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?

  • $\begingroup$ What is your definition of dimension ? If this is Krull dimension, then this is obvious... $\endgroup$ – Roland Jan 3 '18 at 20:13
  • $\begingroup$ Yes, I mean Krull ...Firstly maybe I can say wlog that $X$ affine, because it is covered by affine sets and it's clear if $k$ is algebraically closed... but in general case? $\endgroup$ – KarlPeter Jan 3 '18 at 20:21
  • $\begingroup$ Maybe I can use the fact that every open set has a closed point... $\endgroup$ – KarlPeter Jan 3 '18 at 20:24
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    $\begingroup$ It is even easier than that, if $\eta$ is not closed, then its closure must be $X$ otherwise we would have a chain $\{x\}\subset\overline{\{u\}}\subset X$. So $\eta$ is the generic point and all other points are closed. I used the fact that any closed subset contains a closed point. $\endgroup$ – Roland Jan 3 '18 at 20:42
  • $\begingroup$ Worth mentioning that for existence of closed points you need to use the finite type hypothesis: there are non Noetherian examples where you don't get closed points in a closed subset. $\endgroup$ – Ashwin Iyengar Jan 3 '18 at 21:02

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 $X$.


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