$\tilde \phi:\Gamma (W)\rightarrow \Gamma(V)$ may not extends to all $k(W)$

Question

A problem from Fulton's Algebraic Curves:--

Let $\phi:V\rightarrow W$ be a polynomial map between two affine varieties and $\tilde \phi:\Gamma (W)\rightarrow \Gamma(V)$ be the induced map between co-ordinate rings.Let $P\in V$ and $\phi(P)=Q$. Show that $\tilde \phi$ extends to a unique ring homorphism from $\mathscr O_Q(W)$ to $\mathscr O_P(V)$. Note that $\tilde \phi$ may not extends to all $k(W)$.

My problem is the following : $\tilde \phi:\Gamma(W)\rightarrow \Gamma(V)$ is a ring homorphism between two integral domains and $k(W),k(V)$ are their fields of fractions, so that $\tilde \phi$ can be extended to a field homorphism from $k(W)$ to $k(V)$. Also, $\mathscr O_Q(W)\subseteq k(W)$ and $\mathscr O_P(V)\subseteq k(V)$. So the line written in bold means to say, $\tilde \phi:\Gamma(W)\rightarrow \Gamma(V)$ cannot be extended to a homeomophism from $k(W)$ to $\mathscr O_P(V)$. Am I right?

Answer 1

The line written in bold says that you can't in general extend the morphism $\tilde{\phi}:\Gamma(W)\rightarrow \Gamma(V)$ to a morphism $\psi:K(W)\rightarrow K(V)$. So in order to solve the problem you just have to given an example in which it can't be extended.

There are two reasons why the map shouldn't extend in general.

1. Algebraic reason:

   The universal property of localization tells you the following: Given a morphism of rings $f:A\rightarrow B$ and a multiplicative system $S\subseteq A$. The map $f$ extentends to a morphism $\tilde{f}:S^{-1}A\rightarrow B$ if and only if $f(s)$ is invertible in $B$ for each $s\in S$.

   When $B$ is a field this translate to $f(s)\neq 0$ for every $s\in S$ and when $S=A\setminus\{0\}$ this translate to $f$ being injective. And in general the map $\tilde{\phi}$ is not necessarily injective.

2. Geometric reason:

   The idea is that $K(W)$ is the set of polynomial functions defined over some dense open set in $W$ with values in $k$, and the map $K(W)\rightarrow K(V)$ will be given by precomposition with $\phi: V\rightarrow W$, that is: $$\begin{align} K(W)&\rightarrow K(V) \\ f&\mapsto f\circ \phi \end{align}$$ Hence if the image of $\phi$ is too small then maybe the composition $f\circ \phi$ is not possible because $\text{Im}(\phi)\cap \text{Dom}(f)$ could be empty and there is nothing to compose. For example if $\text{Im}(\phi)$ is a dense set this will never happen.

Finally it happens that $\phi$ has dense image iff $\tilde{\phi}$ is injective and in both cases you can extend the map. But in general neither of this two things happen.

Take for example $V=\mathbb{A}^1$ and $W=\{a\}\in \mathbb{A}^1$. Then the constant map $\phi:V\rightarrow W$ sending everything to $a$ doesn't extend because the map between rings is $$\begin{align} \tilde{\phi}:k[x]&\rightarrow k\\ p(x)&\mapsto p(a) \end{align}$$ and it is not injective.