Points of irreducible, 1 dimensional $k$-scheme $X$ closed Let consider an irreducible, 1 dimensional $k$-scheme $X$, such that the structure morphism $X \to Spec(k)$ is separated and of finite type.
Obviously has $X$ a unique generic point. My question is why and how to see that (all!) other points of $X$ are closed?
 A: Here's another possible way to see this, which is more complicated, but which you may find useful.
Pick a point $x \in X$, and take an affine open neighborhood $Spec(A)$ of $x$ contained in $X$: since $X$ is irreducible, $Spec(A)$ is irreducible as well, and it's also dense in $X$. Since the dimension of $X$ is $1$, the dimension of $Spec(A)$ (as a topological space) can be at most $1$, but this is equal to the Krull dimension of $A$.
Suppose the Krull dimension is $0$. Since $Spec(A)$ is irreducible, the nilradical is prime, so we can assume $Spec(A)$ is reduced, and thus integral. But then $A$ is a dimension $0$ integral domain, so it's a field, so $Spec(A)$ is a point, which must be the generic point since $Spec(A)$ is dense in $X$.
If the Krull dimension is $1$ then every nonzero prime is maximal, so if $x$ isn't the generic point, it's closed in $Spec(A)$. Then use the nice answer of Keenan Kidwell in this question: Closed points of a scheme correspond to maximal ideals in the affines? to conclude that since $X$ is finite type over a field, $\{x\}$ is closed in $X$.
