I want to verify the theorem of Bertini in the specific case of $V = \mathbb V(x^5 + y^5 + z^5 + t^5) \subset \mathbb P^3$. Because there are many different versions of this theorem, below the one I use.

Let $V \subset \mathbb P^n$ be a smooth irreducible projective variety. Consider the set $W = \{ H \in \check{\mathbb P}^n \mid H \cap V \text{ is smooth}\}$ of all hyperplanes in $\mathbb P^n$ intersecting $V$ in a smooth variety. Then W is an open subvariety of $\check{\mathbb P}^n$. That is, the set of hyperplanes having singular intersection with $V$ forms a Zariski-closed subset of $\check{\mathbb P}^n$. (Here $\check{\mathbb P}^n$ denotes the dual of $\mathbb P^n$ and a hyperplane $H = \mathbb V\left(\sum_{i=0}^n a_i x_i\right) \subset \mathbb P^n$ corresponds to the point $[a_0:a_1:\cdots:a_n] \in \check{\mathbb P}^n$.)

An invitation to algebraic geometry by Smith, Kahanpää, Kekäläinen and Traves

In other words, I want to prove that $\{ H \in \check{\mathbb P}^3 \mid H \cap V \text{ is singular} \}$ is closed. I want to do this by finding the equations that describe this set.

My attempt below, inspired by the following statement in this post,

Let $X \subset \mathbb P^n$ be a smooth irreducible projective variety. Let $H \subset \mathbb P^n$ be a hyperplane. The hyperplane section $H \cap X$ is smooth at $p$ if and only if the hyperplane $H$ is not tangent to $X$ at $p$.

So if $H = \mathbb V(ax + by + cz + dt)$ is a hyperplane, then $H \cap V$ is singular if and only if there exists some $p \in H \cap V$ such that $H$ is tangent to $X$ at $p$. Since $H$ and $V$ are smooth, their tangent spaces are two-dimensional, and so $H$ is tangent to $X$ if these tangent spaces coincide. If $p = [p_0:p_1:p_2:p_3] \in H \cap V$, then $$ T_pH = H = \mathbb V(ax + by + cz + dt) \qquad T_pV = \mathbb V\left(p_0^4 x + p_1^4 y + p_2^4 z + p_3^4 t\right). $$ Now $T_pH = T_pX$ if and only if there exists some $\lambda \in \mathbb C \setminus \{0\}$ such that $$ \lambda (ax + by + cz + dt) = p_0^4 x + p_1^4 y + p_2^4 z + p_3^4 t. $$ This is equivalent with requiring $$ \lambda a = p_0^4, \quad\lambda b = p_1^4,\quad \lambda c = p_2^4 \quad\text{ and }\quad \lambda d = p_3^4. $$

To summarize what I have found: a hyperplane $\mathbb V(ax + by + cz + dt)$ belongs to $\{ H \in \check{\mathbb P}^3 \mid H \cap V \text{ is singular} \}$ if and only if there exist $p=[p_0:p_1:p_2:p_3] \in \mathbb P^3$ and $\lambda \in \mathbb C \setminus \{0\}$ satisfying the following system of equations: $$ \begin{align} p_0^5 + p_1^5 + p_2^5 + p_3^5 &= 0 \\ ap_0 + bp_1 + cp_2 + dp_3 &= 0 \\ \lambda a - p_0^4 &= 0 \\ \lambda b - p_1^4 &= 0 \\ \lambda c - p_2^4 &= 0 \\ \lambda d - p_3^4 &= 0 \end{align} $$ From this, I would like to obtain some condition that $[a:b:c:d]$ has to satisfy, but I do not see how I can proceed. The main problem is that these equations still contain the point $p$ and the parameter $\lambda$, so I thought of using the multipolynomial resultant somehow, but the polynomials above are not homogeneous when viewed as polynomials in the variables $p_0,p_1,p_2,p_3,\lambda$.

How can I go from these equations to a condition on the coefficients $[a:b:c:d]$?


1 Answer 1


I believe that such a question is exciting; it is a direct use of the Bertini Theorem. I think that you can use a bit more abstract technique. If you read Theorem 8.18, chapter 2 of the Hartshorne Book, it only remains for you to show that the quintic Fermat surface $\{x^5+y^5+z^5+t^5=0\}$ is smooth, which is not complicated. Moreover, if you want to give explicit prove for this particular case, you can see the proof of the theorem and run it for the Fermat surface.

My best!!

  • $\begingroup$ Hartshorne's proof is very non-constructive - I don't think that one can explicitly get a description of the "bad" hyperplanes via coordinates from it. Maybe this would have been better as a comment? $\endgroup$ Jan 4, 2022 at 21:04
  • $\begingroup$ Thank you. However, I haven't been introduced to the concept of schemes yet (I am following the book 'An invitation to algebraic geometry'), so I wanted to use more basic machinery. $\endgroup$ Jan 5, 2022 at 8:36

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