intersection multiplicity with a hyperplane Let $V$ be an algebraic variety embedded in $\mathbb P^n$. Let $H$ be a hyperplane (that is to say, a variety defined by a single equation of degree $1$) such that for every $x \in H \cap V$, $x$ is a regular point of $V$, and $H$ does not contain the tangent space $T_V,x$. I want to show that the intersection multiplicity of $V$ and $H$ along each irreducible component is $1$, but I did not manage to. Any indications?
 A: I think I found the beginning of a solution. Let $x$ be in a component of intersection multiplicity at least 2. Let us show that if $x$ is regular then $T_x V \subset H$.
First, since everything is local, we can suppose we are in $\mathbb A^n$. After a base change, we can also suppose $H$ is defined by $X_1=0$.
Let $e \in (S/I_V+X)_{\mathfrak P}$ not in the class of $k$ (possible since this module has length at least 2 over $S_{\mathfrak P}$ by assumption).
Then $e^2, e^3, \dots$ is not free.
So we can write $\sum_{i=2}^m a^i e^i=0$ in $(S/I_V+X)_{\mathfrak P}$, with $a_i \in S_{\mathfrak P}$. Multiplying by the denominators, then taking a lift, this gives $ \sum_{i=2}^m b^i e^i=v+X_1g$ with $b_i, g \in S$ and $v \in I_V$.
We can suppose $e(x)=0$ because the class of $e-e(0)$ is not the class of $k$. 
Taking partial derivatives, we find that $\frac{\partial v}{\partial X_j}(x)=0$ except maybe for $j=1$.  Now to complete the proof, we should show that if $x$ is regular, then we can suppose $\frac{\partial v}{\partial X_1}(x) \neq 0$. For now I can't see how to do this.
