# Morphism of varieties $f:X\rightarrow Y$ (Affine)

Im trying to do an exercise from the book Algebraic Curves of Fulton (Exercise $\:6.26^{*}$).

It says:

Let $f:X\rightarrow Y$ be a morphism of affine varieties. Show that $f(X)$ in dense in $Y$ if and only if the homomorphism $\tilde{f}:\Gamma(Y)\rightarrow\Gamma(X)$ is one-to-one.

• This questionand some of its answers are closely related. – tracing Nov 20 '14 at 12:12
• Does either direction seem approachable to you? It's nice to begin somewhere. – Hoot Nov 20 '14 at 23:46
• Yes, I read all chapter 6 of Fulton's book and then it became more natural. You can read it here: math.lsa.umich.edu/~wfulton/CurveBook.pdf For the proof I used proposition 2 and collorary of proposition 7 of chapter 6. – Yesid Nov 21 '14 at 4:00

This is not true. The statement is that $f:\text{Spec}(B)\to\text{Spec}(A)$ is dominant (has dense image), if and only if $\ker(A\to B)$ is contained in $\text{nil}(A)$, the nilradical of $A$. Indeed, consider the map $\text{Spec}(k)\to\text{Spec}(k[x]/(x^2))$ coming from the quotient $k[x]/(x^2)\to k$.

Here's an outline:

1. Show that the closure of the image of $\text{Spec}(B)\to\text{Spec}(A)$ is $V(\ker(A\to B))$. This should just be chasing definitions.
2. Show that $V(I)=A$ if and only if $I\subseteq \text{nil}(A)$. This should also be apparent by definitions.

Let me know if you have trouble!

• My impression is that Fulton's book takes place in the setting of Hartshorne I, so everything is reduced and no schemes. It doesn't change the argument in any serious way, but it might confuse. – Hoot Nov 20 '14 at 7:28
• That is true. Fulton Uses the same notation for a different thing in chapter 6. – Yesid Nov 20 '14 at 12:25
• @Hoot I don't know how one is supposed to know that from the problem statement. I have never read Fulton. :| – Alex Youcis Nov 21 '14 at 10:58
• @Alex Of course! Definitely not claiming that you're to know the setup of every book. – Hoot Nov 21 '14 at 16:43
• I'll have this in mind for further questions. Now I see why is important to many other users of math.stack to write the definitions. – Yesid Nov 22 '14 at 18:19

a) Suppose that $f(X)$ is dense in $Y$ and that $\tilde{f}(\phi)=0$. Then $\phi \circ f=0\Rightarrow \phi=0\ {\rm on} f(X) \Rightarrow \phi=0$ where the last step follows since morphisms are continuous wrt. the Zariski topology.

b) Suppose $X\subset A^m, Y\subset A^n$ and $f(X)$ not dense in $Y$. By definition there exists an open set $O\subset A^n$ with $O\cap Y\not=\emptyset$ and $O\cap f(X)=O\cap Y\cap f(X)=\emptyset$. Since $\emptyset \not=O \not= A^n$ we can write $O=\{\phi_1\not=0\}\cup \cdots \cup \{\phi_s\not=0\}$ for a finite collection $\phi_i$ of polynmials. Hence a polynomail $\phi$ exists with $\{\phi\not=0\}\cap Y\not=\emptyset$ and $\{\phi\not=0\}\cap f(X)=\emptyset$. Then $\phi\in \Gamma(Y), \phi\not=0$ and $\tilde{f}(\phi)=0$. Hence $\tilde{f}$ is not injective.