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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.