# Integral element in the quotient of a polynomial ring

Hello,

I'm writting a "report" (to learn) on algebraic geometry, and was looking to write a proof for the following statement :

Theorem : Let $K$ be an algebraically closed field and $C_1, C_2 \subseteq \mathbb{A} ^2(K)$ two curves given by non-constant polynomials $f, g \in K[x,y]$. Let $\phi : C_1 \to C_2$ be a morphism (polynomial map). Then $\phi$ is either surjective or constant.

The map $\phi$ induces an embedding of the quotient rings,

$\phi^{\ast} : K[C_2] = K[x,y]/(g) \hookrightarrow K[C_1]= K[x,y]/(f)$ given by $h \mapsto \phi \circ h$

and so $K[C_2]$ can be identified as a subring of $K[C_1]$. For a point $P \in C_2$, the proof can be reduced in the statement $\phi^{-1}(P) = \varnothing$ if and only if the $K$-algebra $K[C_1]$ is a module of finite type over $K[C_2]$. (I didn't already defined finite maps, so I can't say $\phi$ is finite so $K[C_1]$ is integral over $K[C_2]$).

If the statement can be reduce to the analogous one but with the fraction fields of the coordinate rings, i.e., the rational polynomials in $K(C_1)$ and $K(C_2)$, it's easier because both fields have transcendence degree one and are finitely generated (over $K$) and the result follows. I don't see how to drag this property with rings.

Any other proof is welcome too !

You call $K[x,y]/(f)$ a "quotient field". But this quotient ring is never a field. – Robin Chapman Jul 16 '10 at 17:43
Did you mean to require $f$ and $g$ to be nonzero? Otherwise, if I'm correctly interpreting your notation, we get a counterexample by taking $u \mapsto x^2 y$, $v \mapsto x y^2$, with $f=g=0$. – Charles Staats Jul 17 '10 at 1:03
If I understood correctly what you are asking, I think that what you want to show is false: if $C_1$ is the hyperbola $xy=1$ and $C_2$ is the $x$-axis $y=0$, then the projection of $C_1$ to $C_2$ is non-constant and misses the origin. – damiano Jul 17 '10 at 17:19