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Consider the affine variety $\mathbb{C}^2$ equipped with Zariski topology. By the question above, I mean why $X:=\mathbb{C}^2\setminus\{(0,0)\}$ cannot be written as

$$ X:=U_f:= \{(x,y)\in \mathbb{C}^2:f(x,y)\neq 0\} \mbox{ for some }f\in\mathbb{C}[x,y]. $$ I am trying to do this without computing the ring of regular functions on $X$. My attempt: Suppose $X$ is a basic open set and $X=U_f$ for some $f\in \mathbb{C}[x,y]$. Writing $x,y,z$ for the coordinates of $\mathbb{C}^3$, I have shown that $X$ is isomorphic to an affine algebraic variety $W=Var(zf-1)\subset \mathbb{C}^3.$ Can this help me arrive at a contradiction?

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    You are looking for a polynomial $f$ so that ${(0,0}$ is the only zero. That this is not possible can be checked quite easily I suppose? –  Jul 16 '16 at 10:07
  • @ArcticChar: Well, it is this easy check that the OP is asking for. – Alex M. Jul 16 '16 at 10:20
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    Try setting $y=\text{ constant }$ and applying the fundamental theorem of algebra. Similarly for $x=\text{ constant }$. – Nefertiti Jul 16 '16 at 10:25

3 Answers3

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A complex bivariate polynomial is either constant or has infinitely many zeroes. This is a very easy exercise. In particular any proper basic open subset has an infinite complement.

MooS
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    To me, you are essentially repeating the question without answering it. (Not my downvote, but I can understand it.) – Alex M. Jul 16 '16 at 10:15
  • @AlexM. To me, that's a rephrasing (not repeating). –  Jul 16 '16 at 10:25
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    Often, a general fact (the one I stated) is easier to prove then a special case (the one the OP wants to prove). So I think my answer is indeed useful, since I gave him this fact, which he was not aware of. And the fact is so easy-to-prove, that I think its suitable to leave the proof to the OP. If you do not agree, feel free to post a detailed proof as an answer. Then I would delete mine.. – MooS Jul 16 '16 at 10:55
  • Full details here: https://math.stackexchange.com/questions/286352/polynomial-with-infinitely-many-zeros – hwong557 Jul 16 '16 at 14:32
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Another solution is the following: as the OP affirms, if $X=\mathbb{C}^2\setminus\{(0,0)\}$ is a basic open then $X$ is (up to regular isomorphism) an affine variety.

The OP excludes the computation of $H^0(X,\mathcal{O}_X)$ (the ring of regular functions on $X$), and then I shall use the Serre's Theorem about the Cohomology of the Affine Varieties (see [P], theorem VII.2.5).

Let $V$ an affine variety, then for any quasi-coherent $\mathcal{O}_V$-module $\mathcal{F}$, one has $\forall k>0,\check{H}^k(\mathcal{U},\mathcal{F})=0$; where $\mathcal{U}$ is an open covering of $V$ by standard open sets.

Let $\mathcal{U}=\{D(x),D(y)\subset\mathbb{C}^2\}$ an affine open covering of $X$, where $D(x)=\{(x,y)\in\mathbb{C}^2\mid x\neq0\},\mathcal{O}_X(D(x))=\mathbb{C}[x,y]_x$ and similarly; considering the Čech cocomplex: \begin{equation} 0\to C^0\left(\mathcal{U},\mathcal{O}_X\right)\stackrel{d_0}{\longrightarrow}C^1\left(\mathcal{U},\mathcal{O}_X\right)\to0 \end{equation} where:

  1. $C^0\left(\mathcal{U},\mathcal{O}_X\right)=\mathcal{O}_X(D(x))\times\mathcal{O}_X(D(y)),\,C^1\left(\mathcal{U},\mathcal{O}_X\right)=\mathcal{O}_X(D(xy))$,

  2. $d^0(f,g)=f_{\displaystyle|D(xy)}-g_{\displaystyle|D(xy)}$.

By definition \begin{equation*} \check{H}^1\left(\mathcal{U},\mathcal{O}_X\right)=\ker d^1_{\displaystyle/\operatorname{Im}d^0}=\mathbb{C}[x,y]_{xy\displaystyle/\operatorname{Im}d^0}, \end{equation*} where $\mathcal{O}_X(D(xy))=\mathbb{C}[x,y]_{xy}$; by construction: \begin{gather*} C^0\left(\mathcal{U},\mathcal{O}_X\right)\ni\left(\frac{f}{y^{\alpha}},\frac{g}{x^{\beta}}\right)\mapsto\frac{xf}{xy^{\alpha}}-\frac{yg}{x^{\beta}y}\in C^1\left(\mathcal{U},\mathcal{O}_X\right); \end{gather*} in other words: \begin{equation*} a,b\in\mathbb{C}[x,y]_{xy},\,[a]=[b]\in\check{H}^1\left(\mathcal{U},\mathcal{O}_X\right)\iff a-b=\sum_{\underline{i}\in(\mathbb{N}_0)^2\setminus\{\underline{0}^2\}}\frac{r_{ij}}{x^iy^j} \end{equation*} where $r_{ij}\in\mathbb{C}$, then: \begin{equation*} \check{H}^1\left(\mathcal{U},\mathcal{O}_X\right)=\bigoplus_{(i,j)\in(\mathbb{N}_0)^2\setminus\{\underline{0}^2\}}x^{-i}y^{-j}\mathbb{C}. \end{equation*}

By Serre's theorem: $X$ can not be an affine variety.

§§§

Bibliography

[P] Perrin D. - Algebraic Geometry. An Introduction. (2008) Springer Verlag.

Armando j18eos
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  • While your technical skill is truly impressive and your mastery of algebraic geometry is obvious, I feel that your approach is overkill for such a simple problem. I believe that most readers would be able to read not even the first line of what you have written, which from a paedagogical point of view is a catastrophy: this is the kind of problem that a beginner student in algebraic geometry will encounter in the first lecture, long before getting to see cohomology or sheaves. In fact, with a professor like you, they will surely quit after the first lecture :) (no offense intended). – Alex M. Jul 16 '16 at 19:07
  • @AlexM. I apologize for my delay! I'm in accord with your opinion; indeed my aim was to post a non elementary proof, after an elementary proof; and as one can read, I wrote:"Another solution is the following..." Pedagogically writing, in my humble opinion: in algebraic geometry there are two type of solutions for any problem: the easy type and the very hard type; as this problem shows. That is it! P.S.: I'm not a professor, dear students don't worry. ;) – Armando j18eos Jul 20 '16 at 12:03
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Let us prove a more general statement: if $K$ is an algebraically closed field, then there is no non-constant $f \in K[X,Y]$ such that $f(x,y) \ne 0 \ \forall (x,y) \ne (0,0) \in K \times K$.

Assume that such an $f = \sum \limits _{i,j = 0} ^{m,n} a_{ij} X^i Y^j$ exists. Group the monomials according to the powers of $Y$ and rewrite $f = \sum \limits _{j=0} ^n p_j (X) Y^j$, with $p_j \in K[X]$.

Let $x \in K \setminus \{0\}$ and consider the polynomial $f(x,Y)$. If $n \ge 1$ and since $K$ is algebraically closed, there exist $y \in K$ such that $f(x, Y)$ has the root $y$. Equivalently, $f(x,y) = 0$. But we had assumed that $f(x,y) \ne 0 \ \forall (x,y) \ne (0,0) \in K \times K$, so we have reached a contradiction, therefore $n = 0$ and $f$ is constant in $Y$, i.e. is just a polynomial in $X$. Write it as $f(X,Y) = \sum \limits _{i = 0} ^m b_i X^i$.

Assume $m \ge 1$.

Since $K$ is closed, $\sum \limits _{i = 0} ^m b_i X^i$ must have at least a root $x \in K$. If $x \ne 0$, this means that $f(x,y) = \sum \limits _{i = 0} ^m b_i x^i = 0 \ \forall y \in K$, which again is impossible for the same reason as above. This means that all the roots of $\sum \limits _{i = 0} ^m b_i X^i$ are $0$ (i.e. $0$ is a root of multiplicity $m$), so $f(X,Y) = b_m X^m$. In turn, this implies that $f(0,y) = 0 \ \forall y \in K$, which again is impossible for the same reason already discussed.

The only possibility, then, is $m = 0$, so that $f$ is a constant polynomial. If $f = 0 \in K$, this again contradicts the fundamental assumption on $f$. It follows then that $f \in K \setminus \{0\}$, i.e. $f$ is a non-zero constant polynomial.

But in this case, $U_f = K \times K$ (the affine space of dimension $2$ over $K$), not $K \times K \setminus \{(0,0)\}$, not what you wanted.

Therefore, $K \times K \setminus \{(0,0)\}$ is not a basic open set.

Alex M.
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