$\mathbb{P}(O\oplus O(-1))\simeq \mathbb{P}^1\times \mathbb{P}^1$?

Question

Let $X=\mathbb{P}^1$. I am looking at $\mathbb{P}(O_X\oplus O_X(-1))$ and can see that it is the blow up of the projective plane at one point. I also see that it is a $\mathbb{P}^1$-bundle over $X$, but can't quite see if it's isomorphic to $\mathbb{P}^1\times \mathbb{P}^1$ or not. Can anyone give some insight?

Answer 1

Any $\Bbb P^1$ in $\Bbb P^1\times \Bbb P^1$ has $0$ self-intersection (trivial normal bundle). The exceptional divisor in the blow-up has self-intersection $-1$ and the hyperplane class has self-intersection $1$.

Answer 2

As something of an aside: to get $\mathbb P^1\times \mathbb P^1$ from $\mathbb P^2$, you blow up the plane at two points, and then blow down the (proper transform of) the line joining the two blown-up points.

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Also, the section of Hartshorne Ch. V discussing ruled surfaces describes quite carefully how the invariants of the line bundle $\mathcal L$ (on some curve, e.g. $\mathbb P^1$) relate to the geometry of $\mathbb P(\mathcal O_C \oplus \mathcal L).$ You might want to look at that section to get more ideas and intuitions about how to investigate this kind of question.