# $\mathbb{A}^2\backslash\{(0,0)\}$ is not affine variety

In our lecture notes we have this example, with the proof why $X = \Bbb{A}^2\setminus \{(0,0)\}$ is not an affine variety:

Let $i:X\hookrightarrow \mathbb{A}^2$ be an inclusion map. We show, that any regular function on $X$ extends uniquely to a regular function on $\mathbb{A}^2$, so that $i^*:k[x,y]\hookrightarrow \mathscr{O}(X)$ is an isomorphism.
Then we have, if $X$ was affine, $i$ is an isomorphism, which is not the case.

So we want to see that $i^*$ is an isomorphism. Therefore note that $\mathscr{O}(X)$ is a subring of the function field $k(x,y)$, that contains $k[x,y]$.
Take $P=(a,b)\in X$. Then $\mathscr{O}_{X,P}=\mathscr{O}_{\mathbb{A}^2,P}=k[x,y]_{\mathfrak{m}_P}$ whith $\mathfrak{m}_P=(x-a,y-b)$ is maximal ideal of $k[x,y]$.
-> Here is my first questions. How did we get this continued equality? I don't see how this is connected... (my commutative algebra knowledge has some gaps.)
But accepting that, we have $\mathscr{O}(X)\subset\bigcap\limits_{P\not=(0,0)}k[\mathfrak{m}_P]$. And this intersection lies inside of k(x,y).
Now we take $f=\frac{g}{h}\in K(x,y)$ with $g,h\in k[x,y]$ coprime.
Then $f\in k[x,y]_{\mathfrak{m}_P}$ iff $h(P)\not=0$. (Use that $k[x,y]$ is a UFD).
-> First, how did we see that $k[x,y]$ is UFD? And then, how did we conclude out of this the above property?
If $h$ is not a unit, then there are infinitely many $P\in \mathbb{A}^2$ such that $h(P)=0$ and thus also such points in $X$, and so $f\not\in\mathscr{O}(X)$.
So we have that $f\in\mathscr{O}(X)$ if $h$ is a unit. So $f\in k[x,y]$.
Thus $k[x,y]=\mathscr{O}(X)$ and hence $X$ not affine.
-> Now the last question, why does follow that $X$ is not affine?

I hope my questions are not too stupid, and that someone feels like helping!
All the best, Luca :)

• This example always bothers me. If you don't have a setting for describing not-necessarily-affine varieties, you don't know what $\mathbb{A}^2 \setminus \{ (0, 0) \}$ is before you determine whether or not it's affine. A priori being an affine variety is a structure and not a property. – Qiaochu Yuan Jun 19 '13 at 0:26
• @QiaochuYuan, Well, an (abstract) variety is a locally ringed space that has an open covering by affine varieties. I think the example makes perfect sense in that context. Or, if you prefer, work with the category of quasi-projective varieties. Again, it makes perfect sense to ask whether one of these is isomorphic to an affine variety or not. – Stephen Jun 19 '13 at 0:32
• @Steve: yes, but I've seen people give this exercise in first courses in algebraic geometry that don't mention either locally ringed spaces or quasi-projective varieties. In such a context I maintain that the exercise is not well-defined. – Qiaochu Yuan Jun 19 '13 at 0:33
• @zyx: that might be okay (it's not obvious to me that it's equivalent to the quasi-projective version of the question), but my point is that without a definition of not-necessarily-projective variety, $\mathbb{A}^2 \setminus \{ (0, 0) \}$ is a priori a set, and a set being an affine variety is a structure, and not a property. (A subset of an affine variety being an affine variety is a property, but that isn't what's being asked for here, since e.g. $\mathbb{A}^1 \setminus \{ 0 \}$ isn't an affine subvariety of $\mathbb{A}^1$ but "is" an affine variety.) – Qiaochu Yuan Jun 19 '13 at 1:30
• In other words, without more machinery the question contains a type error (qchu.wordpress.com/2013/05/28/the-type-system-of-mathematics). – Qiaochu Yuan Jun 19 '13 at 1:31

1. The reason why $\mathcal{O}_{P,X} = \mathcal{O}_{P,\Bbb{A}^2}$ is because for any point $P$ in an affine variety $Y$ and $U \subseteq Y$ open, we have that $\mathcal{O}_{P,U} \cong \mathcal{O}_{P,Y}$. This should essentially be immediate (what is the definition of the local ring at a point?) The second equality comes from Theorem 1.3.2 (c) of Hartshorne.

2. That $k[x,y]$ is a UFD is a standard fact in basic abstract algebra. More generally for any UFD $R$, the polynomial ring in $n$ number of variables over $R$ is also a UFD. For your second question, suppose I take a quotient of two polynomials $f = g/h$ with $h$ and $g$ coprime. Suppose that there is another representation $g'/h'$ of $f$ with $g'$ and $h'$ coprime. Then it will follow that $h'g =g'h$ which means to say that $h| h'g$. By unique factorization, it follows that every prime factor of $h$ divides $h'$ and in fact $h| h'$ since $h$ divides $h'g$ but not $g$.

Similarly we conclude that $h' | h$ and so upto multiplication by a scalar, $h' = h$. Mutadis mutandis the same argument also shows that $g' = g$ and so the representation of $f$ as a quotient of two polynomials is unique upto multiplication by a scalar. Thus whenever we choose an element $f$ of $k(x,y)$, essentially to ask questions about whether a denominator vanishes, it is enough to work with just any representative of $f$ as a quotient of two polynomials with no common factor.

Thus if $f \in k[x,y]_{\mathfrak{m}_P}$ then $f$ can be written as the quotient of two polynomials $g/h$ with $h(P) \neq 0$ (definition of the localization at $\mathfrak{m}_P$). Conversely if $f = g/h$ with $h(P) \neq 0$, then for any other representative $g'/h'$ of $f$, $h'(P) \neq 0$ too for $h'$ is a non-zero scalar multiple of $h$. Thus $f \in k[x,y]_{\mathfrak{m}_P}$.

3. Once you know that $k[x,y] = \mathcal{O}(X)$ it is clear that $i^\ast$ is an isomorphism. But now their is an equivalence of categories between affine varieties over $k$ and finitely generated integral domains over $k$, and so if $X$ is affine the map $i$ has to be an isomorphism too. But this is impossible because it is not even surjective! Thus $X$ cannot be an affine variety.

• Thanks a LOT! :) That is exactly what I needed. – Luca Jun 19 '13 at 1:20
• @Luca No problems dude! I remember struggling with this example too! – user38268 Jun 19 '13 at 3:10

It can help to see the most elementary form on an argument when we are struggling with an abstract one. The most reduced form of this problem is: If a rational function is defined on the punctured plane, then it is defined at the origin as well.

The pole set of a rational function $r$ on the plane is an algebraic subset of the plane. Define $J_r = \{ G\in k[x,y] \mid \overline{G}r\in k[x,y] \}$ where the overline denotes the embedding of $G$ into the function field $k(x,y).$ Then the pole set of $r$ is precisely $V(J_r).$ We show that a rational function can not have $\{(0,0)\}$ as its pole set. If we had $V(J_r)=V(x,y)$ then by the Nullstellensatz, sufficiently large powers of $x$ and $y$ are in $J_r,$ so by the definition of $J_r$ it is possible to write (with no common factors)

$$r = \frac{p_1}{x^s} = \frac{p_2}{y^t}$$

where $p_1,p_2\in k[x,y]$ and this is an equality in the function field. But this can't be, since $y^tp_1 = p_2 x^s$ implies $y$ divides either $p_2$ or $x.$

Another approach is to write $r = \dfrac{f}{g}$ in lowest terms (unique up to scalar multiplication). If $\deg g\geq 1$ then $V(g)$ is an infinite set, while $V(f)\cap V(g)$ is a finite set since $f$ and $g$ have no common factors. Therefore $r$ must have infinitely many poles. Here I have used some elementary factors about plane curves developed in section 1.6 of Fulton's book on Algebraic Curves.

• In the second argument i think you are not using Hilbert Nullstellensatz am I right? – happymath Jun 23 '15 at 13:20
• @happymath Correct. Just some basic and geometrically intuitive facts about plane curves. Newton would have found this question to be trivial, but the modern treatments of algebraic geometry can make it hard for new students to see the geometry behind the algebra (which, frankly, I think is evidenced by the OP's attempt and the other answer here). – Ragib Zaman Nov 20 '19 at 7:43