# Morphism of affine curves and their coordinate rings

Let $\phi\colon\mathcal{C}_f\rightarrow\mathcal{C}_g$ be a morphism of non-singular irreducible affine algebraic curves over an algebraically closed field $k$, with $\mathcal{O}(\mathcal{C}_f),\mathcal{O}(\mathcal{C}_g),k(\mathcal{C}_f), k(\mathcal{C}_g)$ the corresponding coordinate rings and function fields.

i). I want to prove the well-known result that $\mathcal{O}(\mathcal{C}_f)$ is the integral closure of $\phi^*(\mathcal{O}(\mathcal{C}_g))$ in $k(\mathcal{C}_f)$.

This would follow quite simply if every valuation ring of $k(\mathcal{C}_f)$ that contained $\phi^*(\mathcal{O}(\mathcal{C}_g))$ was in fact a DVR of the form $\mathcal{O}_P$ for some point $P\in\mathcal{C}_f$, but I can't quite see this.

ii). I want to go on to prove that $\mathcal{O}(\mathcal{C}_f)$ is a finite $\phi^*(\mathcal{O}(\mathcal{C}_g))$-module of dimension $deg (\phi)$.

A step along the way is to show that if $x\in\mathcal{O}(\mathcal{C}_f)$ then its minimal polynomial over $\phi^*(k(\mathcal{C}_g))$ has coefficients in $\phi^*(\mathcal{O}(\mathcal{C}_g))$. I can show that they lie in $\phi^*(k(\mathcal{C}_g))\cap\mathcal{O}(\mathcal{C}_f)$ and the result would follow if, for example, I could show that if $\phi^*(h)$ for $h\in \mathcal{O}(\mathcal{C}_g)$ was a unit in $\mathcal{O}(\mathcal{C}_f)$ then $h$ would be a unit in $\mathcal{O}(\mathcal{C}_g)$, but I am not sure if this is true.

• Do you require the curves to be smooth? Isn't i) false otherwise? Consider the case $C_f: y^2=x^3+x^2$, $C_g$ the $x$-axis, $\phi(x,y)=x$. Then $y/x$ will be integral over $\phi^*(\mathcal{O}(C_g))$ but not an element of $\mathcal{O}(C_f)$. The problem being that $\mathcal{O}(C_f)$ is not itself integrally closed. Or may be I misunderstood? – Jyrki Lahtonen Aug 3 '16 at 9:07
• This is false without some properness hypothesis. For example, $\phi$ could be an open immersion and then both 1 and 2 will fail. – Mohan Aug 3 '16 at 12:51
• Yes, I am assuming that $\mathcal{C}_f$ is non-singular and thus $\mathcal{O}(\mathcal{C}_f)$ is integrally closed. – Noel Robinson Aug 3 '16 at 15:12