If \(X\) is a scheme of finite type over a field, show that the closed points of \(X\) are dense. Give an example to show that this is not true for arbitrary schemes.
Solution
1. Let \(X\) be of finite type over a field \(k\). Every affine open\[ U=\Spec A\subseteq X \]has \(A\) a finitely generated \(k\)-algebra.
Apply Hartshorne II.3.3(c) to the finite type structure morphism \[ X\longrightarrow\Spec k. \] Since the target is affine, every affine open in \(X\) has coordinate ring finitely generated over \(k\).
2. If \(x\in X\) has residue field \(\kappa(x)\) finite over \(k\), then \(x\) is a closed point of \(X\).
Suppose \(y\) is a specialization of \(x\). Choose an affine neighborhood \[ V=\Spec B \] of \(y\). Open subsets are stable under generization, so \(x\in V\) as well.
Let \(\mathfrak p\subseteq B\) be the prime corresponding to \(x\). By 1, \(B\) is a finitely generated \(k\)-algebra, hence so is \(B/\mathfrak p\). Moreover \[ B/\mathfrak p\subseteq\kappa(x), \] and \(\kappa(x)\) is finite-dimensional over \(k\). Therefore \(B/\mathfrak p\) is finite-dimensional over \(k\). Being a domain, it is a field. Thus \(\mathfrak p\) is maximal in \(B\).
Hence \(x\) has no proper specialization inside \(V\). Since \(y\in V\) is a specialization of \(x\), we obtain \(y=x\). Therefore \(x\) has no proper specialization in \(X\) and is closed.
3. Every nonempty open subset \(W\subseteq X\) contains a closed point of \(X\).
Choose any point of \(W\) and an affine neighborhood \[ U=\Spec A\subseteq W. \] The nonzero ring \(A\) has a maximal ideal \(\mathfrak m\), giving a point \(x\in U\).
By 1, \(A\) is a finitely generated \(k\)-algebra. Zariski’s lemma therefore implies that \[ \kappa(x)=A/\mathfrak m \] is a finite algebraic extension of \(k\). Step 2 now shows that \(x\) is closed in all of \(X\).
4. The closed points of \(X\) are dense.
A subset is dense exactly when every nonempty open subset meets it. This is precisely 3.
5. The conclusion fails for arbitrary schemes.
Let \[ R=k[t]_{(t)} \] be the local ring of the affine line at the origin, and put \[ X=\Spec R. \] The only prime ideals of the discrete valuation ring \(R\) are \[ (0) \qquad\text{and}\qquad (t). \] The only closed point is \((t)\). Its closure is itself: \[ \overline{\{(t)\}}=\{(t)\}\ne X. \] Thus the set of closed points is not dense.
6. Q.E.D.
Step 4 proves density for finite-type schemes over a field, and 5 gives the requested counterexample.