# Algebraic Geometry - Morphisms of Algebraic Sets - Proposition 16 (D&F)

#### Peter

##### Well-known member
MHB Site Helper
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 $$\displaystyle \phi \ : \ V \longrightarrow W$$ 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 ... ... ...

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[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:

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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.

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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 $$\displaystyle \widetilde{\phi} = f \circ \phi$$ 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.

Definitions

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

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