Rational Normal Scroll $S(1,1)$ I have recently learned what a rational normal scroll is and am wondering how I can prove that $S(1,1)$ is a quadric hypersurface. My professor just mentioned and didn't bother to prove it.
My thought of train is as follows: $S(1,1)$ is the union of all lines between two skew lines $X,Y\subset \mathbb{P}^3$ where we connect these two lines with an isomorphism $\varphi$.
At first, I have thought why they can't be contained in a hyperplane: As $\varphi$ is an isomorphism, both lines have to be contained in the union. As a linear subspace this would have to have dimension 3. Contradiction. Thus, there is no hyperplane containing all the lines between the skew lines $X,Y$. However, I fail to see why the scroll has to be a quadric hypersurface. Any help would be greatly appreciated.
 A: Up to automorphisms of $\Bbb P^3$, we may assume the two lines are $\ell_1=[a:b:0:0]$ and $\ell_2=[0:0:c:d]$ with the isomorphism identifying $[a:b:0:0]$ and $[0:0:a:b]$ (why? the cones on the two linear subspaces in $\Bbb A^4$ correspond to two 2-planes in $\Bbb A^4$ meeting only at the origin, so we can choose a basis so that the first two basis vectors span the first plane and the second two span the second; now apply a linear automorphism to shift the points on the second $\Bbb P^1$ so that they line up correctly). Therefore points on our surface are of the form $[ta:tb:sa:sb]$, and it is relatively straightforward to see that no nonzero linear homogeneous polynomial vanishes on all elements of this form while $x_0x_3-x_1x_2$ does vanish, so we have a quadric surface.
A: This follows from the adjunction formula once one can show there is a line in $S(1,1)$ whose normal bundle is the trivial line bundle.
Note there is a natural morphism of varieties $S(1,1) \rightarrow X$ (and also to $Y$ also but we only need one). A point $p \in S(1,1)$ is contained in a unique line $L_p \subset S(1,1)$ which connects two points of the form $q \in X$ and $\varphi(q) \in Y$. The existence and uniqueness of the line is essentially equivalent to $\varphi$ being an isomorphism. One may check that the map $p \mapsto q$ is a morphism and that the fibre over $q$ is a the line $L_p$. It follows that $L_p$ has trivial normal bundle in $S(1,1)$.
Now, $L_p$ is a line $-K_{\mathbb{CP}^3}|_{L_p}= \mathcal{O}(4)|_{L_p}$ has degree $4$. So  using the fact that the normal bundle of $L_p$ in $S(1,1)$ is trivial, and the Whitney sum formula we see that the restriction of the normal bundle of $S(1,1) \subset \mathbb{CP}^3$ to $L_p$ has degree $2$. But then by applying the adjunction formula this is equivelant to $S(1,1)$ being a quadric.
