Let \(P_1, \ldots, P_r, Q_1, \ldots, Q_s\) be distinct points of \(\AA^1\). If \[ \AA^1 \sm \ts{P_1, \ldots, P_r} \cong \AA^1 \sm \ts{Q_1, \ldots, Q_s}, \] show that \(r = s\). Is the converse true?
Solution
Write \(P_i=p_i\in k\) and \(Q_j=q_j\in k\) in the affine coordinate \(x\).
1. The coordinate ring of\[ U=\AA^1\setminus\{P_1,\ldots,P_r\} \]is\[ A_U=k[x,(x-p_1)^{-1},\ldots,(x-p_r)^{-1}]. \]
Put \[ g(x)=\prod_{i=1}^r(x-p_i). \] Then \(U=D(g)\subseteq\Spec k[x]\), so \[ \Gamma(U,\OO_U)=k[x,g^{-1}]. \] Because the linear factors \(x-p_i\) are pairwise coprime, inverting their product is equivalent to inverting each factor separately. This gives the displayed description.
2. The unit group of \(A_U\) is\[ A_U^\times =k^\times\,(x-p_1)^{\ZZ}\cdots(x-p_r)^{\ZZ}, \]and therefore\[ A_U^\times/k^\times\cong\ZZ^r. \]
The polynomial ring \(k[x]\) is a unique factorization domain. An element of its localization at the multiplicative system generated by the \(x-p_i\) is a unit exactly when every irreducible factor of its numerator and denominator belongs to that multiplicative system. Thus every unit has a unique expression \[ c\prod_{i=1}^r(x-p_i)^{n_i}, \qquad c\in k^\times, n_i\in\ZZ. \] Modulo the constant units, the exponent vector \((n_1,\ldots,n_r)\) identifies the quotient with the free abelian group \(\ZZ^r\).
3. If\[ \AA^1\setminus\{P_1,\ldots,P_r\} \cong \AA^1\setminus\{Q_1,\ldots,Q_s\}, \]then \(r=s\).
An isomorphism of the two \(k\)-varieties induces an isomorphism of their coordinate rings and hence of their unit groups. It fixes the subgroup \(k^\times\) of constant units. Step 2 therefore gives an isomorphism \[ \ZZ^r\cong A_U^\times/k^\times \cong A_V^\times/k^\times\cong\ZZ^s. \] Free abelian groups of finite rank are isomorphic only when their ranks agree. Hence \[ \boxed{r=s}. \]
4. More precisely, the two punctured lines are isomorphic if and only if the two \((r+1)\)-point subsets\[ \{\infty,P_1,\ldots,P_r\}, \qquad \{\infty,Q_1,\ldots,Q_r\} \]of \(\PP^1\) are carried to one another by a fractional linear transformation.
Let \(\phi:U\to V\) be an isomorphism. Both \(U\) and \(V\) are nonsingular curves and \(\PP^1\) is projective. Apply Proposition I.6.8 successively at the finitely many omitted points to extend the composite \[ U\xrightarrow{\phi}V\hookrightarrow\PP^1 \] to a morphism \(\bar\phi:\PP^1\to\PP^1\). Doing the same for \(\phi^{-1}\) gives a morphism \(\bar\psi:\PP^1\to\PP^1\). The composites \(\bar\psi\bar\phi\) and \(\bar\phi\bar\psi\) agree with the identity on dense open subsets, hence everywhere. Thus \(\bar\phi\) is an automorphism of \(\PP^1\). By Exercise I.6.6, it is fractional linear.
Since \(\bar\phi\) restricts to a bijection \(U\to V\), it carries the complement of \(U\) in \(\PP^1\) bijectively to the complement of \(V\). This proves necessity. Conversely, any fractional linear transformation carrying one displayed finite subset to the other restricts to an isomorphism of their complements.
5. The converse to \(r=s\) is false in general.
Take \(r=s=3\). For \(\lambda\in k\setminus\{0,1\}\) consider the four-point subset \[ S_\lambda=\{\infty,0,1,\lambda\}\subseteq\PP^1. \] The cross ratio of four ordered points is invariant under fractional linear transformations. Changing the ordering replaces \(\lambda\) by one of the finitely many values \[ \lambda,\quad1-\lambda,\quad\lambda^{-1},\quad (1-\lambda)^{-1},\quad \frac{\lambda}{\lambda-1},\quad \frac{\lambda-1}{\lambda}. \] Because the algebraically closed field \(k\) is infinite, choose \(\mu\in k\setminus\{0,1\}\) outside this finite set. Then no fractional linear transformation carries \(S_\lambda\) to \(S_\mu\). By step 4, \[ \AA^1\setminus\{0,1,\lambda\} \not\cong \AA^1\setminus\{0,1,\mu\}, \] although both have three punctures. Thus equality of the numbers of punctures is necessary but not sufficient.
6. Q.E.D.
Step 3 proves the required invariant, while steps <1>4–<1>5 answer the converse question.