex]

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

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

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

    Peter
     

    Attached Files:

  2. jcsd
  3. Nov 2, 2013 #2
    In general: Let ##R## and ##R^\prime## be rings and let ##I## and ##I^\prime## be ideals of the respective rings. Let ##f:R\rightarrow R^\prime## be a ring morphism.

    Let ##\pi^\prime:R^\prime\rightarrow R^\prime/I^\prime## be the canonical quotient morphism which sends an element ##y## in ##R^\prime## to its equivalence class ##[y]##. So, we consider the map ##\pi^\prime \circ f## which sends ##x\in R## to ##[f(x)]##.

    Now, assume that for all ##x\in I##, we have that ##f(x)\in I^\prime##. This means that ##\pi(f(x)) = 0## for each ##x\in I##. So ##I\subseteq ker(\pi \circ f)##. This implies directly that there is a unique morphism ##F:R/I\rightarrow R^\prime /I^\prime## such that ##F( [x]) = [f(x)]##. Let us prove this. Uniqueness is clear since I have defined ##F## at each point ##R/I## explicitely. To prove existence, assume that ##[x] = [x^\prime]## for some ##x,x^\prime \in R^\prime##. Then ##x-x^\prime \in I##, and thus ##\pi(f(x-x^\prime)) = 0##. Thus ##F([x]) = \pi(f(x)) = \pi(f(x^\prime)) = F([x^\prime])##.
     
  4. Nov 3, 2013 #3
    Thanks R136a1! Appreciate your help

    Now working through your post carefully

    Peter
     
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