Singular points of projective hypersurface and Veronese map

Question

Let $S = \{f = 0\} \subset \mathbb{P}^n$ be a hypersurface of degree $d$. Then $S$ is the intersection of $v(\mathbb{P}^n)$ with a unique hyperplane $H \subset \mathbb{P}^N$, where $v: \mathbb{P}^n \to \mathbb{P}^N$ is the Veronese morphism of degree $d$.

Lamotke writes in The topology of complex varieties after S. Lefschetz, that the regular (he calls them simple, but I think they are the same) points of $S$ are those points, where $H$ and $\mathbb{P}^n$ intersect transversally. Why is that the case? Or are simple points another thing than regular points?

Answer 1

As Parthiv Basu mentioned in the comments, the following is true:

Let $Y_1, Y_2 \subset X$ be smooth varieties, and let $x \in Y_1 \cap Y_2$ be a closed point, such that $\dim Y_1 \cap Y_2 = \dim Y_1 + \dim Y_2 - \dim X$, at $x$. Then $x$ is a regular point in the intersection if and only if $Y_1$ and $Y_2$ intersect transversally at $x$.

This is helpful, because the dimension requirement is always given if $Y_1$ is the image of the Veronese map, and $Y_2$ is a hyperplane, as no hyperplane contains $Y_1$.

Proof:

First observe the equality $$T_x(Y_1 \cap Y_2) = T_x Y_1 \cap T_x Y_2 \subset T_x X.$$ This is true for arbitrary varieties, smooth or not, if $T_x X, \ldots$ are the Zariski tangent spaces. Thus we have the inequality $$\dim(T_x Y_1 \cap T_x Y_2) \geq \dim(Y_1 \cap Y_2) = \dim Y_1 + \dim Y_2 - \dim X = \dim T_x Y_1 + \dim T_x Y_2 - \dim T_x X.$$ So we see that $T_x Y_1 + T_x Y_2 = T_x X$ if and only if we have "$=$" here, which is exactly the criterion for smoothness of $x$ in $Y_1 \cap Y_2$.