Assume \(\operatorname{char}k\ne2\). Let \(Y\) be the algebraic set in \(\PP^3\) defined by the equations \[ x^2 - xz - yw = 0, \qquad yz - xw - zw = 0. \] Let \(P\) be the point \((x,y,z,w) = (0,0,0,1)\), and let \(\varphi\) denote the projection from \(P\) to the plane \(w = 0\). Show that \(\varphi\) induces an isomorphism of \(Y \sm\ts{P}\) with the plane cubic curve \(y^2 z - x^3 + x z^2 = 0\) minus the point \((1,0,-1)\). Then show that \(Y\) is an irreducible nonsingular curve.
This \(Y\) is called the elliptic quartic curve in \(\PP^3\). Since it is defined by two equations it is another example of a complete intersection.
(Characteristic two)
The source does not exclude characteristic \(2\), but the stated conclusion fails there. At the target point \[ Q=[1:0:-1]=[1:0:1] \] the two equations for \(w\) become identities in characteristic \(2\). Consequently every point \[ [1:0:1:w]\in\PP^3 \] lies on \(Y\), so the fibre over \(Q\) is a whole line rather than being empty. Thus the projection isomorphism and the irreducible elliptic-quartic conclusion require \(\operatorname{char}k\ne2\) for these equations.
Solution
Put \[ F_1=x^2-xz-yw,\qquad F_2=yz-xw-zw, \] and let \[ C=V(H)\subseteq\PP^2_{[x:y:z]}, \qquad H=y^2z-x^3+xz^2. \] The projection from \[ P=[0:0:0:1] \] is \[ \varphi([x:y:z:w])=[x:y:z]. \]
1. Projection maps \(Y\setminus\{P\}\) into \(C\).
On \(Y\) the two equations can be rewritten \[ yw=x(x-z),\qquad (x+z)w=yz. \] Multiply the second equation by \(y\) and use the first: \[ y^2z =yw(x+z) =x(x-z)(x+z) =x^3-xz^2. \] Thus \[ H(x,y,z)=0. \] The only point of \(\PP^3\) where the three projection coordinates vanish is \(P\), so the projection is defined on \(Y\setminus\{P\}\) and lands in \(C\).
2. On the cubic \(C\), the simultaneous equations\[ y=0,\qquad x+z=0 \]define the single point\[ Q=[1:0:-1]. \]For every point of \(C\setminus\{Q\}\) there is a unique \(w\) satisfying the two equations of \(Y\).
The first assertion is immediate in projective coordinates. For a point \(q=[x:y:z]\in C\), the required \(w\) must satisfy \[ yw=x(x-z),\qquad (x+z)w=yz. \] The compatibility determinant of these two linear equations in \(w\) is \[ y(yz)-(x+z)x(x-z) =y^2z-x^3+xz^2 =H, \] so they are compatible on \(C\). Away from \(Q\), the two coefficients \(y\) and \(x+z\) are not both zero, so the compatible system has a unique solution.
At \(Q\), one has \[ x(x-z)=1(1-(-1))=2\ne0, \] using \(\operatorname{char}k\ne2\). Thus no \(w\) solves the first equation there. Hence \(Q\) has no preimage and every other point has exactly one.
3. The inverse to projection on \(C\setminus\{Q\}\) is a morphism.
The opens \[ D_+(y),\qquad D_+(x+z) \] cover \(C\setminus\{Q\}\). On the first define \[ w=\frac{x(x-z)}{y}, \] and on the second define \[ w=\frac{yz}{x+z}. \] These are homogeneous degree-one expressions in the sense that scaling \([x:y:z]\) by \(\lambda\) scales the resulting \(w\) by \(\lambda\). Thus they define morphisms to \(\PP^3\). On the overlap they agree because the compatibility equation in step 2 is exactly \[ \frac{x(x-z)}y=\frac{yz}{x+z}. \] The two local morphisms therefore glue to \[ \psi:C\setminus\{Q\}\longrightarrow Y\setminus\{P\}. \] By construction \(\varphi\psi\) is the identity. For a point of \(Y\setminus\{P\}\), its \(w\) already satisfies the same two linear equations, whose solution is unique by step 2, so \(\psi\varphi\) is also the identity. Consequently \[ \boxed{Y\setminus\{P\}\cong C\setminus\{Q\}.} \]
4. The plane cubic \(C\) is nonsingular and irreducible.
Its partial derivatives are \[ H_x=-3x^2+z^2,\qquad H_y=2yz,\qquad H_z=y^2+2xz. \] Suppose all three vanish at a projective point of \(C\). Because \(2\ne0\), the equation \(H_y=0\) gives \(y=0\) or \(z=0\).
If \(y=0\), then \(H_z=2xz=0\). If \(x=0\), then \(H_x=z^2\) forces \(z=0\). If \(z=0\), then the cubic equation \(H=-x^3=0\) forces \(x=0\). Either way there is no projective point.
If \(z=0\), then \(H_z=y^2\) forces \(y=0\), and again \(H=-x^3=0\) forces \(x=0\). This also covers characteristic \(3\). Hence \(C\) has no singular point.
By P-AGH59NONVANISHIRRED, a positive-degree homogeneous plane equation with nowhere-vanishing gradient on its zero set is irreducible. Thus \(C\) is a nonsingular irreducible cubic curve.
5. The algebraic set \(Y\) is irreducible of dimension one.
By steps <1>3–<1>4, the open subset \[ Y\setminus\{P\} \] is irreducible and one-dimensional. Its closure is therefore an irreducible component \(Y_0\) of \(Y\).
Any other irreducible component would have empty intersection with \(Y\setminus\{P\}\), hence would be contained in the singleton \(\{P\}\). But \(\{P\}\) cannot be an irreducible component of an algebraic set cut out by two homogeneous equations in \(\PP^3\). Indeed, its affine-cone prime has height three in \(k[x,y,z,w]\), whereas Krull’s height theorem says that a minimal prime over an ideal generated by two elements has height at most two. Thus no second component exists and \[ Y=Y_0. \] Hence \(Y\) is irreducible, and since the dense open subset \(Y\setminus\{P\}\) is a curve, \(\dim Y=1\).
6. The point \(P\) is nonsingular on \(Y\).
The Jacobian matrix of the two defining quadrics is \[ \begin{pmatrix} 2x-z&-w&-x&-y\\ -w&z&y-w&-x-z \end{pmatrix}. \] At \[ P=[0:0:0:1] \] this becomes \[ \begin{pmatrix} 0&-1&0&0\\ -1&0&-1&0 \end{pmatrix}, \] which has rank two. By step 5, \(Y\) is a projective variety of dimension one. The projective Jacobian criterion P-AGH58JACOBIANRANK requires rank \[ 3-1=2 \] for nonsingularity. Thus \(P\) is nonsingular.
7. The projective curve \(Y\) is nonsingular.
4.Thus every point away from \(P\) is nonsingular.Step 6 proves nonsingularity at \(P\).Therefore\[ \boxed{Y\text{ is an irreducible nonsingular projective curve}.} \]This proves the asserted elliptic-quartic properties required in the exercise.
8. Q.E.D.
Steps <1>1–<1>3 establish the stated projection isomorphism, and steps <1>4–<1>7 prove that \(Y\) is an irreducible nonsingular curve.