Think of \(\PP^1\) as \(\AA^1 \union \ts{\infty}\). Define a fractional linear transformation of \(\PP^1\) by sending \[ x \mapsto \frac{ax + b}{cx + d}, \qquad a,b,c,d \in k, \quad ad - bc \neq 0. \]
Show that a fractional linear transformation induces an automorphism of \(\PP^1\), that is, an isomorphism of \(\PP^1\) with itself. Denote the group of all these fractional linear transformations by \(\PGL(1)\).
Let \(\Aut \PP^1\) denote the group of all automorphisms of \(\PP^1\). Show that \(\Aut \PP^1 \cong \Aut k(x)\), the group of \(k\)-automorphisms of the field \(k(x)\).
Show that every automorphism of \(k(x)\) is a fractional linear transformation, and deduce that \(\PGL(1) \to \Aut \PP^1\) is an isomorphism.
Note: a similar result holds for \(\PP^n\) (II, 7.1.1): every automorphism is given by a linear transformation of the homogeneous coordinates.
Solution
Use homogeneous coordinates \([X:Z]\) on \(\PP^1\), with affine coordinate \(x=X/Z\) on \(Z\ne0\) and \(\infty=[1:0]\).
1. Every matrix\[ M=\begin{pmatrix}a&b\\ c&d\end{pmatrix}\in\operatorname{GL}_2(k) \]defines an automorphism\[ T_M([X:Z])=[aX+bZ:cX+dZ] \]of \(\PP^1\).
The two homogeneous linear forms \(aX+bZ\) and \(cX+dZ\) have no common nonzero zero because \(M\) is invertible. Hence they define a morphism \(T_M:\PP^1\to\PP^1\). The inverse matrix \(M^{-1}\) gives a morphism \(T_{M^{-1}}\) satisfying \[ T_{M^{-1}}T_M=T_MT_{M^{-1}}=\operatorname{id}_{\PP^1}. \] Thus \(T_M\) is an automorphism. On the affine chart it sends \[ x\longmapsto\frac{ax+b}{cx+d}, \] with the usual value \(\infty\) when the denominator vanishes.
Multiplying \(M\) by a nonzero scalar does not change \(T_M\), and two matrices inducing the same projective transformation differ by such a scalar. Thus these transformations form the projective linear group \(\operatorname{PGL}_2(k)\), denoted \(\PGL(1)\) in the source. This proves part (a).
2. Pullback of rational functions gives an isomorphism\[ \Aut(\PP^1)\xrightarrow{\cong}\Aut_k k(x). \]
An automorphism \(\phi:\PP^1\to\PP^1\) induces the \(k\)-automorphism \[ \phi^*:K(\PP^1)=k(x)\longrightarrow k(x), \qquad h\longmapsto h\circ\phi. \] Composition of maps reverses under pullback, so if one uses ordinary composition on both groups the assignment \(\phi\mapsto(\phi^{-1})^*\) is a group homomorphism.
Corollary I.6.12 identifies nonsingular projective curves with their function fields contravariantly: every \(k\)-homomorphism of function fields is induced by a unique dominant morphism of the corresponding nonsingular projective curves. Apply this with both fields equal to \(k(x)\). An automorphism of \(k(x)\) therefore induces a unique dominant morphism \(\psi:\PP^1\to\PP^1\), and its inverse field automorphism induces a morphism inverse to \(\psi\) by uniqueness in the category equivalence. Hence \(\psi\) is an automorphism.
Thus the pullback construction is bijective, and after the harmless inverse convention just noted it is an isomorphism of groups. This proves part (b).
3. Every automorphism of \(\PP^1\) fixing \(\infty\) has the affine form \(x\mapsto ax+b\) with \(a\ne0\).
Let \(\phi\in\Aut(\PP^1)\) satisfy \(\phi(\infty)=\infty\). Then \[ \phi(\AA^1)=\phi(\PP^1\setminus\{\infty\}) =\PP^1\setminus\{\infty\}=\AA^1, \] so \(\phi\) restricts to an automorphism of the affine line. On coordinate rings this gives a \(k\)-algebra automorphism \[ k[x]\longrightarrow k[x]. \] If \(p(x)\) is the image of \(x\), surjectivity provides \(q(x)\) with \(q(p(x))=x\). For nonconstant polynomials degrees multiply under composition, so \[ 1=\deg x=\deg(q\circ p)=(\deg q)(\deg p). \] Hence \(\deg p=1\) and \[ p(x)=ax+b,\qquad a\in k^\times,\ b\in k. \] Therefore \(\phi\) is the fractional linear transformation represented by \(\begin{pmatrix}a&b\\0&1\end{pmatrix}\).
4. Every automorphism of \(\PP^1\) is fractional linear.
Let \(\phi\in\Aut(\PP^1)\) and put \(P=\phi(\infty)\). By step 1 there is a fractional linear transformation \(T\) with \(T(P)=\infty\). Then \(T\circ\phi\) fixes \(\infty\), so step 3 shows that \(T\circ\phi\) is fractional linear. Since \(T^{-1}\) is fractional linear as well, so is \[ \phi=T^{-1}\circ(T\circ\phi). \] Thus every automorphism of \(\PP^1\) lies in \(\PGL(1)\).
5. Every \(k\)-automorphism of \(k(x)\) is fractional linear, and\[ \boxed{\PGL(1)\cong\Aut(\PP^1)\cong\Aut_k k(x)}. \]
By step 2, a field automorphism corresponds to a unique automorphism of \(\PP^1\). Step 4 makes that curve automorphism fractional linear, so its action on the affine coordinate is \[ x\longmapsto\frac{ax+b}{cx+d}, \qquad ad-bc\ne0. \] Conversely step 1 shows that every such formula is induced by an automorphism of \(\PP^1\) and therefore of \(k(x)\). This proves part (c) and both asserted isomorphisms.
6. Q.E.D.
Step 1 proves part (a), step 2 proves part (b), and steps <1>3–<1>5 prove part (c).