3

I am guessing that the closed sets of a curve (i.e. an algebraic variety of dimension 1) are finite . How would one go about proving this? Is it trivial?!

Thanks for any help.

M Davolo
  • 761
  • 4
  • 13

2 Answers2

7

If $X$ is an (irreducible, which may or may not be part of your definition) variety over a field $k$ of dimension $1$ and $F\subseteq X$ is a proper closed subset, then the dimension of $F$ must be strictly less than that of $X$, i.e., $F$ is zero-dimensional. Therefore the irreducible components of $F$ must be points, and since $F$ has finitely many irreducible components (being a Noetherian topological space), and $F$ is equal to the union of its irreducible components, $F$ must be a finite set.

If $X$ is not irreducible, this is not true. For example, $\mathrm{Spec}(k[x,y]/(xy))$, the union of the coordinate axes in $\mathbf{A}_k^2$, is one-dimensional but has two irreducible components, both of dimension one, which are not finite.

Keenan Kidwell
  • 26,770
  • How does one directly infer "$F$ is zero-dimensional" implies "the irreducible components of $F$ must be points"? Don't we need to prove before that the points of $X$ different from the generic point are closed? – Elías Guisado Villalgordo Oct 10 '23 at 10:10
  • The general result is: Let $X$ be a locally Noetherian irred scheme of dimension $\leq 1$ and $\eta\in X$ be its generic point. Then (i) $x\in X-\eta$ is closed, (ii) a proper closed subset $F\subset X$ is locally a finite union of points of $X-\eta$ (whence finite if $X$ is Noetherian). It's already been argued (i) implies (ii). To see $x\in X-\eta$ is closed, we can assume that $X$ is quasi-compact (replace $X$ by an open aff nhbd of $x$), and then use that $\overline{{x}}$ contains a closed point $y$ [ref]; thus $x=y$ as $\dim X\leq1$. – Elías Guisado Villalgordo Oct 29 '23 at 16:34
5

Since any curve is a finite union of affine open subsets, you may assume your curve $C=Spec (A)$ is affine.

a) Assume $A$ is a one dimensional noetherian domain and $\mathfrak p\subsetneq A$ is a prime ideal.
Then $\mathfrak p$ is maximal (because $0\subsetneq \mathfrak p$ is a saturated chain of prime ideals since $dim(A)=1$) and $V(\mathfrak p)$ is a closed point.
Since $A$ is noetherian, a closed subset $V(I)\subset C$ only has finitely many irreducible component and thus is finite.

b) If $A$ is not reduced, the result is false: a counterexample being the curve $C=Spec(\mathbb C[X,Y] /\langle Y^2\rangle)$ which has $C_\text{red}=V(Y)=Spec(\mathbb C[X,Y] /\langle Y\rangle)\subsetneq C$ as a strict closed subscheme of dimension 1.

c) Even if $A$ is reduced but not a domain the result is not true: consider the one dimensional closed subschemes $V(X)\subsetneq V(XY)\subsetneq Spec(\mathbb C[X,Y])$

Georges Elencwajg
  • 156,622