# Are "transverse" hyperplane sections of nondegenerate irreducible projectice varieties always nondegenerate

Let $$X \subseteq \mathbb{P}^n$$ be a irreducible complex projective variety. It is called nondegenerate if it is not contained in a hyperplane in $$\mathbb{P}^n$$.

Assuming $$X$$ is nondegenerate and irreducible, I am wondering whether a hyplernane section $$H \cap X$$ of $$X$$ is also nondegenerate in $$H$$, assuming the intersection of $$X$$ and $$H$$ is "transverse" meaning it has non components with multiplicity greater than one.

I think I have found a proof of this but this seems like a very strong statement and I couldn't find it anywhere in the literature so I'm a bit skeptical.

My proof(?):

Consider the diagram $$\require{AMScd} \newcommand{\O}{\mathcal{O}} \begin{CD} X \cap H @>j>> X\\ @Vi'VV & @ViVV \\ H @>k>> \mathbb{P}^n \end{CD}$$ where all arrows are immersions. We get a corresponding diagram of global sections

$$\begin{CD} H^0(X \cap H, j^* i^* \O(1)) @

Since $$k^*\O(1) = \O_H(1)$$ is the hyperplane bundle of $$H \cong \mathbb{P}^{n-1}$$ and $$j^* i^* \O(1) \cong i'^* \O_H(1)$$ saying that $$X \cap H$$ is nondegenerate is equivalent to $$i'^*$$ being injective.

We know that $$k^*$$ is surjective and has a one-dimensional kernel and since we're assuming $$X \hookrightarrow \mathbb{P}^n$$ is nondegenerate, $$i^*$$ is injective. Finally, consider the short exact sequence of sheaves $$0 \to \mathscr{I}_{X \cap H / X} \otimes i^*\O(1) \to i^*O(1) \to j_* j^* i^* \O(1) \to 0 \label{eq:ses}\tag{1}$$ We have $$X \cap H \subsetneq X$$ by nondegeneracy, so by irreducibility, $$X \cap H$$ is a hypersurface in X and $$\mathscr{I}_{X \cap H / X} \cong \O_X(-[X \cap H]) \cong i^*\O(-1)$$ Therefore the left term in the short exact sequence (\ref{eq:ses}) is $$\O_X$$. (Here I think we either need to work with the scheme-theoretic intersection $$X \cap H$$ or assume that the intersection of $$X$$ and $$H$$ has no components with multiplicity greater than one.) Taking global section of (\ref{eq:ses}) gives us $$0 \to H^0(X, \O_X) \to H^0(X, i^*\O(1)) \xrightarrow{j^*} H^0(X \cap H, j^* i^* \O(1))$$ Since $$H^0(X, \O_X) \cong \mathbb{C}$$ by irreducibility we see that $$j^*$$ has a one-dimensional kernel. This gives us $$1 \geq \dim \ker (j^* i^*) = \dim \ker (i'^* k^*) = 1 + \dim\ker i'^*$$ so $$i'^*$$ must be injective

• Please see Proposition 18.10 of Harris's "Algebraic Geometry: A First Course". Commented Nov 14, 2021 at 1:03
• You can just use Bezout's theorem. If a linear section of codimension $2$ contains $d$ points, then every $d+1$st point of the variety spans a hyperplane that contains $d+1$ points, and thus contains an irreducible component, by Bezout. Commented Nov 14, 2021 at 1:10
• @JasonStarr That's essentially the proof Harris gives. Commented Nov 14, 2021 at 2:02
• @carlos-esparza Are you asking for an evaluation of your proof? I haven't checked the details but the general idea of it seems good. Commented Nov 16, 2021 at 3:42
• @ZachTeitler That was part of the question... though it seems to me that the statement and proof are very similar if not the same as the lemma that you referenced (and I haven't had time yet to look at in detail) Commented Nov 16, 2021 at 3:59

But it's true more generally, for all hyperplane sections when you use the scheme-theoretic intersection. A simple argument is that if $$X \cap H = X \cap H'$$ and if $$h,h'$$ are equations for $$H,H'$$, then $$h/h'$$ is a meromorphic function on $$X$$ with no poles, so it must be constant and $$H=H'$$. Here's a more general statement given as Lemma 8.1 in Buczyński-Landsberg, Ranks of tensors and a generalization of secant varieties:
Lemma: Let $$Y \subset \mathbb{P}W$$ be a connected subvariety, which is not contained in any hyperplane in $$\mathbb{P}W$$. Let $$H \subset W$$ be a hyperplane, which does not contain any irreducible component of $$Y$$ (for example, $$Y$$ is irreducible). Then the scheme $$Z := Y \cap \mathbb{P}H$$ is not contained in any hyperplane in $$\mathbb{P}H$$.
It can fail if $$Y$$ is disconnected, e.g., let $$Y$$ be the union of two skew lines in $$\mathbb{P}^3$$, then a general hyperplane section of $$Y$$ is a set of two points, which is a degenerate subvariety in the hyperplane.
The proof is straightforward. The sequence of ideal sheaves $$0 \to \mathcal{I}_Y \to \mathcal{I}_Y(1) \to \mathcal{I}_{Y \cap H \subset H}(1) \to 0,$$ where the first map is multiplication by a defining equation $$h$$ of $$H$$, is exact because by assumption $$h$$ is a nonzerodivisor on $$Y$$. We have $$H^0(\mathcal{I}_Y(1))=0$$ since $$Y$$ is nondegenerate, so $$H^0(\mathcal{I}_{Y \cap H \subset H}(1))$$ injects into $$H^1(\mathcal{I}_Y)$$. However the defining short exact sequence for $$Y$$ gives the long exact sequence of cohomology $$0 \to H^0(\mathcal{I}_Y) \to H^0(\mathcal{O}_{\mathbb{P}W}) \to H^0(\mathcal{O}_Y) \to H^1(\mathcal{I}_Y) \to H^1(\mathcal{O}_{\mathbb{P}W}) = 0,$$ where $$H^0(\mathcal{O}_{\mathbb{P}W}) \to H^0(\mathcal{O}_Y)$$ is an isomorphism since $$Y$$ is connected. So $$H^1(\mathcal{I}_Y)$$ vanishes and hence so does $$H^0(\mathcal{I}_{Y \cap H \subset H}(1))$$, meaning no nonzero linear form on $$H$$ vanishes on $$Y \cap H$$.