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This is an exercise 2.3 from Harris's AG: A First Course:

Using the result of the exercise 2.2 ($A(\mathbb{A}^2 \setminus \{0\}) = K[x_1, x_2]$) show that $X = \mathbb{A}^n \setminus \{0\}$, $n > 1$, [as a quasi-affine variety] is not isomorphic [as in biregular] to any affine variety.

I have serious issues with this exercise.

I'm going to use the fact that regular mappings are Zariski continuous. I am also assuming that the underlying field $K$ is infinite.

Consider a regular isomorphism $f: X \to Y$ where $Y$ is an affine subvariety of $\mathbb{A}^m$ cut out by polynomials $h_i$, $i \in I$. By (a slight generalization of) ex. 2.2 we know that the components of $f$ are exactly restrictions of polynomials to $X$, we'll denote the extension of $f$ to the whole $\mathbb{A}^n$ as $\hat{f}$. But then $h_i \circ \hat{f} \vert_{\mathbb{A}^1}$ has infinite number of zeros, which means that the image of $0 \in \mathbb{A}^n$ lies in $Y$ as well.

Consider now a line $L \subset \mathbb{A}^n$ passing through both preimages of $\hat{f}(0)$. It is now sufficient to prove that $\hat{f} \vert_{X \cap L}$ is not an isomorphism onto its image.

I don't really know how to proceed. It is clear that $X \cap L$ is isomorphic to a hyperbola $xy = 1$, and thus we can reduce the problem to the one about affine varieties. However, I don't know how to prove that they are not isomorphic without using Euclidean topology (and thus assuming that $K = \mathbb{R}$ or $K = \mathbb{C}$. I guess I could look ahead and use the definition of a singularity, but it's so far ahead in the textbook it would be cheating. I can't think of a nice Zariski closed set that $\hat{f}$ would not pull back onto a Zariski closed set either, and I'm pretty sure one doesn't exist. So I'm left with having to study coordinate rings of the hyperbola and its image, and I don't know how to reason about the latter.

Am I on the wrong track with this exercise? I suspect that I am, yet I don't know any better way to go about it. Is there a hint I could use?

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@GeorgesElencwajg I doubt this something I'll be able to really understand any time soon, if ever :) –  Alexei Averchenko Oct 21 '12 at 6:11
    
Dear Alexei: I, on the contrary, am quite sure you will understand this in less time than you think! Good luck in your study. –  Georges Elencwajg Oct 21 '12 at 8:40
    
@GeorgesElencwajg Thank you for your kind words, I'll definitely try my best! –  Alexei Averchenko Oct 21 '12 at 10:41

1 Answer 1

up vote 2 down vote accepted

Your argument begins by showing that an isomorphism $f:X \cong Y$ with $Y$ affine has to extend to a morphism $\hat{f}:\mathbb A^n \to Y.$

Next, your idea is to choose $x \in X$ such that $\hat{f}(x) = \hat{f}(0)$, and take the line $L$ through $x$ and $0$.

This seems like a good strategy. Perhaps you could use the following to go further: since $f$ is an isomorphism and $L \setminus \{0\}$ is closed in $X$, its image $f(L\setminus\{0\})$ is closed in $Y$. But general topology shows that since $L$ is the closure of $L\setminus \{0\}$, there is an inclusion $\hat{f}(L) \subset$ closure of $f(L\setminus\{0\})$. Thus in fact $\hat{f}(L) = f(L\setminus\{0\}).$

Now $f(L\setminus \{0\})$ is just an isomorphic copy of $L\setminus \{0\}$, so you are reduced to the following question: is there a morphism from $L$ to $L\setminus \{0\}$?

Further remarks: If we compose with $f^{-1}$, then we can eliminate some notation, and then what you are trying to show is that the identity morphism $X \to X$ does not extend to a morphism $\mathbb A^n \to X$. To prove this, you are restricting to a line, i.e. a copy of $\mathbb A^1$, passing through $0$, and showing that the identity $\mathbb A^1 \setminus \{0\} \to \mathbb A^1\setminus \{0\}$ doesn't extend to a morphism $\mathbb A^1 \to \mathbb A^1\setminus \{0\}.$ (So actually you don't need to choose $L$ containing $x$ in the preiamge of $\hat{f}(0)$; as long as it passes through $0$ it should lead to a contradiction.)

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Can I use coordinate rings now? :) I know that $A(\mathbb{A}^1 \setminus \{0\}) \cong \left[\frac{1}{x}\right]K[x] = K(x)$ (by regularly mapping it to the hyperbola), so the question is equivalent to "is there a $K$-algebra morphism $f: K(x) \to K[x]$ that $i \circ f = \mathrm{id}$, where $i: K[x] \to K(x)$ is the inclusion", right? And the answer is no, because the kernel of $f$ cannot be trivial ($K[x] \subset K(x)$ maps to itself, so $f$ cannot be 1-to-1), but $\left[\frac{1}{x}\right]K[x]$ is a field. Also, is this argument cheating? :) –  Alexei Averchenko Oct 18 '12 at 3:33
    
@AlexeiAverchenko: Dear Alexei, Indeed, using coordinate rings at this point is the way to go. (Just as a side remark, $K[x][1/x]$ is not a field; but this doesn't affect the argument.) Regards, –  Matt E Oct 18 '12 at 4:32
    
Isn't $K[x][1/x] \cong K(x)$? I must be missing something. Thanks a lot, anyway :) –  Alexei Averchenko Oct 18 '12 at 6:32
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@AlexeiAverchenko: Dear Alexei, No, in the same way that $\mathbb Z[1/2] \neq \mathbb Q$. One way to think about this is to note that $\mathbb A^1\setminus \{0\}$ has plenty of points, so correspondingly $K[x][1/x]$ must have plenty of maximal ideals. Regards, –  Matt E Oct 18 '12 at 11:57

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