Algebriac Geometry - Morphisms of Algebraic Sets

samedi 2 novembre 2013

I am reading Dummit and Foote: Section 15.2 Radicals and Affine Varieties.

On page 678, Proposition 16 reads as follows: (see attachment, page 678)

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Proposition 16. Suppose [itex] \phi \ : \ V \longrightarrow W [/itex] is a morphism of algebraic sets and [itex] \widetilde{\phi} \ : \ k[W] \longrightarrow k[V] [/itex] is the associated k-algebra homomorphism of coordinate rings. Then

(1) the kernel of [itex] \widetilde{\phi} [/itex] is [itex] \mathcal{I} ( \phi (V) ) [/itex]

(2) etc etc ... ... ...

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[Note: For the definitions of [itex] \phi [/itex] and [itex] \widetilde{\phi} [/itex] see attachment page 662 ]

The beginning of the proof of Proposition 16 reads as follows:

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Proof. Since [itex] \widetilde{\phi} = f \circ \phi [/itex] we have [itex] \widetilde{\phi}(f) = 0 [/itex] if and only if [itex] (f \circ \phi) (P) = 0 [/itex] for all [itex] P \in V [/itex] i.e. [itex] f(Q) = 0 [/itex] for all [itex] Q = \phi (P) \in \phi (V) [/itex]. which is the statement that [itex] f \in \mathcal{I} ( \phi ( V) ) [/itex] proving the first statement.

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My problem concerns the first sentence of the proof above.

Basically I am trying to fully understand what is meant, both logically and notationally, by the following:

"Since [itex] \widetilde{\phi} = f \circ \phi [/itex] we have [itex] \widetilde{\phi}(f) = 0 [/itex] if and only if [itex] (f \circ \phi) (P) = 0 [/itex] for all [itex] P \in V [/itex]"

My interpretation of this statement is given below after I give the reader some key definitions.

Definitions

Definition of Morphism or Polynomial Mapping [itex] \phi [/itex]

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]

Definition of [itex] \widetilde{\phi}[/itex]

[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]

i.e [itex] k[x_1, x_2, ... ... x_n]/\mathcal{I}(W) \longrightarrow k[x_1, x_2, ... ... x_n]/\mathcal{I}(V) [/itex]

[itex] f \rightarrow f \circ \phi [/itex] i.e. [itex] \phi (F) = f \circ \phi [/itex]

Now, to repeat again for clarity, my problem is with the first line of the proof of Proposition 16 which reads:

"Since [itex] \widetilde{\phi} = f \circ \phi [/itex] we have [itex] \widetilde{\phi}(f) = 0 [/itex] if and only if [itex] f \circ \phi (P) = 0 [/itex] for all [itex] P \in V [/itex]"

My interpretation of this line is as follows:

[itex] \widetilde{\phi}(f) = 0 \Longrightarrow f \circ \phi (P) = 0 [/itex]

But [itex] f \circ \phi (P) = 0 [/itex] means that

[itex] f \circ \phi (P) = 0 + \mathcal{I}(V) [/itex]

so then [itex] f \circ \phi \in \mathcal{I}(V) [/itex]

Thus [itex] (f \circ \phi) (P) = 0 [/itex] for all points [itex] P = (a_1, a_2, ... ... , a_n \in V \subseteq \mathbb{A}^n [/itex]

Can someone confirm that my logic and interpretation is valid and acceptable, or not as the case may be - please point out any errors or weaknesses in the argument/proof.

I think some of my problems with Dummit and Foote are notational in nature

Any clarifying comments are really welcome.

Peter

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