# Curves homeomorphic under Zariski topology

Prove any two curves over some field $$k$$ are homeomorphic, where $$k$$ might not be algebraically closed. Curves are defined to be varieties (integral separated scheme of finite type) of dimension $$1$$.

This question came up during discussions with my friends. The algebraically closed case is clear since each curve has the same cardinality as the field and Zariski topology is simply the cofinite topology. However, the non-algebraically closed case seems more difficult. I considered working in the algebraic closure but do not know if that is appropriate. Any help would be appreciated.

• What difference does it make exactly when $k$ is not algebraically closed? Commented Mar 18, 2023 at 12:08
• @reuns When k is algebraically closed, the closed sets are those with a finite number of points. I don't think that is the case for non-algebraically closed field.
– user1055643
Commented Mar 18, 2023 at 12:25
• Why that? For example $y^2=x^3+x$ over $\Bbb{Q}$ or $\Bbb{F}_p$ what are the closed sets? Commented Mar 18, 2023 at 12:52

2. Any positive-dimensional scheme of finite type over a field $$k$$ has $$|\overline{k}|$$ closed points.
Once you know these two facts, your two curves consist of $$|\overline{k}|$$ closed points and one generic point, and the nontrivial closed sets are finite collections of closed points, and thus the topological spaces are homeomorphic.
Proof of 1. Let $$X$$ be our irreducible noetherian scheme of dimension one. Then a nontrivial closed subset is a proper subset of $$X$$, and therefore must be of dimension zero. A dimension zero noetherian topological space is a finite collection of points. $$\blacksquare$$
Proof of 2. Consider the base change $$X_{\overline{k}}\to X$$. I claim this map is finite-to-one on closed points. We can compute the fiber over a closed point $$x\in X$$ by computing the base change of $$\operatorname{Spec} \kappa(x)\to X$$ by $$X_{\overline{k}}\to X$$, which reduces to computing the base change of $$\operatorname{Spec} \kappa(x) \to \operatorname{Spec} k$$ by $$\operatorname{Spec} \overline{k} \to \operatorname{Spec} k$$. But this is just the spectrum of $$\kappa(x)\otimes_k \overline{k}$$, and by Zariski's lemma $$\kappa(x)$$ is a finite extension of $$k$$. Therefore this tensor product is finite-dimensional as a $$\overline{k}$$-vector space, and its spectrum is finitely many points.
As $$\overline{k}$$ is infinite and base change to the algebraic closure preserves dimension, it suffices to show the claim when $$k=\overline{k}$$. Cover our scheme $$X$$ by finitely many affine opens $$X_i$$; then $$X_i\hookrightarrow \Bbb A^{n_i}_k$$ is a closed immersion and the cardinality of $$X_i$$'s closed points is at most the cardinality of $$\Bbb A^{n_i}_k$$'s closed points, which is $$|k|^{n_i}=|k|$$. So $$|\text{closed points of } X| \leq |k|$$. For the reverse direction, we may assume $$X$$ is reduced (since this doesn't change the topology), let $$U$$ be an irreducible affine open contained in a positive-dimensional irreducible component and let $$f\in k[U]$$ be a nonconstant element. Then $$f$$ determines a dominant morphism $$U\to\Bbb A^1_k$$, so it surjects on to a non-empty open subset of $$\Bbb A^1_k$$. But such an open set contains all but finitely many closed points of $$\Bbb A^1_k$$, and any closed point in $$U$$ must map to a closed point of $$\Bbb A^1_k$$, so $$U$$ must have at least $$|k|$$ closed points. Thus $$|\text{closed points of }X|=|k|$$. $$\blacksquare$$