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In a previous post, I have defined a generalized principal open set: $$D_{f_1,\dots,f_n}=\mathbb{C}^m\setminus \{f_1 = \dots = f_n =0\}$$ where $f_1,\dots,f_n$ are polynomials in $m$ variables. This is clearly a quasi-protective variety, and the question is when is it affine. (As mentioned below, these are just zariski-open sub-spaces of $\mathbb{C}^n$; The definition was a bit more general in the previous post).

In the previous post, in one of the solutions it was claimed that in non-degenerate cases (like $f_1 = \dots = f_n$) this is not an affine variety; the proof uses a vast theorem, that I do not want to use.

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Here I give the precise proposition, including an easy proof:

if $\gcd(f_1,\dots,f_n)=1$ then $D_{f_1,\dots,f_n}$ is not an affine variety

Proof:

We prove that if $\gcd(f_1,\dots,f_n)=1$ then $D_{f_1,\dots,f_n}$ is not an affine variety. On the contrary, Assume $D_{f_1,\dots,f_n}$ is affine.

Look at the inclusion map $i:D_{f_1,\dots,f_n} \to \mathbb{C}^m$: this induces a map $i_*:\mathbb{C}[\mathbb{C}^m]\to \mathbb{C}[D_{f_1,\dots,f_n}]$, where $\mathbb{C}[\cdot]$ is the coordinate ring.

Let $\frac{p}{q} \in \mathbb{C}[D_{f_1,\dots,f_n}]$; Thus $q$ is nonzero on $D_{f_1,\dots,f_n}$, that is $$\{q=0\} \subseteq \{f_1 = \dots = f_n=0\}$$

This assertion turns out to be not that trivial - since $p,q$ may have common zeros, and thus we must look at all the representation of the form $\frac {p'}{q'}$, with $p',q'$ coprime and such that $p'q=pq'$ on $V$ - but by the Identity theorem it is also true on $\mathbb{C}^n$. Since the polynomial ring is a UFD, we have that each prime factor $q_i$ of $q$ divides pq', and can not divide $p$ so must divide $q'$, and thus $q|q'$. Similarly $q'|q$. so $q=\alpha q'$ with $\alpha\in \mathbb{C}$, which means they have the same zeros.

Thus $$\langle q \rangle \supseteq \langle f_1,\dots,f_n \rangle$$

So each $f_i$ can be divided by $q$, thus $q \mid \gcd(f_1,\dots,f_n)=1$ so $q$ is constant, meaning $\frac{p}{q}$ is a polynomial, so is defined on the whole $\mathbb{C}^m$, and thus

$$\mathbb{C}[D_{f_1,\dots,f_n}] = \mathbb{C}[\lambda_1,\dots,\lambda_m]$$

Thus the map $i_*:\mathbb{C}[\lambda_1,\dots,\lambda_m] \to \mathbb{C}[\lambda_1,\dots,\lambda_m]$ is the identity, and since the functor $f \mapsto f_*$ is a $1-1$ bijection (since $D_{f_1,\dots,f_n}$ is affine), we get that $i:D_{f_1,\dots,f_n} \to \mathbb{C}^m$ should be an isomorphism, but is not onto.

Is this proof true?

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Mike
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  • "Generalised principal open set" is a funny name for this. As it turns out, every Zariski open set is of this form. – Zhen Lin Aug 04 '15 at 15:49
  • @ZhenLin Sure, this is the definition. Is the proof true? – Mike Aug 04 '15 at 15:58
  • @ZhenLin Yes, maybe it's better to just call it an open set :) – Hoot Aug 04 '15 at 17:44
  • Maybe it's okay because you're working in affine space? Everything should live in the function field $k(x_1, \dots, x_n)$ and because you haven't taken a quotient maybe the question of whether a given representative actually "evaluates" to the function is not so difficult. – Hoot Aug 04 '15 at 17:57
  • Slight nitpick: you might have to worry about taking the radical of $(q) $ (just removing multiple factors) right? I don't think it affects the proof much. – Hoot Aug 04 '15 at 18:04
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    One basic issue here. You write $p/q\in \mathbb{C}[V_{f_1,f_2,\ldots}$ and then say $q$ has no zeroes on this open set. While correct (am assuming that $p/q$ is in reduced form) you still need to say a bit more. For example, wherever $q$ has a zero in this open set, $p$ may also have a zero and they may cancel each other (as often happens in such situations). So, you have to say how UFD properties are used. – Mohan Aug 04 '15 at 21:21
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    Your notation clashes with the standard notatation used in in every book $V(f_1, \dots,f_n )={f_1 = \dots = f_n =0}$. You should write $D(f_1, \dots,= f_n)=\mathbb C^n\setminus V(f_1, \dots,f_n )$. – Georges Elencwajg Aug 05 '15 at 09:33

1 Answers1

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Your highlighted "precise proposition" is unfortunately not true.
For example $D(zx,zy)=\mathbb C^3\setminus \{zx=zy=0\}$ is not affine although $\gcd (zx,zy)=z\neq 1$.

Georges Elencwajg
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  • You are right; the direction I have regarded as the easy one turns to be flase. I have edited the post - Is this direction true? And what can be said about the other direction - can an explicit degenerecy condition be given? – Mike Aug 05 '15 at 21:54
  • Yes, the new highlighted statement is correct, unless $D(f_1,...,f_n)$ is empty (and thus affine!). However I didn't check your proof. – Georges Elencwajg Aug 06 '15 at 07:34
  • I have edited the proof so I hope it is true now. Can you go over it? Thanks – Mike Aug 07 '15 at 10:35