How is the Map $\tilde{\varphi}$ Well-Defined in Algebraic Geometry?

[Math Amateur]

I am trying to gain an understanding of the basics of elementary algebraic geometry and am reading Dummit and Foote Chapter 15: Commutative Rings and Algebraic Geometry ...

At present I am focused on Section 15.1 Noetherian Rings and Affine Algebraic Sets ... ...

I need someone to help me to fully understand the reasoning/analysis behind the statements following the Definition of morphisms (or polynomial maps) of algebraic sets (which eventually leads to the statement of Theorem 6, Section 15.1) ... ...

On page 662 (in Section 15.1) of D&F we find the following relevant text ... :
View attachment 4770
View attachment 4771In the above text, after the Definition of morphisms of algebraic sets, we find the following:

" ... ... Thus \(\displaystyle F \circ \phi \in \mathcal{I} (V) \). It follows that \(\displaystyle \phi\) induces a well defined map from the quotient ring \(\displaystyle k[ x_1, \ ... \ ... \ , x_n ]/ \mathcal{I} (W)\) to the quotient ring \(\displaystyle k[ x_1, \ ... \ ... \ , x_n ]/ \mathcal{I} (V)\):

\(\displaystyle \tilde{ \phi } \ : \ k[W] \rightarrow k[V]\)

\(\displaystyle f \mapsto f \circ \phi \)

... ... "
My question is as follows:

Can someone please explain exactly (formally and rigorously) how the above statement follows: that is, how/why is it true?

Hope someone can help ...

Peter

*** NOTE ***

The above notes from Dummit and Foote displayed above, constitute the proof of the first part of Theorem 6, Section 15.1 which reads as follows:View attachment 4772

 

[Deveno]

What is it you don't understand? The definition of the mapping $\tilde{\varphi}$, or how it is well-defined?

The well-defined issue is easier to dispatch: since $\tilde{\varphi}$ is defined on an equivalence class (coset of functions), one has to verify that:

$\tilde{\varphi}(F) = \tilde{\varphi}(G)$ if $F - G \in \mathcal{I}(W)$.

Now $\tilde{\varphi}(F) = F\circ \varphi + \mathcal{I}(V)$ and $\tilde{\varphi}(G) = G\circ \varphi + \mathcal{I}(V)$.

To assert that $\tilde{\varphi}(F) = \tilde{\varphi}(G)$ is to say that:

$\tilde{\varphi}(F) - \tilde{\varphi}(G) = 0_{k[V]} = 0 + \mathcal{I}(V) = \mathcal{I}(V)$.

Thus we seek to show that:

$F - G \in \mathcal{I}(W) \implies [ F\circ \varphi + \mathcal{I}(V)] - [G\circ \varphi + \mathcal{I}(V)] = \mathcal{I}(V)$, which is true when:

$(F \circ \varphi) - (G \circ\varphi) \in \mathcal{I}(V)$.

However, $(F \circ \varphi) - (G \circ\varphi) = (F - G)\circ \varphi$, and since

$F - G$ annihilates $W$ (which is what it MEANS to be an element of $\mathcal{I}(W)$), and $\varphi(V) \subseteq W$, it is clear that $(F-G) \circ \varphi$ annihilates $V$, that is:

$(F - G)\circ \varphi \in \mathcal{I}(V)$.

Thus it doesn't matter which "$F$" we use to define $f = F + \mathcal{I}(W)$, as $\tilde{\varphi}$ is constant on the entire coset.

********************

The contravariant nature of $\tilde{\varphi}$ is often expressed as saying $\tilde{\varphi}(f)$ is a pullback of $f$ along $\varphi$, for just as $\varphi$ "goes forward" from $V \to W$, $\tilde{\varphi}$ "pulls back" from $k[W] \to k[V]$. This kind of construction is *very* common in mathematics; for example, in differential geometry, where one uses it to relate tangent spaces of two manifolds that have a smooth mapping between them. This kind of construction is characterized by *pre-composition*, and shows up in various disguises in many places.

 

[Math Amateur]

What is it you don't understand? The definition of the mapping $\tilde{\varphi}$, or how it is well-defined?

The well-defined issue is easier to dispatch: since $\tilde{\varphi}$ is defined on an equivalence class (coset of functions), one has to verify that:

$\tilde{\varphi}(F) = \tilde{\varphi}(G)$ if $F - G \in \mathcal{I}(W)$.

Now $\tilde{\varphi}(F) = F\circ \varphi + \mathcal{I}(V)$ and $\tilde{\varphi}(G) = G\circ \varphi + \mathcal{I}(V)$.

To assert that $\tilde{\varphi}(F) = \tilde{\varphi}(G)$ is to say that:

$\tilde{\varphi}(F) - \tilde{\varphi}(G) = 0_{k[V]} = 0 + \mathcal{I}(V) = \mathcal{I}(V)$.

Thus we seek to show that:

$F - G \in \mathcal{I}(W) \implies [ F\circ \varphi + \mathcal{I}(V)] - [G\circ \varphi + \mathcal{I}(V)] = \mathcal{I}(V)$, which is true when:

$(F \circ \varphi) - (G \circ\varphi) \in \mathcal{I}(V)$.

However, $(F \circ \varphi) - (G \circ\varphi) = (F - G)\circ \varphi$, and since

$F - G$ annihilates $W$ (which is what it MEANS to be an element of $\mathcal{I}(W)$), and $\varphi(V) \subseteq W$, it is clear that $(F-G) \circ \varphi$ annihilates $V$, that is:

$(F - G)\circ \varphi \in \mathcal{I}(V)$.

Thus it doesn't matter which "$F$" we use to define $f = F + \mathcal{I}(W)$, as $\tilde{\varphi}$ is constant on the entire coset.

********************

The contravariant nature of $\tilde{\varphi}$ is often expressed as saying $\tilde{\varphi}(f)$ is a pullback of $f$ along $\varphi$, for just as $\varphi$ "goes forward" from $V \to W$, $\tilde{\varphi}$ "pulls back" from $k[W] \to k[V]$. This kind of construction is *very* common in mathematics; for example, in differential geometry, where one uses it to relate tangent spaces of two manifolds that have a smooth mapping between them. This kind of construction is characterized by *pre-composition*, and shows up in various disguises in many places.

Hi Deveno,

Apologies for the short absence ... back on the Internet now ...

Just reaquainting with with the theorem ... and working through your post ...

Thanks for your help ... it is much appreciated ...

Peter