Recall that a curve is rational if it is birationally equivalent to \(\PP^1\). Let \(Y\) be a nonsingular rational curve which is not isomorphic to \(\PP^1\).
Show that \(Y\) is isomorphic to an open subset of \(\AA^1\).
Show that \(Y\) is affine.
Show that \(A(Y)\) is a unique factorization domain.
Solution
Let \(K=K(Y)\). Since \(Y\) is rational, \[ K\cong k(t)=K(\PP^1). \]
1. The curve \(Y\) is isomorphic to an open subset of the unique nonsingular projective curve with function field \(K\).
For a nonsingular curve, each local ring at a closed point is a discrete valuation ring of its function field. The construction in Hartshorne I.6 therefore sends a point \(P\in Y\) to the corresponding valuation ring \(\OO_{Y,P}\) in the abstract nonsingular curve \(C_K\). The argument preceding Theorem I.6.9 identifies every nonsingular affine curve with the open subset of \(C_K\) consisting of its local valuation rings. Covering \(Y\) by affine opens and using uniqueness of the valuation center shows that these identifications agree on overlaps. Hence they glue to an open immersion \[ Y\hookrightarrow C_K. \]
Theorem I.6.9 identifies \(C_K\) with a nonsingular projective curve, uniquely determined by \(K\). Since \(K\cong K(\PP^1)\) and \(\PP^1\) is nonsingular and projective, this projective model is \(\PP^1\). Thus \(Y\) is isomorphic to an open subset of \(\PP^1\).
2. After an automorphism of \(\PP^1\), the image of \(Y\) is contained in \(\AA^1\).
If the open immersion of step 1 were surjective, then \(Y\cong\PP^1\), contrary to the hypothesis. Choose a point \(Q\in\PP^1\setminus Y\). An automorphism of \(\PP^1\) carries \(Q\) to the point at infinity \(\infty=[1:0]\). After composing the open immersion with this automorphism, \[ Y\subseteq\PP^1\setminus\{\infty\}=\AA^1 \] as an open subset. This proves part (a).
3. Every nonempty open subset of \(\AA_k^1\) is affine.
Because \(k\) is algebraically closed, every proper closed subset of \(\AA^1\) is finite. Thus the complement of the open subset representing \(Y\) has the form \[ \{a_1,\ldots,a_m\} \] for distinct \(a_i\in k\), possibly with \(m=0\). Put \[ q(t)=\prod_{i=1}^m(t-a_i), \] with \(q=1\) when \(m=0\). Then \[ Y\cong D(q)\subseteq\Spec k[t] \cong\Spec k[t,q^{-1}]. \] Hence \(Y\) is affine, proving part (b).
4. The affine coordinate ring \(A(Y)\) is a unique factorization domain.
By step 3, \[ A(Y)\cong k[t,q^{-1}]. \] The polynomial ring \(k[t]\) is a principal ideal domain and hence a unique factorization domain. A localization of a unique factorization domain is again a unique factorization domain: factor an element before localization and discard precisely those irreducible factors that become units. Therefore \(k[t,q^{-1}]\), and hence \(A(Y)\), is a unique factorization domain. This proves part (c).
5. Q.E.D.
Steps <1>1–<1>2 prove part (a), step 3 proves part (b), and step 4 proves part (c).