# Algebriac Geometry - Morphisms of Algebraic Sets

1. Nov 3, 2013

### Math Amateur

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 $\phi \ : \ V \longrightarrow W$ is a morphism of algebraic sets and $\widetilde{\phi} \ : \ k[W] \longrightarrow k[V]$ is the associated k-algebra homomorphism of coordinate rings. Then

(1) the kernel of $\widetilde{\phi}$ is $\mathcal{I} ( \phi (V) )$

(2) etc etc ... ... ...

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[Note: For the definitions of $\phi$ and $\widetilde{\phi}$ see attachment page 662 ]

The beginning of the proof of Proposition 16 reads as follows:

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Proof. Since $\widetilde{\phi} = f \circ \phi$ we have $\widetilde{\phi}(f) = 0$ if and only if $(f \circ \phi) (P) = 0$ for all $P \in V$ i.e. $f(Q) = 0$ for all $Q = \phi (P) \in \phi (V)$. which is the statement that $f \in \mathcal{I} ( \phi ( V) )$ 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 $\widetilde{\phi} = f \circ \phi$ we have $\widetilde{\phi}(f) = 0$ if and only if $(f \circ \phi) (P) = 0$ for all $P \in V$"

My interpretation of this statement is given below after I give the reader some key definitions.

Definitions

Definition of Morphism or Polynomial Mapping $\phi$

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

Definition of $\widetilde{\phi}$

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

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

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

i.e $k[x_1, x_2, ... ... x_n]/\mathcal{I}(W) \longrightarrow k[x_1, x_2, ... ... x_n]/\mathcal{I}(V)$

$f \rightarrow f \circ \phi$ i.e. $\phi (F) = f \circ \phi$

Now, to repeat again for clarity, my problem is with the first line of the proof of Proposition 16 which reads:

"Since $\widetilde{\phi} = f \circ \phi$ we have $\widetilde{\phi}(f) = 0$ if and only if $f \circ \phi (P) = 0$ for all $P \in V$"

My interpretation of this line is as follows:

$\widetilde{\phi}(f) = 0 \Longrightarrow f \circ \phi (P) = 0$

But $f \circ \phi (P) = 0$ means that

$f \circ \phi (P) = 0 + \mathcal{I}(V)$

so then $f \circ \phi \in \mathcal{I}(V)$

Thus $(f \circ \phi) (P) = 0$ for all points $P = (a_1, a_2, ... ... , a_n \in V \subseteq \mathbb{A}^n$

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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2. Nov 3, 2013

### R136a1

I don't really get your interpretation, so I'll give my own.

First, you need to be careful about the quotient maps. So I'm gonna write the map $\tilde{\phi}$ as

$$\tilde{\phi}( [f] ) = [ f\circ \phi ]$$

So, assume that $\tilde{\phi}([f])= 0$, then $[f\circ \phi] = 0$. This implies that $f\circ \phi\in \mathcal{I}(V)$. Thus for all $P\in V$, we have that $f(\phi(P)) = 0$.

The way to interpret the coordinate rings $k[X_1,...,X_n]/\mathcal{I}(V)$ is that they are exactly the polynomial maps $V\rightarrow k$. We identify two polynomial maps if they are equal for all points on $V$. This makes this proof a bit more natural, and removes the need to rely too much on equivalence classes. Indeed, let $f:W\rightarrow k$ be a polynomial map, then $\tilde{\phi}(f) = f\circ \phi$ is a polynomial map $V\rightarrow k$. If $f\circ \phi=0$ as such a polynomial map, then that means that $f(\phi(P)) = 0$ for each $P\in V$.

Personally, I think Dummit and Foote presentation of algebraic geometry is pretty awful. If you're really interested in the subject, you should really get another book. I highly recommend "Algebraic Geometry: A First Course" by Harris.

3. Nov 3, 2013

### Math Amateur

Thanks for the helpful post, R136a1.

I followed the first part of your post - most clear and helpful

However, I did not follow the sentence:

"The way to interpret the coordinate rings k[X1,...,Xn]/I(V) is that they are exactly the polynomial maps V→k. "

Can you clarify? How is a coordinate ring a polynomial map (or am I being too literal?)

By the way, I have the book by Harris - just thought it might be too advanced for me and was going to tackle it after Dummit and Foote - but maybe I should switch now?

A book that I have found helpful and not too advanced is the following:

"Ideals, Varieties and Algorithms; An Introduction to Algebraic Geometry ans Commutative Algebra"

By Davidf Cox, John Little and Donald O'Shea

Do you know the book? Do you have an opinion of it?

Peter

4. Nov 3, 2013

### R136a1

The two are not literally the same of course, but they're isomorphic.
I mean the following, if $V\subseteq k^n$ is an algebraic set, then a polynomial function $f:V\rightarrow k$ is just given by a polynomial $k[X_1,...,X_n]$. The difference however is that two polynomial functions are regarded as equal if they agree on $V$.

For example, take $V = \{(0,0)\}$ in $\mathbb{C}^2$. Then $f(X,Y) = X+Y+1$ and $g(X,Y) = X^2 + YZ + 1$ are equal as polynomial functions on $V$. Indeed, for any $P\in V$, we have $f(P) = g(P)$ because only $P = (0,0)$ is in $V$.

More formally, any polynomial function in $k[X_1,...,X_n]$ restricts to a polynomial function on $V$. But we have introduced a new equality, namely that $f \equiv g$ if $f(P) = g(P)$ for all $P\in V$. This is a different equality from the usual equality on $k[X_1,...,X_n]$.

Now, it is true claim that $f\equiv g$ if and only if $f-g\in \mathcal{I}(V)$. Check it yourself!

So the quotient $k[X_1,...,X_n]/\equiv$ is the same as $k[X_1,...,X_n]/\mathcal{I}(V)$. Of course, $\equiv$ is very geometrical. But the latter equivalence using ideals is very algebraic and allows us to use the tools of algebra.

The book by Cox is nice. If you like it, then you should try it. Other good books are