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I need to prove the following lemma:

Lemma: Let $f_i\in \mathbb{C}[x_1,\dots,x_m]$ s.t. $\gcd(f_1,f_2,\dots,f_n)=1\quad(1<n\le m)$. Prove that the variety $V=\mathbb{A}_\mathbb{C}^m\setminus\{f_1=0,f_2=0,\dots,f_n=0\}$ cannot be affine.

I don't know from where to start. Can someone supply a complete proof (or counterexample) for this lemma please?

Wenchao Shang
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    Are you sure you have the hypotheses correct? If $\text{gcd} (f_{1}, \ldots, f_{n}) = 1$, then $1$ is contained in $(f_{1}, \ldots, f_{n})$ and then the locus $f_{1} = \cdots = f_{n} = 0$ is empty. – Siddharth Venkatesh Aug 14 '15 at 10:58
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    Also, in general, this looks like the standard question of showing that an affine variety minus a codimension $\ge 2$ closed subset cannot be affine. This follows from the algebraic Hartog's lemma (rational functions defined away from a codimension $\ge 2$ set extend to regular functions). – Siddharth Venkatesh Aug 14 '15 at 10:59
  • @SiddharthVenkatesh . The locus is ${f_i=0}$ which is not empty. I don't know Hartog's lemma. Can one find a direct proof which doesn't depend on lemmas like that? – Wenchao Shang Aug 14 '15 at 11:04
  • @WenchaoShang Ah I see, it's the complement of the union of $f_{i} = 0$. One possible way of doing a direct proof would be to sidestep Hartog's lemma and directly show that global regular functions on such a variety are the same as polynomials $k[x_{1}, \ldots, x_{m}]$ and then use the fact that if this variety were then affine, the inclusion of the variety into $\mathbb{A}^{m}$ would need to be an isomorphism. – Siddharth Venkatesh Aug 14 '15 at 11:09
  • @SiddharthVenkatesh If it's the complement of the union of $f_i = 0$ this seems false. Can't you just invert the functions $f_i$ then? – user148177 Aug 14 '15 at 11:13
  • @user148177 I mixed up my complement and union. It's the union of the complements or the complements of the intersections. My bad sorry. – Siddharth Venkatesh Aug 14 '15 at 11:14
  • Ah yeah, I was also confused – user148177 Aug 14 '15 at 11:16
  • @SiddharthVenkatesh I don't fully understand... Can you please write a formal answer using the isomorphism you mentioned above please? – Wenchao Shang Aug 14 '15 at 11:16
  • Well, I'm still not super convinced. If the gcd is 1, then the intersection of the zero sets should be empty... I still think there is a problem with the above statement. Maybe he means something like, $f_1, \ldots, f_r$ is a regular sequence? – user148177 Aug 14 '15 at 11:18
  • @WenchaoShang I don't understnad your notation. Are you deleting the union of those hyperplanes, or the intersection? If you are deleting the union, then you get something affine. If you are deleting the intersection, well those don't have gcd 1, and if they did, you'd be deleting nothing. – user148177 Aug 14 '15 at 11:20
  • I mean the intersection. For example $V=\mathbb{A}_\mathbb{C}^3\setminus\begin{cases}x=0\y=0\z=0\end{cases}$. The gcd is 1 and I remove some points. – Wenchao Shang Aug 14 '15 at 11:22
  • My bad, $k[x,y]$ is not a Euclidean domain – user148177 Aug 14 '15 at 11:27
  • So what's their gcd? Is the lemma correct suppose ${f_i}$ are regular sequence? – Wenchao Shang Aug 14 '15 at 11:28
  • Here's a counterexample then. Take $k[x, y]$ and remove the intersection of the zero sets of $x$ and $x-1$. They have gcd 1. However, the have empty intersection. – user148177 Aug 14 '15 at 11:28
  • I think gcd is the wrong thing here. Should be something like regular sequence. – user148177 Aug 14 '15 at 11:29
  • And if we limit ourselves to $\gcd=1$ but with non-trivial intersection? Then it makes sense? – Wenchao Shang Aug 14 '15 at 11:30
  • Yes, I think that's true. Lemme see if there's an elementary proof. – user148177 Aug 14 '15 at 11:33
  • No. This is fine as long as your intersections are nonempty. Your polynomial algebra is still a UFD. The gcd condition is good enough. Regular sequences also work but there you need to use Hartog's lemma and a codimension criterion. There might be an easier way to prove that but I can't think of one off the top of my head. – Siddharth Venkatesh Aug 14 '15 at 11:38
  • Have you seen http://math.stackexchange.com/q/1384228/127490 – Hoot Aug 14 '15 at 16:09

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Sorry for being so confusing in the comments. Guess that's what happens when you try to answer a question so late at night.

I'll first answer using Hartog's lemma.

Hartog's Lemma: Let $X$ be a normal algebraic variety. Let $U$ be an open subset of $X$ such that $Z$, the complement of $U$, has codimension $\ge 2$ in $X$. Then, any function that is regular on $U$, extends to a regular function on $X$.

Corollary: If $X$ is a normal affine algebraic variety, and $U$ is an open subset whose complement has codimension $\ge 2$, then $U$ is not affine.

Proof of Corollary: By Hartog's lemma, $\mathbb{C}[U] \cong \mathbb{C}[X]$. The inclusion of $U$ into $X$ corresponds to the restriction map $\mathbb{C}[X] \rightarrow \mathbb{C}[U]$. This map is always an injection as opens are dense. But in this case, this map is also a surjection. Hence, the map at the level of rings is an isomorpism. If $U$ were affine, then the inclusion of $U$ into $X$ would have to be an isomorphism. This is clearly false. Hence, $U$ is not affine.

Here's a way to sidestep this issue of Hartog's lemma using gcd's in this case (I still wanted to state it for your benefit since it's an important fact that you should know). This method will basically work as long as $X$ is the spectrum of a UFD. The really crucial thing in the corollary is that any regular function on $U$ extends to a regular function on $X$. So, let $p$ be a regular function on $U$. Let $U_{i}$ be $\mathbb{C}[X] \backslash \{f_{i} = 0\}$, so that

$$U = \bigcup_{i} U_{i}.$$

Restrict $p$ to $U_{1}$ to get a regular function on $U_{1}$, which must be of the form $p_{1}/f_{1}^{a}$, with $p_{1}, f_{1}$ relatively prime and $a \in \mathbb{Z}$. Do the same for $U_{2}$ to get an expression $p_{2}/f_{2}^{b}.$ Then,

$$\frac{p_{1}}{f_{1}^{a}} = \frac{p_{2}}{f_{2}^{b}}$$

(as long as $U_{1} \cap U_{2}$ is nonempty) and hence

$$p_{1}f_{2}^{b} = p_{2}f_{1}^{a}.$$

This implies that $a$ must be nonpositive, because neither $p_{1}$ not $f_{2}$ share any prime factors in common with $f_{1}$ (the gcd condition). Hence, we can write

$$p = p_{1}f_{1}^{-a}$$

and as $-a$ is nonnegative, this is clearly a function that is regular on all of $X$. Hence, we have shown that any function on $U$ extends to one on $X$ and the corollary still holds.

Siddharth Venkatesh
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  • Thanks. How the extension of a function on $U$ to $X$ proves the lemma? – Wenchao Shang Aug 14 '15 at 11:39
  • The corollary only needs that the map from functions on $X$ to functions on $U$ is surjective (it is always injective for an irreducible variety (I assumed that variety was irreducible). It does not need the codimension criterion. That's only used to apply Hartog's lemma. The extension of functions property implies that the map on functions is an isomorphism. But maps between affine varieties are determined by the maps on their functions. So isomorphism on functions would imply isomorphism of varieties but that can only happen if $U = X$. – Siddharth Venkatesh Aug 14 '15 at 11:44
  • A small comment: your answer currently only handles the case $n = 2$. If $n > 2$ then $f_1$ and $f_2$ may not have g.c.d. $1$; rather, all you know is that their g.c.d. is coprime to the g.c.d. of $f_3, \dots , f_n$. Of course, to handle the case of general $n$ just involves a minor change in your argument. (One way is to consider a particular prime dividing $f_1$. If $q$ is such a prime, then there is at least one $f_i$ that $q$ doesn't divide, and so looking at the equation $p_1 f_i^{a_i} = p_i f_1^{a_1}$, we see that if $a_1 > 0$ then $q$ divides $p_1$, contradicting that ... – tracing Aug 14 '15 at 12:56
  • ... $p_1$ and $f_1$ are coprime. Thus we see that $a_1$ is non-positive. Similarly, all $a_i$ are non-positive, as you wanted.) – tracing Aug 14 '15 at 12:57
  • Oops. You're right. I didn't really elaborate on the $n > 2$-case since it was essentially the same argument. Thanks for adding the comment to fix that. – Siddharth Venkatesh Aug 14 '15 at 13:03
  • No worries; actually, the argument in my comment is also imperfect :) . (We can't assume that $p_1$ and $f_1$ are coprime, just that that $p_1$ is not divisible by any positive power of $f_1$. But treating each prime power factor $q^i$ of $f_1$ separately, we see that for each of them, $q^{a_1}$ divides $p_1$, and hence $f_1^a$ divides $p_1$. This then gives that $a_1 =0$.) – tracing Aug 14 '15 at 13:18