I would like some help with an exercise from Reid's paper of algebraic geometry.

Prove that the intersection of a hypersurface $V \subset \mathbb{P}^n$ (not a hyperplane) with the tangent hyperplane $T_PV$ is singular in P.

Maybe I'm using the wrong approach. I'm trying to prove this by taking partial derivatives and evaluate them in the point P.

Thank you!!


2 Answers 2


First note that the hypersurface $V$ must be regular (=non-singular) at $P$, else its Zariski tangent space $T_P(V)$ would be $\mathbb P^n$ and not a hyperplane.
Next, since the problem is local at $P$ we can suppose, upon replacing $V$ by $W=V\cap \mathbb A^n$, that we have a hypersurface $W\subset \mathbb A^n_{x_1,...,x_{n}}$ containing the smooth point $P=(0,\cdots,0)$ and that the tangent hyperplane $T_P(W)$ is given by $x_1=0$.
This means that the the polynomial defining $W=Z(f)$ is of the form $$f(x_1,\cdots,x_n)=x_1+f_2(x_1,\cdots,x_n)+\cdots+f_r(x_1,\cdots,x_n)$$ where $r\geq 2$, $f_i(x_1,\cdots,x_n)$ is homogeneous of degree $i$ and $f_r(x_1,\cdots,x_n)\neq 0$.
But then the intersection $W\cap T_P(W)$ is the hypersurface $Z(f_0)\subset T_P(W)=\mathbb A^{n-1}_{0,x_2,...,x_n}$ of zeros of the polynomial $$f_0(x_2,\cdots,x_n)=0+f_2(0,x_2,\cdots,x_n)+\cdots+f_r(0, x_2,\cdots,x_n)$$ and that hypersurface is singular because the displayed polynomial $f_0(x_2,\cdots,x_n)$ has zero as linear term.


The statement in the exercise is in fact not true as it stands. The easiest counterexample is the intersection of a smooth curve with its tangent: it is a point and therefore smooth.

A specific example is the parabola $y=x^2$ with tangent $y=0$ at the origin. Of course one would like to count this intersection point with multiplicity two, and therefore consider it as singular.

The correct statement in the affine case is that the restriction of the defining function to the tangent plane has a critical point. A proof can be given with techniques of multivariable calculus, namely Lagrange multipliers: the gradient of the function is the normal to the tangent plane, and for a linear function the gradient is constant and equal to the normal vector, so the two gradients in question are linearly dependent. Using the gradient supposes that we are in the real case. An illustration for the argument is the fact the height function on a sphere has a maximum at the north pole. Related to the original question, depending on your point of view the intersection of a sphere with its tangent plane is a point or consists of two complex conjugate lines.

For any field, let $f$ be the defining function of the hypersurface and $l$ the defining function of the tangent plane (in the projective case $f$ and $l$ are homogeneous polynomials). If we define the intersection by the ideal $(f,l)$ with non-reduced structure, then at the intersection point the Jacobian matrix of $(f,l)$ has rank one, as the rows are equal. But the zero set $V(f,l)$ can be non-singular.


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