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This is question $5$ from Shafarevich's book page $66$. Let $X=\mathbb{P}^{2} \setminus x$ where $x$ is a point. Prove that $X$ is not isomorphic to affine nor a projective variety. How to prove this?

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hmm...global sections are regular functions on $\mathbb{A}^3$ minus a line which are $k^*$ invariant, these are probably just constants. So that'll tell you it's not affine. –  uncookedfalcon Apr 22 '12 at 23:27
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Questions worded like this bother me. Being an affine or projective variety isn't a property of a set, and it isn't obviously a property of a subset of a projective variety until you have some reasonable theory of non-projective varieties already in place. It's extra structure that you define and the problem as worded doesn't make it clear what compatibility it wants between that extra structure and the existing structure of the problem. A more careful statement would be "show that there does not exist an affine or projective variety $X$ together with a morphism $X \to \mathbb{P}^2$ which... –  Qiaochu Yuan Apr 22 '12 at 23:33
    
...is injective and misses exactly one point." –  Qiaochu Yuan Apr 22 '12 at 23:33
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oh yeah and for projectivity, if it were isomorphic to a projective $X$, consider $X \rightarrow \mathbb{P}^2 - pt \rightarrow \mathbb{P}^2$, the image of any such map has to be closed. –  uncookedfalcon Apr 22 '12 at 23:45
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er yeah, as per Qiaochu's comment, I'm assuming the "reasonable theory of non-projective varieties", namely you're taking $O_{\mathbb{P}_2 - pt}$ to just be the restriction of $O_{\mathbb{P}^2}$ hey (where $O$'s the sheaf of reg functions)? –  uncookedfalcon Apr 22 '12 at 23:50
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4 Answers 4

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Okay yeah yeah I think the stuff in the comments works. Here's how to see the only regular functions on $\mathbb{P}^2 - pt$ are $k$, where $k$ is your algebraically closed ground field (your argument for why that would imply $\mathbb{P}^2 - pt$ is just a single point is legit, I think).

(Global) regular functions on $\mathbb{P}^2 - pt$, by definition, are regular functions on $\mathbb{A}^3 - line$ which are $k^*$ invariant.

Let $f$ be such a function, introduce coordinates $x,y,z$, where the line is $\{x = 0\} \cap \{y = 0\}$ (the $z$ axis). On $\{x \neq 0 \}$, $f$ must be of the form $p(x,y,z,x^{-1})$, and on $\{y \neq 0\}$, $f$ must be of the form $p(x,y,z,y^{-1})$, where $p$ just means ``a polynomial in".

Suppose for a contradiction $f = \sum_k p_k(y,z)x^k$ on $\{x \neq 0\}$ had terms with $x$ to a negative power.

Now on $\{ y \neq 0\}$, we have an equality of functions $$x^{-k}p_{-k}(y,z) + \ldots = p_0(y, y^{-1},z) + p_1(y,y^{-1},z)x + \ldots$$Now multiply by through by $x^{k}$, now we have something of the form $$p_{-k}(y,z) + \ldots = p_0(y^{-1}, y,z)x^k + \ldots$$Now just evaluate at any point of the form $(0,y \neq 0,z)$, we get 0 on the RHS every time, which tells us the LHS must be the 0 polynomial in $y,z$, as desired.

So this tells us $\mathbb{P}^2 - pt$ ain't affine, and as to why it ain't projective, as in the comments, in general if we have $X \rightarrow Y$, $X$ proper, and $Y$ separated, then $X \rightarrow Y$ has closed image.

Were $\mathbb{P}^2 - pt$ isomorphic by $i$ to some projective variety $X$, consider $$X \xrightarrow{i} \mathbb{P}^2 - pt \rightarrow \mathbb{P}^2$$so $\mathbb{P}^2 - pt$ is closed, a contradiction.

Well, there's something annoying more to be said. We already know $\mathbb{P}^2 - pt$ is open, why can't it be clopen? Well, $\mathbb{P}^2$ has the quotient topology, so we're asking, is $\mathbb{A}^3$ connected? You can see this in various ways, one would be that if it weren't we could write $\mathbb{A}^3 = V(I) \coprod V(J)$, both nonempty, the $LHS$ has (Krull) dimension 3, the RHS has dim $\leqslant 2$.

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You can use the fact that if a polynomial $F \in k[X_{1}, X_{2}, \ldots, X_{n}], \#k = \infty$, vanishes in the complementary of a finite subset of $k^{n}$, then $F = 0$ (proof by induction on $n$). Therefore, you cannot have $\mathbb{P}^{2} \setminus \{ x \}$ as a set of common zeroes of polynomials.

I think it is true that the only regular functions on $\mathbb{P}^{2} \setminus \{ x \}$ are the constant ones, and this might be proved using the above fact about polynomials.

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I am not sure about the projective variety but you should be able to show that a line without a point is not an affine variety, and similarly an affine space without a point is not an affine variety. The view that a projective space is 'glued' from affine space, and projective variety is "glued" by affine varieties may help but I cannot write a proof explicitly.

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Dear Changwei, The complelement of any positive number of points in $\mathbb P^1$ is an affine variety, and so some of the statement you have written are false. Regards, –  Matt E Apr 23 '12 at 2:43
    
Dear Matt E: I mean an affine line, not in $\mathbb{P}^{1}$. As you note I do not know how to deal with the projective cases. –  Kerry Apr 23 '12 at 2:46
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Dear @Changwei, The complement of a point in $\mathbf{A}^1_k$ is an affine variety. For instance, $\mathbf{A}^1_k-\{0\} \cong \text{Spec } k[x,y]/(xy-1)$. In fact every open subset of $\mathbf{A}^1_k$ is an affine variety. –  Bruno Joyal Apr 23 '12 at 3:07
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Dear @Bruno:I see. Thank you. –  Kerry Apr 23 '12 at 3:12
    
Dear @ChangweiZhou, you are welcome. (I only repeated what Matt said. ;)) –  Bruno Joyal Apr 23 '12 at 3:14
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You can use topological generalities to solve the problem (at least showing non-projectiveness), assuming for instance we are over the complex numbers.

$\mathbb A^2\setminus\{x\}$ is not compact in the classical topology (take the open cover $(\mathbb R\times (-i,i))_i$) so not a projective variety (i.e. not a closed subset of some projective space). Working with the Zariski topology for compactness is not so simple. Rather, using the Zariski topology, it is not a complete variety, since a canonical embedding in $\mathbb P^2$ is open (and not all of $\mathbb P^2$) and therefore not closed since $\mathbb P^2$ is irreducible.

$\mathbb P^2\setminus\{(0,0)\}$ is not compact in the classical topology (take the open cover $(((\infty\cup\mathbb R\setminus\{0\})\times\mathbb R)\cup((\infty\cup\mathbb R)\times (\infty,-1/i)\cup(1/i,\infty))\cup\{\infty\times\infty\})_i$), and using the Zariski topology, it is not complete, taking for instance the canonical embedding into $\mathbb P^2$, which is irreducible. You can do the same with any point (adding some notation) or take an isomorphism to $\mathbb P^2\setminus\{x\}$.

Showing that they are not affine is more difficult, since for instance $\mathbb A^1\setminus\{x\}$ is affine. You must show that you can only remove at most codimension 1 subsets from affine varieties to obtain varieties that are still affine. But there are other arguments, I think that of uncookedfalcon could be ok, but I did not really read it.

EDIT: I tried to correct the example of open cover for $\mathbb P^2\setminus\{(0,0)\}$, that is tricky.

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