Reading about Bertini's theorem in a book of algebraic algebraic geometry, I see the statement:

Theorem: If $\Lambda$ is a linear system on a nonsingular variety X, then any general member $D \in \Lambda$, Sing$(D) \subset B_{\Lambda}$.

My question starts with how to see a divisor $D=m_1D_1+\cdots +m_kD_k$ as a subvariety of X. After all, singular points are defined in varieties.

My second question is whether the following statement may be a consequence of Bertini's theorem, but I don't know how to justify it:

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 $p$.



1 Answer 1


In your situation, a divisor $D$ is locally cut out by a single equation: for any point $p\in X$ we can find an open neighborhood $U\subset X$ of $p$ so that $D\cap U = V(f_{D})$. The sum of divisors $D_1+D_2$ is then the closed subscheme locally given by $V(f_{D_1}\cdot f_{D_2})$.

Part 2 has no relation to Bertini: all you need to do is to observe that if $H$ isn't tangent to $X$ at $p$, then $T_p(X\cap H)=T_pX \cap T_pH = T_pX\cap H$, and so this intersection is $\dim_p X-1$ dimensional, just like $X\cap H$ is of dimension $\dim X-1$ at $p$. So the dimension of the tangent space at $p$ is the dimension of the variety at $p$, and it's smooth.


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