Some questions on the proof via Terracini's lemma and dimension that the secant variety of the twisted cubic is $\Bbb P^3$

Question

I have found an argument in Four lectures about secant varieties by Enrico Carlini, Luke Oeding and Nathan Grieve on how to prove that the secant variety of the twisted cubic in projective space is indeed the entire space. From what I understand, it goes something like this:

Consider the map $\phi:\mathbb P^1\to \mathbb P^3, \ [s:t]\mapsto [s^3,s^2t,st^2,t^3]$ and set $\phi(\mathbb P^1)=C$ the twisted cubic. Now for a generic point $p$ on the secant variety $\dim S(C)=\dim T_p(S(x))$ (why is this true, aren't we assuming non-singularity here?). Now because of Terracini's lemma we can choose the appropriate $p_1,p_2\in C$ and see, that: $\dim S(C)=\dim T_p(S(C))=\dim \langle T_{p_1}(C),T_{p_2}(C)\rangle$. Now if we could show that these two tangent lines do not lie in the same hyperplane, then the dimension of the linear span must be 3 and we are finished.

Now to the part where I really don't understand the argument. Take a hyperplane $H\subset \mathbb P^3$. Then the intersection $C\cap H$ is defined by the degree 3 (why?) homogeneous polynomial $g(s,t)$ defining $\phi^{-1}(H)$ in $\mathbb P^1$. Now if $H\supset T_{p_1}(C)$, then $g$ has a double root (i think this is just equivalent to the tangency condition?). However, $g$ is smooth (why, and what does it matter?), therefore not both tangent lines can lie in the hyperplane ($g$ can not have two different double roots?). I am very confused with this whole last part but would be extremely glad, if somebody could put lights on the parts I marked with question marks. Thanks in advance.

Answer 1

1. For a regular local ring $(R,m)$, we have $\dim_{R/m} m/m^2=\dim R$. For a (closed) point on a variety, we have that $\dim_x X = \dim \mathcal{O}_{X,x}$. Since the local dimension of a variety at a point is the dimension of a variety, we have that for any regular (closed) point on a variety, $\dim T_xX = \dim_x X = \dim X$. As the regular locus is open and dense, a generic (closed) point of a variety $X$ has tangent space of dimension equal to $\dim X$. So we're assuming non-singularity of the point $x\in X$, but that's okay, since that's true on a dense open subset.

2. Well, what is the pullback of a hyperplane in $\Bbb P^3$ to $\Bbb P^1$? If we have the hyperplane $ax+by+cz+dw=0$ for $a,b,c,d\in k$ not all zero, then pulling back to $\Bbb P^1$ along $\phi$ gives us $as^3+bs^2t+cst^2+dt^3=0$ on $\Bbb P^1$. So $g(s,t)$ is homogeneous of degree three.

3. Yes, $H\supset T_{p_1}C$ is equivalent to $g$ having a double root at $p_1$. If $H$ didn't contain the tangent space, then $H\cap C$ would be smooth at $p$ since the Jacobian would be of maximal rank, and $p_1$ would be a single root of $g$. But it does, hence $p_1$ is at least a double root of $g$.

4. "However, $g$ is smooth..." no, it need not be, but that doesn't cause any problems for our argument, since as you point out, it can't be the case that $g$ has two double roots - it's only of degree three, after all. So since $g$ can't have double roots at both $p_1$ and $p_2$, no choice of $H$ can contain the tangent lines $T_{p_1}C$ and $T_{p_2}C$, and we're done.