Let \(k\) be an algebraically closed field and let \[ \operatorname{Specm}k[x] \] be the maximal spectrum with its Zariski topology. Show that this topology is the cofinite topology.
Solution
1. The points of \(\operatorname{Specm}k[x]\) are exactly\[ \mathfrak m_a=(x-a), \qquad a\in k. \]
By the weak Nullstellensatz, every maximal ideal of the finitely generated \(k\)-algebra \(k[x]\) is the kernel of evaluation at a point \(a\in k\). Hence it has the form \[ (x-a). \] Conversely, \[ k[x]/(x-a)\cong k \] is a field, so every \((x-a)\) is maximal.
2. Every proper Zariski-closed subset of\(\operatorname{Specm}k[x]\) is finite.
Let \[ Z=V_{\max}(I) \] be a proper closed subset. Since \(k[x]\) is a principal ideal domain, write \[ I=(f). \] Properness of \(Z\) excludes \(I=(0)\), because \[ V_{\max}(0)=\operatorname{Specm}k[x]. \] Thus \[ f\ne0. \]
By step 1, \[ \mathfrak m_a\in V_{\max}(f) \quad\Longleftrightarrow\quad f\in(x-a) \quad\Longleftrightarrow\quad f(a)=0. \] A nonzero polynomial has only finitely many roots. Therefore \(Z\) is finite.
3. Every finite subset of \(\operatorname{Specm}k[x]\) is Zariski closed.
Let \[ F = \{\mathfrak m_{a_1},\ldots,\mathfrak m_{a_r}\}. \] Put \[ g(x) = \prod_{i=1}^r(x-a_i). \] For \(a\in k\), \[ g\in\mathfrak m_a \quad\Longleftrightarrow\quad g(a)=0 \quad\Longleftrightarrow\quad a\in\{a_1,\ldots,a_r\}. \] Hence \[ V_{\max}(g)=F. \] The empty set is also closed, being \(V_{\max}(1)\).
4. The Zariski closed subsets are exactly the whole space and the finitesubsets.
Step 2 shows every proper closed subset is finite. Step 3 shows every finite subset is closed. Together with the whole space, these are exactly the closed subsets.
5. Therefore the Zariski topology on \(\operatorname{Specm}k[x]\) is thecofinite topology.
The cofinite topology is defined by declaring the whole space and the finite subsets to be the closed sets. Step 4 gives exactly this family.
Equivalently, the correspondence between affine varieties and their coordinate rings identifies \[ \operatorname{Specm}k[x]\cong\AA^1_k, \] and this is the special case of the cofinite topology on an irreducible curve.
6. Q.E.D.
Step 5 is the required conclusion.