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Is there a field k such that $Spec(k)\times \mathbb P^1$ is affine?

Probably I have no technical background to claim any good idea on this. However, I think the answer should be negative since $Spec(k)=\{(0)\}$ and topologically thinking $Spec(k)\times \mathbb P^1$ homeomorphic to $\mathbb P^1$ and we know that the scheme $\mathbb P^1$ is not affine.

How to correctly show this, and is my explanation so stupid?

Arctic Char
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  • What does $\mathbb{P}^1$ mean? Projective space over what base scheme? – Douglas Molin Jan 31 '22 at 20:31
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    I suppose by $\mathbb{P}^1$ you mean $\mathbb{P}^1_{\mathbb{Z}}$, so $\mathrm{Spec}(k) \times_{\mathbb{Z}} \mathbb{P}^1_{\mathbb{Z}} = \mathbb{P}^1_k$ which is never affine because the global sections of $\mathbb{P}^1_k$ is $k$, so if it is affine, it is trivial but $\mathbb{P}^1_k$ has a lot of other points! – Alexey Do Jan 31 '22 at 20:53

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Note that over an infinite field $k$, $\mathbb{P}^1_k$ is homeomorphic to $\mathbb{A}^1_k=\text{Spec } k[x]$. Being affine is not a topological property.

The best way, in my opinion, to see that the projective line is not affine is to compute the global regular functions. There are no non-constant global functions over $\mathbb{P}^n_k$ (see for example, this question), so if it were affine, it would necessarily be equal to $\text{Spec } k$.

Sergey Guminov
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