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While learning commutative algebra and basic algebraic geometry and trying to understand the structure of results (i.e. what should be proven first and what next) I came to the following question:

Is it possible to prove that $\mathbb A^2-point$ is not an affine variety, if you don't know that the polynomial ring is a unique factorisation domain?

It seems to me, that this question has some meaning, since when we define affine variety, we don't need to use the fact that the polynomial ring is an UFD. Don't we?

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If $X:=\mathbb{A}^2−0\subset \mathbb{A}^2$ is an affine variety then assume that $X=V(p_1,\ldots,p_k)$ for some polynomials $p_i$'s. The zero locus of any polynomial $p(x,y)$ is of codimension $1$ and since $X\subset V(p_i)$ this implies that $p_i(x,y)\equiv 0$, a contradiction! –  Somnath Basu Mar 5 '11 at 22:53
    
Somnath, thanks for your comment. The only thing, is that you use the word "codimension". And it seems to me, that in order to define codimension you need to prove some amount of theorems. Will it be easier than to prove that polynomials are UFD? –  aglearner Mar 5 '11 at 23:19
    
As an interesting sidenote, $\mathbb{A}^2-0$, $is$ an affine variety over a field where $-1$ is not a square.. –  J.C. Ottem Mar 5 '11 at 23:44
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@J.C. Ottem: $\mathbb{A}^n \setminus 0$ is always affine over a non-algebraic closed field. –  Guillermo Mantilla Mar 5 '11 at 23:59
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$A^2-0$ is not an affine algebraic variety over any field, by the same argument that works over an algebraically closed field; or more simply because an affine algebraic variety stays affine when you extend the base field. Ottem only constructs an affine variety V and a bijection on k-rational points --- the bijection is not an isomorphism of algebraic varieties. –  mephisto Mar 6 '11 at 5:55
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5 Answers

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Yes, you can do it over any field.

First, it is enough to show $\mathcal O(Y) = k[x,y]$ ($Y=A^2-0$). If that is true and $Y$ is affine, then the embedding $Y \to A^2$ must correspond to some $k$-algebra map $k[x,y] \to k[x,y]$, which is absurd.

The key point now, as in Guillermo's post, is to show that $R_{(x)} \cap R_{(y)}= R$ ($R=k[x,y]$). It will follow from the

Fact: $(x^m, y^n)$ form a regular sequence on $R$ for all $m,n>0$

Indeed, if $f/x^m =g/y^n$, then $fy^n=0$ modulo $x^m$, so $f=hx^m$ and we are done.

The above Fact is elementary. For example you can induct on $m$. Clearly $m=1$ is OK. Now if $m>1$, use the short exact sequence:

$$0 \to R/{(x^{m-1})} \to R/(x^m) \to R/(x) \to 0$$

and Snake Lemma to conclude that $y^n$ is regular on the middle term as well.

Note that I used $x,y$ abstractly and all you need is that the elements $(x,y)$ form a regular sequence on a commutative Noetherian ring $R$ to start with. Then the proof shows that $\mathcal O(Y) =R$ if $Y=\text{Spec}(R) - V(x,y)$. In fact, more general results are true for ideals of depth at least $2$. If you are interested, it will be a good motivation to learn about depth and regular sequences.

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Thank you for the answer! Could you please expand a bit lines 2-3? How to get absurd from the fact that embedding $Y\to \mathbb A^2$ corresponds to a $k$-algebra map $k[x,y]\to k[x,y]$? –  aglearner Mar 6 '11 at 12:10
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Here's a short argument over $\mathbb{C}$.

If $\mathbb{A}^2-\{0\}$ were affine, then a standard application of Morse theory [Milnor, chap 1, sect 7] would show that $H^i(\mathbb{A}^2-\{0\},\mathbb{Z})=0$ for $i>2$. But it's homotopic to $S^3$.

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This is nice! If only I could assume that my students know what is topology... –  aglearner Mar 6 '11 at 23:03
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I'd say yes. I'll work over $\mathbb{C}$ to make my life easier. First it is enough to show that the ring of regular functions on $Y:=\mathbb{A}^2(\mathbb{C})−0$ is isomorphic $\mathbb{C}[x,y]$. If Y were affine, the identity map from $\mathbb{C}[x,y]$ to itself would induce an isomorphism between $Y$ and $X:=\mathbb{C}^{2}$ which is impossible.(Polynomial maps are continuous in the usual topology, but $X$ is simply connected and $Y$ is not.)

Now let $Y_{1}:=\mathbb{A}^2(\mathbb{C}) \setminus \{x=0\}$ and $Y_{2}:=\mathbb{A}^2(\mathbb{C}) \setminus \{y=0\}$. Then $\mathcal{O}(Y)=\mathcal{O}(Y_1)\cap \mathcal{O}(Y_2)=\mathbb{C}[x,y]_{(x)} \cap \mathbb{C}[x,y]_{(y)}$.

On the other hand $\mathbb{C}[x,y]_{(x)} \cap \mathbb{C}[x,y]_{(y)}=\mathbb{C}[x,y]$. At first I thought that one needs unique factorization for the last equality, but the only thing one needs is that $\displaystyle (x^{i}y^{j})_{i,j \in \mathbb{N}}$ is a linearly independent subset of $\mathbb{C}[x,y]$. So my point is that $\left( x^{m}y^{n}=x^{p}y^{q} \rightarrow (m,n)=(p,q)\right)$ follows if we know the ring is a UFD, but it also follows trivially from linearly independence.

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Guillermo thnaks. Just a nitpick, $\mathbb A^2(\mathbb C)-0$ is simply connected, you need to use $\pi_3(Y)=\mathbb Z$. –  aglearner Mar 6 '11 at 10:11
    
@aglearner: my mind sometimes pictures $\mathbb{C}^2$ as a "real" plane. Good catch. –  Guillermo Mantilla Mar 6 '11 at 10:44
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The abstract reason for this is that $\mathbb A^2$ satisfies Serre's property $(S_2)$ (which follows from being smooth) and hence every regular function is determined in codimension $1$. In other words, the point is that a regular function on $\mathbb A^2\setminus \{0\}$ is a rational function on $\mathbb A^2$ and the locus of indeterminacy of a rational function on $\mathbb A^2$ is of pure codimension $1$ (think of meromorphic functions on $\mathbb C^2$), so if you leave out something of codimension at least $2$, then you cannot get new regular functions. This implies that the natural injection of the ring of regular functions on $\mathbb A^2\setminus \{0\}$ into the ring of regular functions on $\mathbb A^2$ is an isomorphism, but then they can't be both affine as then they would have to be isomorphic. (Note that I did not just say that their rings of regular functions are isomorphic, but that the isomorphism is induced by the embedding).

More generally this argument shows that if $X$ is an affine variety of dimension at least $2$ and $P\in X$ is a closed point such that $\mathrm{depth}_P\mathscr O_X\geq 2$ (this is automatic if for example $X$ is normal, or a complete intersection in something smooth, or Cohen-Macaulay), then $X\setminus\{P\}$ is not affine.

This is an instance of Hartogs' theorem on extending holomorphic functions on normal complex analytic spaces. See this MO question for more.

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We can easily see that the function field of $\mathbb{A}^2_k-(0,0)$ is still $k(x,y)$. So the ring of functions is of the form $f/g$ where $f$ and $g$ are polynomials. But any polynomial in 2 variables will vanish at a codim 1 sub-variety, i.e. cannot vanish at exactly 1 point. This is the Krull dimension theorem. But if you think this is too much, from the fact that $k$ is algebraically closed, you can see that $g$ must vanish at more than 1 point: for each $x$ you can solve for $y$. Thus the ring of functions on $\mathbb{A}^2_k-(0,0)$ is $k[x,y]$. Thus, if it's affine, it must be isomorphic to $\mathbb{A}^2_k$ through the identity map. But it's not. So we are done.

Another way that uses Cohomology is the follows: using $\check{\mathrm{C}}\mathrm{ech}$ cohomology, we can show that $H^1(\mathbb{A}_k^2-\{0\}, \mathcal{O}_X)$ is infinite dimensional. But if our space is in fact affine, then this must vanish, due to Serre's criterion for affineness.

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