Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $k$ be an algebraically closed field. Let $X$ be the quasi-afine variety $X = A^2 - \{0,0\}$. I know from complex-analytic arguments that the ring of regular functions on $X$ is equal to $k[x,y]$. How can I prove this algebraically?


share|cite|improve this question
up vote 5 down vote accepted

The answer of Plop is correct, but I'll try to explain it a little bit more. I know how confused you are. I'm assuming you are using Scheme Theory.

By definition, $\mathbb{A}^2_k=\text{Spec }k[x,y]$. The origin of the plane is given by the maximal ideal $(x,y)$. Let $X=\text{Spec }k[x,y] \setminus (x,y)$. Then $X=D(x) \cup D(y)$. We got an affine cover for $X$. By the definition of the sheaf, a function on $\mathcal{O}_{\mathbb{A}^2_k}(X)$ can be represented as an ordered pair of a function in $\mathcal{O}_{\mathbb{A}^2_k}(D(x))$ and the other in $\mathcal{O}_{\mathbb{A}^2_k}(D(y))$ wich coincides in the intersection $\mathcal{O}_{\mathbb{A}^2_k}(D(x)\cap D(y))$. The 'coincides' means that they are equal after applying the restrinction map.

Now, we know that $\mathcal{O}_{\mathbb{A}^2_k}(D(x))=k[x,y]_x=k[x,y,1/x]$ and $\mathcal{O}_{\mathbb{A}^2_k}(D(y))=k[x,y]_y=k[x,y,1/y]$. Since $D(x)\cap D(y)=D(xy)$, we get $\mathcal{O}_{\mathbb{A}^2_k}(D(x)\cap D(y))=k[x,y]_{xy}=k[x,y,1/(xy)]$.

What you have to do now is to show that if you have a function $f\in k[x,y,1/x]$ and $g\in k[x,y,1/y]$ wich coincides in $k[x,y,1/(xy)]$, they must be equal and belong to $k[x,y]$. Remember that, in this case, the restriction map is a localisation map.

This is an important example because it's one of the first (and maybe the simpler) examples of a non-affine scheme! Can you see why?

share|cite|improve this answer
Thanks (for both you and Plop). I now understand... – the L Feb 2 '11 at 8:34

The ring of regular functions on $A^2 \setminus \{ x=0 \}$ is $k[x,y,x^{-1}]$ and the ring of regular functions on $A^2 \setminus \{ y=0 \}$ is $k[x,y,y^{-1}]$, and $A^2 \setminus \{0\}$ is the union of these open subsets, so its ring of regular functions is the intersection of the two, which is equal to $k[x,y]$.

share|cite|improve this answer
Plop, I don't understand why this ring is the intersection of these two rings. The definition of a regular function is local, so we know that there is a local representation on each of these open subsets, but it may happen that a regular function on $X$ has two different representations on each of these open sets. How do you know that this ring is the given intersection? how do you know that the two "representations" of such a function coincide? – the L Feb 1 '11 at 21:33
This comes from the sheaf axioms To give a function on $A^2 \setminus \{0 \}$ is the same as to give functions on a cover, which agree on the intersections. – Plop Feb 1 '11 at 21:58

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.