Algebriac Geometry - Morphisms of Algebraic Sets

[Math Amateur]

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

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

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

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

-------------------------------------------------------------------------------------------------------D&F then go on to define a map between the quotient rings k[W] and k[V] as follows: (see attachment page 662)-------------------------------------------------------------------------------------------------------
Suppose F is a polynomial in $$k[x_1, x_2, ... ... x_n]$$.

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

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

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

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

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

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

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

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

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

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

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

Peter

 

[jvicens]

Hello Peter,

I can definitely help explain the logic behind why \phi induces a well-defined map from the quotient ring k[x_1, x_2, ... ... x_n]/\mathcal{I}(W) to the quotient ring k[x_1, x_2, ... ... x_n]/\mathcal{I}(V).

First, let's review the definitions of quotient rings and polynomial maps. The quotient ring k[x_1, x_2, ... ... x_n]/\mathcal{I}(W) is the set of all cosets of polynomials in k[x_1, x_2, ... ... x_n] modulo the ideal \mathcal{I}(W). This means that we are essentially taking all the polynomials in k[x_1, x_2, ... ... x_n] that vanish on the algebraic set W and considering them as equivalent elements in the quotient ring. Similarly, the quotient ring k[x_1, x_2, ... ... x_n]/\mathcal{I}(V) is the set of all cosets of polynomials in k[x_1, x_2, ... ... x_n] modulo the ideal \mathcal{I}(V).

Now, a polynomial map \phi: V \rightarrow W is defined as a map that takes points in V to points in W by applying a set of polynomials {\phi}_1, {\phi}_2, ... , {\phi}_m to the coordinates of the point. In other words, for any point (a_1, a_2, ... a_n) \in V, \phi maps it to the point ({\phi}_1 (a_1, a_2, ... a_n), {\phi}_2 (a_1, a_2, ... a_n), ... {\phi}_m (a_1, a_2, ... a_n)) \in W.

Now, let's consider a polynomial F \in k[x_1, x_2, ... ... x_n] that vanishes on the algebraic set W, meaning F({\phi}_1, {\phi}_2, ... , {\phi}_m) = 0 for all points in W. This also means that F \circ \phi vanishes on the algebraic set V, since for any point (a_1, a_2, ... a_n) \in