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On the affine line, the classical Zariski topology is cofinite

Statement

Let $k$ be an algebraically closed field. A subset of $A_{k}^{1}$ is Zariski-closed if and only if it is either all of $A_{k}^{1}$ or a finite subset. Equivalently, the Zariski topology on $A_{k}^{1}$ is the cofinite topology.

Facts & Assumptions

Given: An algebraically closed field $k$ .

[L1]

Closed subsets of $A_{k}^{1}$ are exactly zero loci of subsets of $k [t]$ (Zero loci in affine space are the closed sets of the classical Zariski topology).

[L2]

A nonzero polynomial of degree $n$ over an integral domain has at most $n$ distinct roots (A nonzero polynomial of degree $n$ over an integral domain has at most $n$ distinct roots).

Proof

technique · direct

1.1

Let $C \subseteq A_{k}^{1}$ be Zariski-closed and proper. By [L1], $C = V (S)$ for some $S \subseteq k [t]$ . Because $C \neq A_{k}^{1}$ , some polynomial $f \in S$ is nonzero. Then $C \subseteq V (f)$ , and [L2] says $V (f)$ is finite. Hence every proper closed subset is finite.

L1L2givenchoose

1.2

Conversely, if $F = {a_{1}, \dots, a_{r}} \subseteq k$ is finite, then $F = V ⁣ ((t - a_{1}) \dots (t - a_{r})) .$ Also $\emptyset = V (1)$ and $A_{k}^{1} = V (\emptyset)$ by the definition of zero locus. So every finite subset and the whole line are Zariski-closed.

L1algebra

2.1

Steps 1.1 and 1.2 are exactly the statement that the closed sets are the finite subsets together with the whole space, that is, the Zariski topology on $A_{k}^{1}$ is cofinite.

step 1.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

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Sources

Donu Arapura, Notes on Basic Algebraic Geometry, cofinite-topology discussion in §1.3 (standard reference, not scraped)
J. S. Milne, Algebraic Geometry, Chapter 2c (standard reference, not scraped)