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Let $X_{2,3}\subset\mathbb{P}^n$, with $n\geq 5$, be a complete intersection of a quadric $X_2$ and a cubic $X_3$ containing a $2$-plane $H$. Assume $X_2$ and $X_3$ to be general among the hypersurface of the same degree containing $H$. In particular $X_2$ and $X_3$ are smooth.

Question: If $n = 5$ is $X_{2,3}$ necessarily singular? Is $X_{2,3}$ smooth for $n\geq 6$?

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2 Answers 2

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If $X \subset \mathbb{P}^n$ is a non-degenerate, smooth complete intersection variety of dimension at least $3$, then the restriction map $$\operatorname{Pic}(\mathbb{P}^n) \to \operatorname{Pic}(X)$$ is an isomorphism by Grothendieck-Lefschetz theorem. So, we get $\operatorname{Pic}(X) = \mathbb{Z}$, generated by the hyperplane section.

This implies that $X$ contains no linear spaces of codimension $1$; in particular, the threefold $X_{2,3} \subset \mathbb{P}^5$ is necessarily singular.

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If $n=5$ then let $\mathbb P^5$ have coordinates $x_0,\ldots,x_5$ and suppose the plane is $H=\mathbb P^2_{(x_0:x_1:x_2)}$. The two equations of $X$ are necessarily of the form $$ \begin{pmatrix} A_1 & B_1 & C_1 \\ D_2 & E_2 & F_2 \end{pmatrix}\begin{pmatrix} x_3 \\ x_4 \\ x_5 \end{pmatrix} = 0 $$ for some polynomials $A_1,\ldots,F_2\in\mathbb C[x_i]$ of the indicated degree. The scheme $Z\subset \mathbb P^5$ defined by the $2\times 2$ minors of the $2\times 3$ matrix has codimension 2, and thus $Z\cap H$ is nonempty (generically given by some points). Since both equations have order of vanishing $\geq2$ along $(Z\cap H)\subset X$, the 3-fold $X$ must be singular there.

In dimension $n\geq6$ the scheme $Z$ (obtained by the analogous argument) has codimension $n-3\geq3$ so in the general case $Z\cap H=\emptyset$ and the $n$-fold $X$ is smooth (along $H$ at least). Essentially, since the matrix never drops rank we can always use the equations to eliminate two of the variables $x_3,\ldots,x_n$, showing that $X$ is smooth of dimension $n$ at every point along $H$.

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    $\begingroup$ It's not true that both equations have order of vanishing $\geq 2$ at these points but rather that some linear combination does (but this still implies a singularity). $\endgroup$
    – Will Sawin
    Jan 18 at 12:58
  • $\begingroup$ Sorry, yes good point. That is what I meant to say. $\endgroup$
    – Tom Ducat
    Jan 18 at 14:55

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