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Let $X$ be a scheme integral scheme of dimension 1. If $X$ is affine, then it is clear that every non-generic point is closed. I wonder if this is true in general. If not, is it true if we suppose that $X$ is a curve? (A variety of dimension 1.)

Gabriel
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    What first comes to mind is this: if $p \in X$ is not closed, then its closure is a closed set for which $p$ is then basically by definition the generic point. But 1) that almost seems too easy, and 2) it doesn't use the dimension $1$ assumption. Do you have reason to believe this doesn't hold in higher dimensions? – Tabes Bridges Mar 27 '21 at 17:46
  • It's not clear what sense you're using "general" in here. If you mean for arbitrary integral schemes of dimension one, then this is trivial from the definition of dimension as the maximum length of a chain of specializations. – KReiser Mar 27 '21 at 18:32
  • @TabesBridges it is surely not true that this result is true for arbitrary schemes. Not even for affine ones: the rings for which every non-zero prime ideal is maximal are precisely those of dimension one. – Gabriel Mar 27 '21 at 21:23
  • @KReiser I do indeed mean arbitrary integral schemes of dimension one. Since I am not really used to arguing about specializations, would you mind explaining this a little further? – Gabriel Mar 27 '21 at 21:24
  • I'd say @TabesBridges meant that $p$ is the generic point of the closure of ${p}$, Gabriel is asking if for an integral scheme of dimension $1$ the closure is either ${p}$ or $X$, and the answer is that we get the chain of closed irreducible sets $X\supset \overline{{ p}}\supset {a}$ (for $a$ a closed point $\in \overline{{ p}}$) and dimension $1$ gives either $\overline{{ p}}=X$ or $\overline{{ p}}={a}$. – reuns Mar 28 '21 at 01:03

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If $x\in\overline{\{y\}}$, we say $x$ is a specialization of $y$ or $y$ is a generalization of $x$ and denote this $y\leadsto x$. Observe that in a scheme, a chain of specializations is in one-to-one correspondence with a chain of inclusions of irreducible closed subsets: in one direction, we take the closure of the points, while in the other, we take the generic points of the irreducible closed subsets. Therefore the dimension of a scheme is exactly the number of specializations in a maximal chain of nontrivial specializations.

Now suppose $X$ is irreducible of dimension one with generic point $\eta_X$. If $\overline{\{x\}}=\{x\}$, $x$ is closed. If $\overline{\{x\}}\neq\{x\}$, then let $y$ be some element distinct from $x$ in $\overline{\{x\}}$, and we get a chain of specializations $\eta_X\leadsto x\leadsto y$. As $x\neq y$, we must have $x=\eta_X$ otherwise we contradict the fact that $X$ has dimension one.

KReiser
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