- Thread starter
- #1

- Jun 22, 2012

- 2,918

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)

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

(1) the kernel of \(\displaystyle \widetilde{\phi} \) is \(\displaystyle \mathcal{I} ( \phi (V) ) \)

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

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

[Note: For the definitions of \(\displaystyle \phi \) and \(\displaystyle \widetilde{\phi} \) see attachment page 662 ]

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

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

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

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 \(\displaystyle \widetilde{\phi} = f \circ \phi \) we have \(\displaystyle \widetilde{\phi}(f) = 0 \) if and only if \(\displaystyle (f \circ \phi) (P) = 0 \) for all \(\displaystyle P \in V \)"

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

Definition. A map [TEX] \phi \ : V \rightarrow W [/TEX] is called a morphism (or polynomial map or regular map) of algebraic sets if

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

[TEX] \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)) [/TEX]

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

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

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

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

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

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

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

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

\(\displaystyle \widetilde{\phi}(f) = 0 \Longrightarrow f \circ \phi (P) = 0 \)

But \(\displaystyle f \circ \phi (P) = 0 \) means that

\(\displaystyle f \circ \phi (P) = 0 + \mathcal{I}(V) \)

so then \(\displaystyle f \circ \phi \in \mathcal{I}(V) \)

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

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

Any clarifying comments are really welcome.

Peter

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

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

**Proposition 16.**Suppose \(\displaystyle \phi \ : \ V \longrightarrow W \) is a morphism of algebraic sets and \(\displaystyle \widetilde{\phi} \ : \ k[W] \longrightarrow k[V] \) is the associated k-algebra homomorphism of coordinate rings. Then(1) the kernel of \(\displaystyle \widetilde{\phi} \) is \(\displaystyle \mathcal{I} ( \phi (V) ) \)

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

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

[Note: For the definitions of \(\displaystyle \phi \) and \(\displaystyle \widetilde{\phi} \) see attachment page 662 ]

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

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

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

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 \(\displaystyle \widetilde{\phi} = f \circ \phi \) we have \(\displaystyle \widetilde{\phi}(f) = 0 \) if and only if \(\displaystyle (f \circ \phi) (P) = 0 \) for all \(\displaystyle P \in V \)"

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

DefinitionsDefinitions

**Definition of Morphism or Polynomial Mapping \(\displaystyle \phi \)**Definition. A map [TEX] \phi \ : V \rightarrow W [/TEX] is called a morphism (or polynomial map or regular map) of algebraic sets if

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

[TEX] \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)) [/TEX]

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

**Definition of \(\displaystyle \widetilde{\phi}\)**[TEX] \phi [/TEX] induces a well defined map from the quotient ring [TEX] k[x_1, x_2, ... ... x_n]/\mathcal{I}(W) [/TEX]

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

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

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

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

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

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

**My interpretation of this line is as follows:**\(\displaystyle \widetilde{\phi}(f) = 0 \Longrightarrow f \circ \phi (P) = 0 \)

But \(\displaystyle f \circ \phi (P) = 0 \) means that

\(\displaystyle f \circ \phi (P) = 0 + \mathcal{I}(V) \)

so then \(\displaystyle f \circ \phi \in \mathcal{I}(V) \)

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

**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

Last edited: