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If $k$ is algebraically closed, it's known that $\mathbb{A}^2\setminus0$ is not an affine scheme (see, e.g. this). But what about the scenario when $k$ is not algebraically closed?
My thought: in the case of alg. closed fields, the argument works in the spirit of "there is no polynomial in two variables that vanishes only at $(0,0)$, therefore, any rational function on $\mathbb{A}^2\setminus0$ extends to $\mathbb{A}^2$", but for example, for $k = \mathbb{R}$ there is a polynomial $x ^ 2 + y ^2$ and it turns out that $1/(x^2+y^2)$ does not extend to the whole plane.

Dmitry
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    Not all points are real on $\Bbb{A^2_R}$, $x^2+y^2$ vanishes at the maximal ideal $(x^2+1,y-1)\subset \Bbb{R}[x,y]$ – reuns Dec 22 '22 at 18:01
  • Base changing to an algebraic closure doesn't change affineness. So if it were affine over a non-algebraically closed field, it would also be affine over an algebraically closed field. – Legendre Dec 24 '22 at 02:09
  • where does it require $k$ algebraicly closed in the linked post? – onRiv Dec 24 '22 at 05:57

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