Is $\mathbb{C}^{n+1}$ \ $\{0\}$ isomorphic to $\mathbb{P}^n\times \mathbb{C}^*$?

Question

Is $\mathbb{C}^{n+1}$ \ $\{0\}$ isomorphic to $\mathbb{P}^n\times \mathbb{C}^*$?

Where $\mathbb{P}^n$ is $n$-dimensional complex projective space, $\mathbb{C}^*=\mathbb{C}$ \ $\{0\}$.

My idea:

1. Under the classical topology, the $\pi_1(\mathbb{C}^{n+1})=\{e\}$ but $\pi_1(\mathbb{P}^n\times \mathbb{C}^*)=\mathbb{Z}$. hence their classical topologies are not homeomorphic. The regular map is clearly continuous, hence if they are isomorphic(as two varieties), they will be homeomorphic, this can't be true.
2. The above idea is effective but not algebraic geometrical. Is there some algebraic geometrical way to prove $\mathbb{C}^{n+1}$ \ $\{0\}$ is not isomorphic to $\mathbb{P}^n\times \mathbb{C}^*$?

(I haven't learned about "scheme", can somebody tell me in basic language?)

Answer 1

$\mathbb{C}^{n+1}$ \ $\{0\}$ is the tautological line bundle (with the zero section removed), whereas $\mathbb{P}^n\times \mathbb{C}^*$ is the trivial line bundle (with the zero section removed). The tautological line bundle is known to be non-trivial. For example, for $n=1$ one gets the Hopf bundle if one considers vectors of unit length.