Covering projective variety with open sets $U_i$ such that $\pi^{-1}(U_i) \cong U_i \times \Bbb{A}^1$: How to improve geometric intuition?

Question

I am looking at exercise II 6.3 of Hartshorne. In the first part, he asks to show the following. If $V \subseteq \Bbb{P}^n$ is a projective variety (over some field $k$), let $X = C(V)$ denote its affine cone in $\Bbb{A}^{n+1}$, and let $\overline{X}$ be the projective closure of $X$ in $\Bbb{P}^{n+1}$. Then we can cover $V$ with open sets $U_i$ so that under the projection $\pi : \overline{X} - \{P\} \to V$, we have $\pi^{-1}(U_i) \cong U_i \times_k \Bbb{A}^1$.

In the case $V= \Bbb{P}^n$, I can see by writing down equations that we can just take $U_i$ to be the standard affine open sets. But what about more general projective varieties?

At the moment I have zero idea on how to approach such a problem geometrically without writing down equations. I am ok when it comes to bashing the commutative algebra, but for more geometric things like this I am quite bad. In general, how does one approach a problem like this? How can I improve my geometric intuition?

Answer 1

Basically it suffices to treat the case $V=\mathbb{P}^n$. The rest comes via pullback. So if $V \subseteq \mathbb{P}^n$ (you didn't write that, but I think you assume this), then almost by definition $$\begin{array}{c} C(V) & \rightarrow & C(\mathbb{P}^n) \\ \downarrow && \downarrow \\ V & \rightarrow & \mathbb{P}^n \end{array}$$ is a pullback square. The same is true for the projective closures $$\begin{array}{cc} C(V) & \rightarrow & C(\mathbb{P}^n) = \mathbb{A}^{n+1} \\ \downarrow && \downarrow \\ \overline{C(V)} & \rightarrow & \overline{C(\mathbb{P}^n)} = \mathbb{P}^{n+1} \end{array}$$ and the desired map $$\begin{array}{c} \overline{C(V)} \setminus \{\infty\} & \rightarrow & V \\ \downarrow && \downarrow \\ \overline{C(\mathbb{P}^n)} \setminus \{\infty\} & \rightarrow & \mathbb{P}^n \end{array}$$ In order to construct it, you just have to verify that the composition $\overline{C(V)} \setminus \{\infty\} \to \mathbb{P}^n$ has image in $V$, i.e. that the defining equations are satisfied. But of course this comes because we start from $C(V)$. This is just a sketch, but the detailed calculation is not hard.