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For which $n$ is $\mathbb{A}^n(k)\setminus \{0\}$ an affine variety? I think for $n=0$ and $n=1$ it is.

  • For $n=0$, take the ideal $\mathfrak{a}:=(1)$ in $k[T]$. Then $V(\mathfrak{a})=\emptyset$ should be isomorphic to $\mathbb{A}^n(k)\setminus \{0\}$.

  • For $n=1$, take the ideal $\mathfrak{a}:=(T_1T_2-1)$ in $k[T_1,T_2]$. Then $V(\mathfrak{a})$ should be isomorphic to $\mathbb{A}^n(k)\setminus \{0\}$. The isomorphism is given by $f:V(\mathfrak{a})\to \mathbb{A}^n(k)\setminus \{0\}$, $(x,y)\mapsto x$.

  • For $n>1$ probably not, but I don't have a proof.

So my questions are:

  1. Is it correct what I did for the cases $n\in\{0,1\}$?

  2. How do I prove that $f$ in case $n=1$ is an isomorphism of spaces of functions?

  3. What about the case for $n>1$?

Edit: As already noted in my comment: $\mathbb{A}^0(k)\setminus \{0\}$ is not affine because the empty set is not irreducible.

Mike Pierce
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principal-ideal-domain
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  • To be affine variety means to be isomorphic to a space with functions associated to an irreducible affine algebraic set. It does not depend on the dimension of the space in which the irreducible affine algebraic set lives. But now I realize that the empty set is not irreducible and therefore $\mathbb{A}^0(k)\setminus {0}$ can't be affine. – principal-ideal-domain Jul 28 '14 at 21:18
  • Here are three proofs of non affineness. – Georges Elencwajg Aug 01 '14 at 07:46

3 Answers3

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For $n=0$ you get the empty set.

For $n=1$, we have that $\mathbb{A}^1-0$ is isomorphic to the affine variety $V(xy-1)\subset\mathbb{A}^2$ via the map: $$\begin{array}{lrcl}\phi:&\mathbb{A}^1-0&\longrightarrow&V(xy-1)\\&z&\longmapsto&\left(z,\frac{1}{z}\right)\end{array}$$

Finally, for any $n>1$ we have that $\mathbb{A}^n-0$ is not affine. Indeed in this case we have that the ring of regular functions $\Gamma(\mathbb{A}^n-0)\cong\mathbb{C}[x_1,\ldots,x_n]$, and thus if it were affine it should be isomorphic to $\mathbb{A}^n$. However this is not possible.

Daniel Robert-Nicoud
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    Why is it not possible? There are several ways of showing this, but none which are immediately obvious. To make your argument more correct, I think you should say that the inclusion $\mathbb{A}^n-{0}\to\mathbb{A}^n$ induces an isomorphism on global sections which, if $\mathbb{A}^n-{0}$ were affine (since $\mathbb{A}^n$ is) would imply that the inclusion is an isomorphism, which it is not. The only ways I can think of showing that $\mathbb{A}^n$ and $\mathbb{A}^n-{0}$ aren't isomorphic are cohomological (de Rham or sheaf,say). – Alex Youcis Jul 29 '14 at 02:55
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    @AlexYoucis Actually you can also directly deduce that $\mathbb{A}^n-0$ is not affine by noticing that if it were, then the maximal ideals of its ring of functions would be in bijection with the points of the variety (canonically, taking the common zero locus), but the ideal $(x_1,\ldots,x_n)$ gives a contradiction. – Daniel Robert-Nicoud Jul 29 '14 at 15:19
  • I think it's a bit trickier than you say. While this is an intuitive idea, how do you prove it? You are implicitly assuming that the isomorphism between sections of punctured n-space and the polynomial ring is by restriction from sections of n-space--right? Otherwise how are you rigorously justifying that '(x_1,...,x_n) gives the empty set'. But, if you assume this, then you are done already, by the anti-equivalence. Showing that it's the restriction which is an iso is the hard part. – Alex Youcis Jul 29 '14 at 18:35
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    @AlexYoucis Yeah, you are right. +1 to your answer, and a recommendation to everyone reading this to look at Alex's answer for a formal treatment of the ideas exposed intuitively in my own. – Daniel Robert-Nicoud Jul 29 '14 at 20:34
  • That $\Gamma(\mathbb{A}^n-0)=\mathbb{C}[x_1,\ldots,x_n]$ is a trivial and down-to-earth calculation based on the fact that the ring of regular functions on the affine variety $D(x_i)\subset \mathbb A^n_\mathbb C$ (defined by $x_i\neq0$) is the polynomial ring on the variables $x_j/x_i; (j\neq i)$. – Georges Elencwajg Aug 01 '14 at 07:54
  • @AlexYoucis: taking $V$ gives a bijection between radicals of the coordinate ring and subvarieties of $\mathbf{C}^2-{0}$, now $V(x,y)={0}$ is not contained in $V$. Isn't this found in any book? – A B C Aug 10 '14 at 20:46
  • @Polynomialring That's true of any AFFINE variety. – Alex Youcis Aug 18 '14 at 01:01
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I would like to add a different way of phrasing Daniel Robert-Nicoud's argument.

Suppose that $\mathbb{A}^n-\{0\}$ were affine. Consider the inclusion $\mathbb{A}^n-\{0\}\hookrightarrow\mathbb{A}^n$. The induced map on global sections is just the restriction map $\mathcal{O}_{\mathbb{A}^n}(\mathbb{A}^n)\to\mathcal{O}_{\mathbb{A}^n}(\mathbb{A}^n-\{0\})$--this is obviously injective. But it is surjective, since the image, when both are thought of as embedded in $K(\mathbb{A}^n)$, is just the copy of $k[x_1,\ldots,x_n]$ sitting inside the elements of $K(\mathbb{A}^n)$ of the form

$$\displaystyle \bigcap_{\mathfrak{p}\in\mathbb{A}^n-\{0\}}k[x_1,\ldots,x_n]_\mathfrak{p}$$

But, this in particular, is an intersection over all codimension $1$ primes, and since $k[x_1,\ldots,x_n]$ is Noetherian and normal, this is just $k[x_1,\ldots,x_n]$ (this is sometime's called 'Algebraic Hartog's Lemma'). Thus, the restriction map is also surjective.

So, the induced map $\mathcal{O}_{\mathbb{A}^n}(\mathbb{A}^n)\to\mathcal{O}_{\mathbb{A}^n}(\mathbb{A}^n-\{0\})$ is an isomorphism, which if $\mathbb{A}^n-\{0\}$ were affine would imply (since both are affine and the anti-equivalence of categories between affine schemes and rings) that the original map, the inclusion $\mathbb{A}^n-\{0\}\hookrightarrow\mathbb{A}^n$ is an isomorphism--but this is nonesense.

Honestly, even though that looks really technical, that is the only proof (that I know of) that is simultaneously clean, and rigorous. It's kind of an annoying problem to do 'precisely' when you don't have a lot of machinery. The quickest proofs, as other answerers have said, is cohomological.

Alex Youcis
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  • Can you explain the surjectivity of $\mathcal{O}{\mathbb{A}^n}(\mathbb{A}^n)\to\mathcal{O}{\mathbb{A}^n}(\mathbb{A}^n-{0})$ again with different words? Unfortunately, I don't know Algebraic Hartog's Lemma (and what codimension 1 primes are). I also don't see where you used $n>1$. – principal-ideal-domain Jul 29 '14 at 08:36
  • @principal-ideal-domain Do you know much commutative algebra? This is the statement that for a normal domain, the intersection of the localization at all height one (height is another word for codimension) primes is just the original ring. (here's a screen cap from Matsumura: http://puu.sh/avI9w/18a6933455.jpg). I used $n>1$ to say that all codimension $1$ primes are in $\mathbb{A}^n-{0}$. For $n=1$, the prime $(x,y)$ IS codimension $1$, so this argument doesn't work. – Alex Youcis Jul 29 '14 at 08:43
  • @principal-ideal-domain This theorem, about the intersection of codimension 1 primes, essentially is the statement that 'a function with no poles on a big enough set can't have any poles at all'. The technical translation of 'big enough' is the complement of a codimension $\geqslant 2$ point. For $n\geqslant 2$, the origin has codimension $n$, and so in particular, codimension bigger than $2$. Since any function on $\mathbb{A}^n-{0}$ is essentially a function on $\mathbb{A}^n$ with a possible pole at the origin, we see that it has no pole on a 'big enough set' as defined above, and so – Alex Youcis Jul 29 '14 at 08:44
  • can't have a pole at all--i.e. is an element of $k[x_1,\ldots,x_n]$. The intuition for alg. Hartog's is relatively simple, and I can explain it if you like.

    Unfortunately, you kind of need all of this fancy language to make Daniel's proof rigorous, but it's essentially summarized as follows: "the inclusion map gives rise to the restriction map on functions. Since functions can't have only a pole at the origin (their pole sets are always at least codim 1), the restriction map is a surjection, but it's also an injection, so an iso. If punctured $n$-space were affine, this iso on coordinate

     – Alex Youcis Jul 29 '14 at 08:50
  • rings would imply the inclusion is an iso, which it isn't--contradiction." – Alex Youcis Jul 29 '14 at 08:50
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For $n\geq 2$, $H^1(\mathbb{A}^n\backslash \{0\}, \mathcal{O})$ is infinite dimensional (so this must vanish ). By Serre's criterion for affineness, this is not affine.

Hamou
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