# Bertini Theorem & Hyperplane section $H \cap X$ smooth

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 $$p$$.

Thanks

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.