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)) @<j^*<< H^0(X, i^*\O(1))\\
@Ai'^*AA & @Ai^*AA \\
H^0(H, k^*\O(1)) @<k^*<< H^0(\mathbb{P}^n, \O(1))
\end{CD}
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
 A: This is a bit of a folk theorem. Harris (Algebraic Geometry, Proposition 18.10 and Exercise 18.11) states it for general hyperplane sections, but actually proves it for all generically transverse hyperplane sections.
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 $Y \cap H$.
