A nodal plane curve need not be a projection of a nonsingular space curve

The partial derivatives are \[ F_x=4x^3+yz^2, \qquad F_y=4y^3+xz^2, \qquad F_z=2xyz. \] Because \(\operatorname{char}k\ne2\), both \(2\) and \(4\) are nonzero in \(k\).

First suppose \(z=0\). A singular point would satisfy \[ x^4+y^4=0, \qquad 4x^3=0, \qquad 4y^3=0, \] which forces \[ x=y=0, \] impossible in projective space. Thus no point at infinity is singular.

Now work on the affine chart \(z=1\). At a singular point, \[ y+4x^3=0, \qquad x+4y^3=0. \] Multiplying the first equality by \(x\) and the second by \(y\) gives \[ xy+4x^4=0, \qquad xy+4y^4=0. \] Hence \[ x^4=y^4=-\frac{xy}{4}. \] Substituting in \[ xy+x^4+y^4=0 \] gives \[ \frac{xy}{2}=0. \] Since \(2\ne0\), one has \(xy=0\). The two derivative equations then force \[ x=y=0. \] Thus \(P\) is the unique singular point.

On the affine chart \(z=1\), a local equation at \(P\) is \[ f(x,y)=xy+x^4+y^4. \] Its lowest-degree nonzero homogeneous part is \[ xy. \] Therefore the tangent cone is \[ V(xy)=V(x)\cup V(y), \] the union of two distinct lines. Hence \(P\) is an ordinary double point, that is, a node.

First, \(F\) is squarefree. Indeed, if an irreducible polynomial \(G\) occurred in \(F\) with multiplicity at least \(2\), then \(G\) would divide each partial derivative of \(F\). Every point of the positive-dimensional curve \(V(G)\) would then be singular on \(C\), contradicting step 1, which found only the single singular point \(P\). Thus \(C\) is reduced.

Step 1 says that different irreducible components of \(C\) could meet only at \(P\). Step 2 says that at \(P\) there are exactly two smooth local branches meeting transversely, so if \(C\) were reducible it would have exactly two irreducible components \[ C_1,\ C_2 \] meeting only at \(P\), with \[ I_P(C_1,C_2)=1. \]

By Bézout’s theorem, \[ \deg C_1\deg C_2 = \sum_{Q\in C_1\cap C_2}I_Q(C_1,C_2) =1. \] Thus both components would be lines, making \[ \deg C=2, \] contrary to the quartic equation. Hence \(C\) is irreducible. Since \(F\) is a single irreducible equation in the polynomial ring over a field, and we have already shown it squarefree, \(C\) is integral.

An integral plane quartic has arithmetic genus \[ p_a(C) = \frac{(4-1)(4-2)}2 =3. \] By step 1, \(P\) is its only singularity, and by step 2 it is a node. Exercise IV.1.8 gives \[ \delta_P=1 \] and \[ p_a(C) = g(\widetilde C)+\delta_P. \] Therefore \[ g(\widetilde C)=3-1=2. \]

The smooth curve \(X\) is normal. Since \(\pi\) is birational onto the integral curve \(C\), the universal property of normalization factors it as \[ X \longrightarrow \widetilde C \longrightarrow C. \] The first map is a birational morphism between nonsingular projective curves. Such a morphism is an isomorphism. Hence \[ X\cong\widetilde C. \] By step 4, \[ g(X)=2. \]

Projection from a point outside \(X\) is defined by a three-dimensional subspace of \[ H^0(X,\mco_X(1)), \] and the pullback of the hyperplane bundle on the target plane is \[ \pi^*\mco_C(1)\cong\mco_X(1). \] Since \(\pi\) is birational, it has degree one on function fields, so degrees of line bundles are preserved. More explicitly, a nonconstant morphism of projective curves is finite, and therefore \[ \deg\mco_X(1) = \deg\mco_C(1). \] The right-hand side is the degree of the plane quartic, namely \(4\). Therefore \[ \deg X=4. \]

Steps <1>5–<1>6 would produce a nonsingular curve in \(\PP^3\) with \[ \deg X=4, \qquad g(X)=2. \] But Exercise IV.3.6(a) classifies nonsingular degree-\(4\) curves: in \(\PP^3\) their genus is either \(0\) or \(1\), never \(2\). This contradiction proves that no such projection exists.

Steps <1>1–<1>4 show that the displayed curve is an integral plane quartic with exactly one node. Steps <1>5–<1>7 show that it cannot arise as the birational projection of any nonsingular curve in \(\PP^3\). Hence it is the required counterexample.