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Algebriac Geometry - Morphisms of Algebraic Sets

Peter

Well-known member
MHB Site Helper
Jun 22, 2012
2,918
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:

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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]

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


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Suppose F is a polynomial in [TEX] k[x_1, x_2, ... ... x_n] [/TEX].

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

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

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

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

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

It follows that [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]

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

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My problem is, how exactly does it follow (and why?) that [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] ?

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

Peter