Let \[ C=V(f)\subseteq\PP^2_\CC \] be a plane projective curve, where \[ f\in\CC[x,y,z]\sm\{0\} \] is irreducible and homogeneous. Let \[ P=[a:b:c]\in C. \]
Show that \(P\) is singular if and only if \[ f_x(P)=f_y(P)=f_z(P)=0. \]
If \(P\) is smooth, show that the tangent line to \(C\) at \(P\) has equation \[ x f_x(P)+y f_y(P)+z f_z(P)=0. \]
Solution
Let \[ d=\deg f. \]
1. The point \(P\) is smooth exactly when\[ \rank \begin{pmatrix} f_x(P)&f_y(P)&f_z(P) \end{pmatrix} = 1. \]
The curve \(C\) is a hypersurface in the smooth surface \(\PP^2_\CC\), so its codimension is \(1\). Choose a nonzero coordinate of \(P\); after permuting coordinates, assume \(c\ne0\) and scale the representative so that \[ P=[a:b:1]. \] On the chart \(z=1\), the equation is \[ g(x,y)=f(x,y,1)=0. \] The Jacobian criterion PR-MORJAC says that \(P\) is smooth exactly when \[ \bigl(g_x(P),g_y(P)\bigr) = \bigl(f_x(P),f_y(P)\bigr) \] is nonzero.
Euler’s identity at \(P\) gives \[ a f_x(P)+b f_y(P)+f_z(P)=d f(P)=0. \] Thus \[ f_x(P)=f_y(P)=0 \quad\Longrightarrow\quad f_z(P)=0. \] The converse is immediate. Hence the affine Jacobian is nonzero exactly when the homogeneous row vector \[ \begin{pmatrix} f_x(P)&f_y(P)&f_z(P) \end{pmatrix} \] has rank \(1\).
2. The point \(P\) is singular if and only if\[ \boxed{ f_x(P)=f_y(P)=f_z(P)=0. } \]
A one-row matrix has rank less than \(1\) exactly when all of its entries vanish. By step 1, rank less than \(1\) is exactly the singular condition. This proves (a).
3. If \(P\) is smooth, the kernel of the homogeneous differential\[ df_P:\CC^3\longrightarrow\CC \]is the two-dimensional vector subspace\[ \left\{ (u,v,w)\in\CC^3: u f_x(P)+v f_y(P)+w f_z(P)=0 \right\}. \]
The differential of the homogeneous polynomial \(f\) at a representative \((a,b,c)\) of \(P\) is \[ df_P(u,v,w) = u f_x(P)+v f_y(P)+w f_z(P). \] Since \(P\) is smooth, step 1 says that this linear functional is nonzero. Therefore its kernel is a vector subspace of codimension one in \(\CC^3\), hence has dimension two, with the displayed equation.
4. The radial vector\[ (a,b,c) \]belongs to \(\ker df_P\).
Euler’s identity for a homogeneous polynomial of degree \(d\) is \[ x f_x+y f_y+z f_z=d f. \] Evaluating at \(P\) gives \[ a f_x(P)+b f_y(P)+c f_z(P) = d f(P) = 0, \] because \(P\in C=V(f)\). Hence \((a,b,c)\in\ker df_P\).
5. The projectivization of \(\ker df_P\) is the tangent line to \(C\) at\(P\).
The affine cone over \(C\) is \[ \widehat C=V(f)\subseteq\AA^3_\CC. \] At the nonzero representative \((a,b,c)\) of a smooth projective point, the Zariski tangent space to the cone is \[ T_{(a,b,c)}\widehat C = \ker df_P. \] Step 4 shows that this tangent plane contains the radial line through \((a,b,c)\). The projective tangent space is the image of this plane under the quotient by the radial direction: \[ \PP(\ker df_P)\subseteq\PP^2_\CC. \] Because \(\ker df_P\) is two-dimensional by step 3, its projectivization is a projective line: the tangent line to \(C\) at \(P\).
6. The tangent line at a smooth point \(P\) has equation\[ \boxed{ x f_x(P)+y f_y(P)+z f_z(P)=0. } \]
Step 3 gives exactly this homogeneous linear equation for \[ \ker df_P. \] Step 5 identifies the projectivization of that kernel with the tangent line. Step 4 verifies explicitly that the line passes through \(P\). This proves (b).
7. Q.E.D.
Step 2 proves the singularity criterion, and step 6 proves the tangent-line formula.