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It is Exercise 4.3.2 in the book An Invitation to Algebraic Geometry.

Show that the ring $\mathcal{O}_V(U)$ of regular functions on the punctured plane $U=\mathbb{A}^2\backslash\{(0,0)\}$ is the polynomial ring $\mathbb{C}[x,y]$ (where I think $V=\mathbb{A}^2$). Conclude that this quasi-projective variety is not affine.

My try is as follows.

Obviously we have $\mathbb{C}[x,y]\subset \mathcal{O}_V(U)$.

Now let $f\in\mathcal{O}_V(U)$, by definition we have for any $p$ in $U$, we can choose a neighborhood $U_p$ of $p$ such that $$ \left.f\right|_{U_p}=\frac{h_p}{k_p} $$ for some polynomial $h_p$ and $k_p$ and $k_p(p)\neq 0$.

Because Zariski topology is compact, we can select finitely many such neighborhood, indexed by $i=1,\cdots,n$.

Because $k_i$c cannot simultaneously vanish on $U$, we must have $$ \mathbb{V}(k_1,\cdots,k_n)\subset U^C=\{(0,0)\} $$

If $\mathbb{V}(k_1,\cdots,k_n)=\varnothing$, then $(k_1,\cdots,k_n)=(1)$, then we can select $$ 1=\sum_{j}l_jk_j $$ for some $l_j$, and the rest is easy.

My problems lies in the case $\mathbb{V}(k_1,\cdots,k_n)=\{(0,0)\}$ which follows $$ (k_1,\cdots,k_n)=(x,y) $$ and I don't know how to move on...

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  • $\begingroup$ If you don't get responses I would search around -- this has definitely been asked about before. Still, I wouldn't vote to close this question (even if I could) because your work deserves its own response. By the way, I think you want $k_p$ to be nonzero on all of $U_p$. $\endgroup$
    – Hoot
    Commented Apr 24, 2014 at 20:20
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    $\begingroup$ math.stackexchange.com/a/122826/3217 $\endgroup$ Commented Apr 24, 2014 at 20:26
  • $\begingroup$ @Hoot, thanks. indeed I want some one to help with my approach:) $\endgroup$
    – hxhxhx88
    Commented Apr 24, 2014 at 21:28

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So, your method is interesting. It is not how I approached the problem when I did it first. $\def\O{\mathcal{O}}$ $\def\Spec{\mathop{Spec}}$

I think the main ingredient you are missing is imposing the compatibility condition that on $U_{i} \cap U_{j}$

$$\frac{h_{i}}{k_{i}} = \frac{h_{j}}{k_{j}}.$$

But it is hard to get a handle on the compatibility condition if you don't impose some controls on the open sets $U_{i}, U_{j}.$ So instead, maybe you can approach the problem like this.

Suppose $f \in \O_{V}(U)$. Then, we can cover $U$ by the affines $$U_{1} = (V)_{x} = \Spec \mathbb{C}[x, y]_{x}, U_{2} = (V)_{y} = \Spec \mathbb{C}[x, y]_{y}.$$

These are the open sets where $x$ and $y$ are respectively nonzero. Restricting $f$ to each of these open affines give $g_{1} \in k[x, y]_{x}, g_{2} \in k[x, y]_{y}.$ Hence, for some integer $m$, we have $$f|_{U_{1}} = \frac{p_{1}(x, y)}{x^{m}}, f|U_{2} = \frac{p_{2}(x, y)}{y^{m}}.$$

Now, on $U_{1} \cap U_{2} = \Spec \mathbb{C}[x, y]_{x, y}$, these expressions must agree. So, we must have $$\frac{p_{1}}{x^{m}} = \frac{p_{2}}{y^{m}} \Rightarrow y^{m}p_{1} = x^{m}p_{2}.$$ Since $k[x, y]$ is a unique factorization domain, this implies that $x^{m}$ must divide $p_{1}$ and $y^{m}$ must divide $p_{2}$. Hence, $f = \frac{p_{1}}{x^{m}} \in \mathbb{C}[x, y]$

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  • $\begingroup$ Hi, I am still confused. Why must $f$ be a rational function on $U_1=\mathbb{A}\backslash\mathbb{V}(x)$? We only know given a point $p$, $f$ is rational on some neighborhood of $p$, maybe sufficiently small. Why won't $U_1$ be too big? $\endgroup$
    – hxhxhx88
    Commented Apr 24, 2014 at 22:36
  • $\begingroup$ So if $f$ is a regular function on $U$, it must be a regular function on any open subset of $U$ and thus in particular on $U_{1}.$ You can restrict regular functions to open sets to get regular functions on the open set. Then, I use the fact that regular functions on $\mathbb{A}^{2} \backslash V(x)$ are exactly $\frac{p(x, y)}{x^{m}}$ where $p$ is a polynomial. $\endgroup$ Commented Apr 25, 2014 at 0:59
  • $\begingroup$ Got it, thanks! $\endgroup$
    – hxhxhx88
    Commented Apr 25, 2014 at 11:33
  • $\begingroup$ @SiddharthVenkatesh how exactly do you use UFD at the end? $\endgroup$
    – user166467
    Commented Jul 17, 2016 at 12:33
  • $\begingroup$ @monomorphic Without any UFD hypothesis, what we get is that $y^m p_1 = x^m p_2$ (this requires only that the polynomial ring is a domain). To go from here to the conclusion that $x^m$ divides $p_1$ and $y^m$ divides $p_2$ requires that the polynomial ring is a UFD because $x$ and $y$ being irreducibles doesn't give you this information if you aren't in a UFD. Actually, all that I'm really using is that the polynomial ring is normal. You can show that if you have a normal affine variety then any open set of codimension 2 or greater has the same ring of global functions as the original variety. $\endgroup$ Commented Jul 30, 2016 at 17:33

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