Zariski Topology on the Affine Line as a Cofinite Example

Polynomial Equations Define a Topology

On the affine line over a field, polynomial equations give a way to specify sets of points: a point belongs to the set when it makes the required polynomial equations vanish. Over an infinite field, this construction leads to a particularly simple topology. Its closed sets are just the finite subsets and the entire line. We will derive this description and use it to check two separation properties directly.

Fix an infinite field \(K\). We use the set \(K\) as the affine line of \(K\)-valued points, and give it the Zariski topology defined below. A polynomial is an element of the polynomial ring \(K[t]\).

The empty family of equations has zero set \(K\), while the equation \(1=0\) has zero set \(\varnothing\). Thus both \(K\) and \(\varnothing\) are Zariski-closed. To identify all the other closed sets, we need a basic fact about roots of polynomials.

Proof. We use induction on \(d\). A nonzero constant polynomial has no roots, so the result holds for \(d=0\). Suppose \(d\geq 1\), and let \(f\) be a nonzero polynomial of degree \(d\). If \(f\) has no roots, the assertion holds. Otherwise choose a root \(a\). The polynomial division algorithm gives \(f(t)=(t-a)g(t)+r\), where \(r\) is constant. Substituting \(t=a\) gives \(0=f(a)=r\), so \(f(t)=(t-a)g(t)\), with \(g\) of degree \(d-1\). If \(b\ne a\) is another root of \(f\), then \(0=(b-a)g(b)\). Since \(K\) is a field and \(b-a\ne0\), it follows that \(g(b)=0\). By induction, \(g\) has at most \(d-1\) distinct roots. Including \(a\), \(f\) has at most \(d\) distinct roots. \(\square\)

Classifying the Closed Sets

Proof. Let \(C=Z(S)\) be Zariski-closed. If every polynomial in \(S\) is the zero polynomial, every point satisfies every equation, so \(C=K\). Otherwise \(S\) contains a nonzero polynomial \(f\). Every point of \(C\) is a root of \(f\), so \(C\subseteq Z(\{f\})\). The proposition says that \(Z(\{f\})\) is finite; hence \(C\) is finite.

Conversely, let \(A\subseteq K\) be finite. If \(A\) is empty, then \(A=Z(\{1\})\). If \(A=\{a_1,\ldots,a_n\}\) is nonempty, form \(p(t)=\prod_{i=1}^{n}(t-a_i)\). For any \(x\in K\), the product \(p(x)\) is zero exactly when at least one factor \(x-a_i\) is zero, which holds exactly when \(x\in A\). Thus \(A=Z(\{p\})\) is closed. Finally, \(K=Z(\{0\})\) is closed. These cases give precisely the stated family. \(\square\)

This classification also verifies that the specified closed sets do form a topology. A finite union of finite sets is finite, and a union involving \(K\) is \(K\). For an arbitrary intersection, if every set being intersected is \(K\), the result is \(K\); if one of them is finite, the intersection is a subset of that finite set and is finite. Thus the family contains \(\varnothing\) and \(K\), is closed under finite unions, and is closed under arbitrary intersections, as required for closed sets of a topology.

Open Sets and Separation

The complement of a finite set is called cofinite. The classification says that the open sets in this topology are exactly \(\varnothing\) and the cofinite subsets of \(K\). In particular, every nonempty open set is cofinite. An open neighborhood of \(x\) is an open set containing \(x\), so every such neighborhood excludes at most finitely many points of \(K\).

Proof. Recall that a space is \(T_1\) if, for any two distinct points \(x,y\), there is an open set containing \(x\) but not \(y\). Given \(x\ne y\), the set \(K\setminus\{y\}\) is open because its complement is finite. It contains \(x\) and excludes \(y\), so the space is \(T_1\).

A space is Hausdorff if every pair of distinct points has disjoint open neighborhoods. Let \(x,y\in K\) be distinct, and let \(U,V\) be any open neighborhoods of \(x,y\), respectively. Both are nonempty, so each has finite complement. The complement of their intersection is \(K\setminus(U\cap V)=(K\setminus U)\cup(K\setminus V)\), which is finite. Since \(K\) is infinite, \(U\cap V\) cannot be empty: if it were empty, its complement would be all of \(K\), which is not finite. Thus every such pair of neighborhoods intersects. No disjoint open neighborhoods exist for \(x\) and \(y\), so the space is not Hausdorff. \(\square\)

Why the Infinite-Field Hypothesis Matters

The proof that a nonzero polynomial has finitely many roots does not by itself guarantee that its zero set is a proper subset: that conclusion also uses the infinitude of \(K\). The hypothesis matters for the separation conclusion as well. A common mistake is to conclude that the Zariski topology on the affine line is never Hausdorff, without checking whether the field is infinite.

For comparison, let \(K=\mathbb{F}_3=\{0,1,2\}\), the field with three elements. Each singleton \(\{a\}\) is the zero set of \(t-a\). Every subset of \(\mathbb{F}_3\) is a finite union of singletons, so every subset is closed; its complement is therefore closed as well, making every subset open. This is the discrete topology, and distinct points have disjoint singleton neighborhoods. Hence the affine line over this finite field is Hausdorff. The cofinite classification and the failure of Hausdorffness proved above are specifically conclusions for an infinite field.