# Thread: Algebriac Geometry - Morphisms of Algebraic Sets

1. ## Algebriac Geometry - Morphisms of Algebraic Sets

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 $\displaystyle \phi \ : V \rightarrow W$ is called a morphism (or polynomial map or regular map) of algebraic sets if

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

$\displaystyle \phi(( a_1, a_2, ... a_n)) = ( {\phi}_1 ( a_1, a_2, ... a_n) , {\phi}_2 ( a_1, a_2, ... a_n), ... ... ... {\phi}_m ( a_1, a_2, ... a_n))$

for all $\displaystyle ( a_1, a_2, ... a_n) \in V$

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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 $\displaystyle k[x_1, x_2, ... ... x_n]$.

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

since $\displaystyle {\phi}_1, {\phi}_2, .......... , {\phi}_m$ are polynomials in $\displaystyle x_1, x_2, ... ... x_n$.

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

since $\displaystyle \phi (( a_1, a_2, ... a_n)) \in W$.

Thus $\displaystyle F \circ \phi \in \mathcal{I}(V)$

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

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

$\displaystyle \widetilde{\phi} \ : \ k[W] \rightarrow k[V]$

$\displaystyle f \rightarrow f \circ \phi$

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

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

Peter