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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 $U_{i}$.

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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.

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