# Classification of points on an irreducible reduced scheme of finite type over $k$ of dimension $1$.

Suppose that $$X$$ is an irreducible reduced scheme of finite type over $$k$$ of dimension $$1$$. Meaning that the biggest chain of distinct irreducible closed subsets $$X_{0}\subset X_{1}\subset ... \subset X_{n}=X$$ is $$1$$.

I want to show that all the points (except for the unique generic point) of X are closed points. So let $$x\in X$$ be a non-generic point.

Note that it is enough to show that if we consider an affine open covering $$\{U_{i}\}_{i\in I}$$ of $$X$$ the set $$\{x\}$$ is closed in the sets $$U_{i}$$ for all $$i\in I$$ which contain $$x$$.

Since $$\dim(X)=1$$ we know that $$\dim(U_{i})\leqslant 1$$ for all $$i$$. But when $$\dim(U_{i})=0$$ we are actually able to show that it only contains the generic point, and thus from now on we can assume that $$\dim(U_{i})=1$$. In this case we find that $$U_{i}=\operatorname{Spec}(R_{i})$$ only contains the zero ideal and all the other prime ideals are maximal. And notice that maximal ideals are closed points.

From here I am stuck. I think that if I can show that for a point $$x\in X$$ which is not the generic point and for an open affine neigbhourhood $$U_{i}$$ of $$x$$, $$x$$ can only correspond to one of the maximal ideals. Since then I can conclude that it is closed in $$U_{i}$$.

First, note that if $$X$$ is a finite type $$k$$-scheme, a point $$x \in X$$ is closed iff its residual field is finite over $$k$$. So if is enough to show it when $$X$$ is affine, ie the spectrum of a finitely generated $$k$$-algebra $$A$$ which is an integral domain (any open subscheme of an integral scheme is integral).
By Noether normalization and as $$X$$ has dimension $$1$$, there is a finite injective morphism $$k[X] \rightarrow A$$. So if $$p$$ is a nonzero prime ideal of $$A$$, with inverse image $$q$$ in $$k[X]$$, then the quotient $$k[X]/q \rightarrow A/p$$ is finite injective.
If $$q \neq 0$$, then $$k[X]/q$$ is a finite dimensional integral $$k$$-algebra so is a finite field extension of $$k$$. So $$A/p$$ is a finite dimensional integral $$k$$-algebra so is a field so $$p$$ is maximal.
If $$q=0$$, then the morphism $$k[X] \rightarrow A/p$$ is finite injective. In particular, if $$f \in k[X]$$ has image in $$p$$, then $$f=0$$. Let now $$a \in p$$ be nonzero, we know (by Cayley-Hamilton) that there exists a monic polynomial of lowest degree $$\Pi$$ with coefficients in $$k[X]$$ that vanishes at $$a$$.
As $$A$$ is an integral domain, $$\Pi(0) \neq 0$$ (because $$\Pi(T)/T$$ would work). So $$\Pi(0) \in k[X] \cap aA \subset k[X] \cap p=q$$ and is nonzero, a contradiction.