Embedding singular divisors into smooth varieties

Question

Let $X=X_{d_1}\cap\cdots \cap X_{d_m}\subset \mathbb{P}^n$ be a smooth complete intersection of smooth hypersurfaces $X_{d_i}$ of degree $d_i$ with $1<d_1\leq ...\leq d_m$ and $m>1$. For simplicity, we assume that $\dim X=3$.

Let $S\subset X$ be a singular hyperplane section, then $S=\mathbb{P}^{n-1}\cap X_{d_1}\cap\cdots \cap X_{d_m}$ is also a complete intersection.

Question: Is it always true that we can find a smooth hypersurface $Y\subset \mathbb{P}^4\subset \mathbb{P}^n$ of degree $d_1$ containing $S$?

If the answer is not, then how about assuming $X$ to be very general and replacing the degree of $Y$ by other $d_i$, $1\leq i\leq m$?

Answer 1

The answer is negative. For instance, consider a complete intersection $X$ of type $(2,3,3)$ defined as follows. Let $V_2 \subset V_n$ be a 2-dimensional vector subspace in an $n$-dimensional vector space. Set $$ V_2^\perp := \mathrm{Ker}(V_n^\vee \to V_2^\vee). $$ Let $F_2 \in S^2(V_2^\perp) \subset S^2V_n^\vee$ and $F_3',F_3'' \in S^3V_n^\vee$ be general. Then the intersection $$ X = \{F_2 = F_3' = F_3'' = 0 \} \subset \mathbb{P}(V_n) $$ is smooth. Now we can choose $F_1 \in V_2^\perp$ in such a way that $$ S = X \cap \{F_1 = 0\} $$ is singular. Then any quadric through $S$ will be given by equation of the form $aF_2 + F_1\cdot G_1$ and will be singular at the point $\mathbb{P}(V_2) \cap \{G_1 = 0\}$.