This is Exercise 5.8 from Gathmann's notes on Algebraic Geometry, and I'm having a bit of trouble for (a) and (b): page 41 of http://www.mathematik.uni-kl.de/~gathmann/class/alggeom-2014/main.pdf

(a) asks you to show that any morphism $f:\mathbb{A}^1\backslash \{0\}\rightarrow\mathbb{P}^1$ can be extended to a morphism from $\mathbb{A}^1$ to $\mathbb{P}^1$.

(b) asks you to show that not all morphisms $f:\mathbb{A}^2\backslash \{(0,0)\}\rightarrow\mathbb{P}^1$ can be extended to a morphism from $\mathbb{A}^2$ to $\mathbb{P}^1$.

I suspect for (b) that the morphism that takes $(x,y)$ to $(x:y)$ can't be extended, but I'm not exactly sure how to prove this. Also, is the fact that $\mathcal{O}_{\mathbb{A}^2}(\mathbb{A}^2\backslash \{(0,0)\})=\mathcal{O}_{\mathbb{A}^2}(\mathbb{A}^2)$ useful in any way?

I would prefer to have an answer that doesn't involves schemes or anything too advanced like that.

  • 2
    $\begingroup$ For (a), I think the idea should be (I'm being very loose, but this works in general) to factor. In the sense that if $t$ is a coordinate on $\mathbb A^1$ then near $0$ your map looks something like $t \mapsto [t^2 + t, t^3 -t^2]$. As it stands this formula does not determine a map at $0$. But away from $0$ it's the same as $[t + 1, t^2 - t]$, and that formula does say something at $0$. $\endgroup$
    – Hoot
    Jul 2, 2016 at 16:22

1 Answer 1


Let $\phi: k^2 \setminus \{(0,0)\} \rightarrow \mathbb{P}^1$ be the map $\phi(x,y) = \overline{(x,y)}$. If you try to extend $\phi$ to a map $k^2 \rightarrow \mathbb{P}^1$, it won't even be continuous.

Suppose $\phi$ did extend to a function on $k^2$. Let $U_0 = \{ \overline{(x,y)} \in \mathbb{P}^1 : x \neq 0\}$, and similarly define $U_1$. Then $U_0, U_1$ form an open cover $\mathbb{P}^1$. Suppose that $\phi(0,0)$ lies in $U_0$. Then the preimage of $U_0$ is $$\{ (x,y) \in k^2 : x \neq 0 \} \cup \{(0,0) \}$$ whose complement in $k^2$ is $$E = \{ (0,y) : y \neq 0\}$$ Now $E$ should be closed in $k^2$. But notice $E$ is contained in $\{0\} \times k$, a closed set in $k^2$ which is homeomorphic to the affine line $k$ via $(0,y) \mapsto y$. The image of $E$ in $k$ should then be closed in $k$. But closed sets in $k$ are either finite sets or the whole space, contradiction.


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