The linear system of conics in \(\PP^2\) with two assigned base points \(P_1\) and \(P_2\) (4.1) determines a morphism \(\psi\) of \(X^{\prime}\) (which is \(\PP^2\) with \(P_1\) and \(P_2\) blown up) to a nonsingular quadric surface \(Y\) in \(\PP^3\), and furthermore \(X^{\prime}\) via \(\psi\) is isomorphic to \(Y\) with one point blown up.
Solution
After a projective change of coordinates, take \[ P_1=[0:0:1], \qquad P_2=[0:1:0] \] in \(\PP^2\) with coordinates \([x_0:x_1:x_2]\).
1. The vector space of conics through \(P_1\) and \(P_2\) has basis\[ x_0^2, \qquad x_0x_1, \qquad x_0x_2, \qquad x_1x_2. \]
A general conic is a linear combination of \[ x_0^2, x_1^2, x_2^2, x_0x_1, x_0x_2, x_1x_2. \] Vanishing at \(P_1\) forces the coefficient of \(x_2^2\) to be zero, and vanishing at \(P_2\) forces the coefficient of \(x_1^2\) to be zero. The four remaining monomials are linearly independent and give the displayed basis.
2. The corresponding rational map is\[ \phi:\PP^2\dashrightarrow\PP^3, \qquad [x_0:x_1:x_2] \longmapsto [x_0^2:x_0x_1:x_0x_2:x_1x_2]. \]Its base locus is exactly \(\{P_1,P_2\}\).
All four coordinates vanish precisely when \[ x_0=0, \qquad x_1x_2=0. \] On the line \(x_0=0\), this gives exactly \[ [0:0:1]=P_1 \qquad\text{or}\qquad [0:1:0]=P_2. \]
3. Let\[ \pi:X'=\operatorname{Bl}_{\{P_1,P_2\}}\PP^2\longrightarrow\PP^2 \]be the blowup. The rational map \(\phi\) lifts to a morphism\[ \psi:X'\longrightarrow\PP^3. \]
If \(H\) denotes the pullback of the class of a line and \(E_1,E_2\) the two exceptional curves, then the strict transforms of the conics through the base points form the complete linear system \[ |2H-E_1-E_2|. \] Blowing up the two simple base points removes the base locus, so this linear system is base-point free and defines the asserted morphism.
4. The image of \(\psi\) is contained in the quadric\[ Y=V(y_0y_3-y_1y_2)\subseteq\PP^3. \]
Write the four target coordinates in the order \[ [y_0:y_1:y_2:y_3] = [x_0^2:x_0x_1:x_0x_2:x_1x_2]. \] Then identically \[ y_0y_3 = x_0^2x_1x_2 = (x_0x_1)(x_0x_2) = y_1y_2. \] Thus every point in the image satisfies the quadric equation.
5. The quadric \(Y\) is nonsingular.
For \[ F=y_0y_3-y_1y_2, \] the four partial derivatives are \[ y_3, \qquad -y_2, \qquad -y_1, \qquad y_0. \] They vanish simultaneously only at the zero vector, which is not a point of \(\PP^3\). Hence \(Y\) is nonsingular.
6. The morphism \(\psi\) is birational onto \(Y\).
On the dense open set \(x_0\ne0\), scale so \(x_0=1\). Then \[ \phi([1:x_1:x_2]) = [1:x_1:x_2:x_1x_2]. \] Conversely, on the dense affine open \(y_0\ne0\) of \(Y\), the quadric equation determines \[ y_3=\frac{y_1y_2}{y_0}, \] so projection to \[ [y_0:y_1:y_2] \] recovers the original point of \(\PP^2\). Thus \(\psi\) is an isomorphism on dense opens and is birational. Since its image is a closed irreducible surface contained in the irreducible surface \(Y\), the image is all of \(Y\).
7. Let \(L\subseteq X'\) be the strict transform of the line\[ x_0=0 \]through \(P_1\) and \(P_2\). Then\[ L\sim H-E_1-E_2, \qquad L^2=-1, \]and \(\psi\) contracts \(L\) to\[ p=[0:0:0:1]\in Y. \]
The class formula is the standard strict-transform formula, and therefore \[ L^2 = H^2+E_1^2+E_2^2 = 1-1-1 =-1, \] using the usual intersection convention \[ H^2=1, \qquad E_i^2=-1, \qquad H\cdot E_i=E_1\cdot E_2=0. \]
At a point \([0:x_1:x_2]\) of the original line away from the two base points, one has \(x_1x_2\ne0\), so \[ \phi([0:x_1:x_2])=[0:0:0:1]=p. \] Hence its strict transform is contracted to \(p\). Equivalently, \[ (2H-E_1-E_2)\cdot(H-E_1-E_2)=2-1-1=0. \]
8. Projection from \(p\) defines the rational inverse\[ \rho:Y\dashrightarrow\PP^2, \qquad [y_0:y_1:y_2:y_3] \longmapsto [y_0:y_1:y_2]. \]
The map is defined away from \(p\), the unique point of \(Y\) at which \[ y_0=y_1=y_2=0. \] On \(y_0\ne0\), step 6 shows directly that it is inverse to \(\phi\).
On the quadric, the two ruling lines through \(p\) are \[ \{y_0=y_1=0\} \qquad\text{and}\qquad \{y_0=y_2=0\}. \] Projection contracts them to \[ [0:0:1]=P_1 \qquad\text{and}\qquad [0:1:0]=P_2, \] respectively. Thus \(\rho\) is exactly the birational inverse whose two exceptional images are the base points of \(\phi\).
9. The graph of \(\rho\) is simultaneously\[ \operatorname{Bl}_pY \]and\[ \operatorname{Bl}_{\{P_1,P_2\}}\PP^2=X'. \]
For projection of a projective variety from a point \(p\), the closure of the graph is the blowup of the variety at the base point of that projection. Hence the graph of \(\rho\) is \(\operatorname{Bl}_pY\).
On the other hand, the inverse rational map \(\phi\) is defined by the conic linear system whose only base points are \(P_1,P_2\). Blowing them up gives \(X'\) and resolves \(\phi\) to \(\psi\). The pairs \[ (\psi(q),\pi(q))\in Y\times\PP^2, \qquad q\in X', \] form the closure of the graph of \(\rho\), because on the common dense open where \(\phi\) and \(\rho\) are inverse they give exactly that graph. Therefore \[ X'\cong\operatorname{Bl}_pY. \]
10. Consequently the conic system gives the required morphism andidentification:\[ \boxed{ X'\xrightarrow{\psi}Y\subseteq\PP^3, \qquad X'\cong\operatorname{Bl}_pY.} \]
Steps <1>3–<1>6 construct the morphism onto the nonsingular quadric. Steps <1>7–<1>9 identify its unique contracted \((-1)\)-curve and show that the morphism is the blowdown inverse to blowing up the point \(p\in Y\).
11. Q.E.D.
Steps <1>1–<1>10 prove both assertions of the exercise.