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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?)

clgdj
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    If $\mathbb{A}^{n+1}\setminus 0$ and $\mathbb{P}^n\times\mathbb{A}^{}$ were isomorphic as functors, then this would also entail $\mathbb{R}^{2}\setminus 0\cong \mathbb{R}P^1\times\mathbb{R}^{}.$ But the left-hand side is connected, while the right-hand side is not, it's two disjoint cylinders. But I suppose this is another classical topology argument, like your #1, not algebraic. – ziggurism Nov 28 '23 at 13:35
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    I guess we should look at the coordinate rings. The coordinate ring of $\mathbb{C}^{n+1}\setminus 0$ is $\mathbb{C}[x_0,\cdots,x_n]$ while the coordinate ring of $\mathbb{P}^{n}\times\mathbb{C}^{*}$ is $\mathbb{C}[t,t^{-1}]$ – ziggurism Nov 28 '23 at 13:45
  • @ziggurism how to compute the coordinate ring of $\mathbb{P}^n \times \mathbb{C}^*$? – clgdj Nov 28 '23 at 13:58
  • The coordinate ring of a product is the tensor product. The coordinate ring of $\mathbb{P^n}$ is just $k$. – ziggurism Nov 28 '23 at 15:02
  • I meant to say $\mathbb{C}$ instead of $k$ – ziggurism Nov 28 '23 at 15:05
  • @ziggurism I only know the coordinate ring of a product is the tensor product is true for two affine varieties. Is this true for any two varieties? – clgdj Nov 28 '23 at 15:15
  • yes https://math.stackexchange.com/questions/4588523/is-the-tensor-product-of-coordinate-rings-always-a-coordinate-ring – ziggurism Nov 29 '23 at 03:36
  • Of course, perhaps you can also say that $\mathbb{C}^*$ is an affine variety, and $\mathbb{P}^n$ has a well-known coordinate ring of only constants. – ziggurism Nov 29 '23 at 03:38

1 Answers1

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$\mathbb{C}^{n+1}$ \ $\{0\}$ 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.

Mikhail Katz
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