Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I read this exercise:

Prove that the set $S = \{ (n, 2^n, 3^n ) \mid n \in \mathbb{N} \}$ is dense in $\mathbb{C}^3$ with Zariski topology.

I have seriously thought about it, but I do not manage to solve the problem. Besides I cannot answer the simpler question if $ \{ (n, 2^n) \mid n \in \mathbb{N} \}$ is Zariski-dense in $\mathbb{C}^2$. However, using Artin's theorem about independence of characters, I can prove that $\{ (2^n, 3^n ) \mid n \in \mathbb{N} \}$ is Zariski-dense in $\mathbb{C}^2$.

Can someone give me a hint?

share|cite|improve this question
This is a neat question, I'm looking forward to seeing the answer. Just two half-baked ideas I thought may help: it seems that an analytic property of polynomials might be necessary to prove that $V(S)=\{0\}$ - perhaps there is an argument similar to the one I made here that works? And, for the other idea, $f\in V(S)$ is equivalent to the statement that $$f(x-(n-1),2^{n-1}y,3^{n-1}z)\in V(1,2,3)\text{ for all }n\in\mathbb{N},$$ perhaps we can prove that if $f\neq0$ this is impossible? – Zev Chonoles Aug 20 '11 at 18:51

1 Answer 1

up vote 17 down vote accepted

If $x^i y^j z^k$ is any monomial, then substituting $(n, 2^n, 3^n)$ gives $n^i (2^j 3^k)^n$. In particular, the growth rates of all such monomials are distinct and totally ordered (first by $j \log 2 + k \log 3$, then by $i$) by unique factorization.

It follows that if $f(x, y, z)$ is a polynomial, there is a unique term $f_{ijk} x^i y^j z^k$ in $f$ of maximal growth rate, and taking $n \to \infty$ it follows that $f$ cannot vanish on $S$.

Edit: Andrea asks in the comments for a more algebraic proof. Here's one based on finite differences. For any sequence $a_n, n \ge 0$ define the shift operator

$$S(a_0, a_1, a_2, ...) = a_1, a_2, a_3, ....$$

If $f(x, y, z)$ is a nonzero polynomial, let $a_n = f(n, 2^n, 3^n)$. This is a sum of terms of the form $n^i (2^j 3^k)^n$ as above. All of these terms satisfy linear recurrence relations, which is another way of saying that they are all annihilated by operators of the form $p(S)$ where $p$ is some polynomial. In particular,

$(S - \lambda)^m$ annihilates precisely terms of the form $n^d \lambda^n$ where $d < m$.

By repeatedly applying such operators we can eliminate all terms in $a_n$ where $2^j 3^k$ does not have its maximal value, then eliminate all remaining terms where $i$ does not have its maximal value. The resulting sequence is nonzero, which implies that the original sequence must have been nonzero.

Edit: Here is a third proof which perhaps makes the underlying idea a little clearer. As above, if $f$ is a nonzero polynomial, let $a_n = f(n, 2^n, 3^n)$. Now consider

$$A(z) = \sum_{n \ge 0} a_n z^n.$$

This is a rational function (exercise) with a pole of order $i+1$ at $\frac{1}{2^j 3^k}$ whenever $i$ is maximal such that $x^i y^j z^k$ is a nonzero term in $f$ (exercise). In particular, it has at least one pole, so is necessarily nonzero.

share|cite|improve this answer
I would have preferred a more algebraic proof, but it seems to me that it works. I would have never thought of using analysis in this problem... Thanks! – Andrea Aug 20 '11 at 19:22
@Andrea: I included a more algebraic proof. It only works in characteristic $0$; indeed the statement is false in positive characteristic, since $x^p - x, y^p - y, z^p - z$ always vanish in characteristic $p$. – Qiaochu Yuan Aug 20 '11 at 19:39
Oh, wonderful! If you discover other proofs, write them! This would be very interesting for me. Could Artin's theorem on independence of characters be useful in some way, like in proving that $\{ (2^n, 3^n ) \}_{n \in \mathbb{N}}$ is Zariski-dense? – Andrea Aug 20 '11 at 20:09
@Andrea: I think the underlying idea behind the second proof is a little deeper than independence of characters (or rather it generalizes a basic fact in a different direction than independence of characters). The idea is that the space of all functions satisfying a linear recurrence with constant coefficients breaks up into a direct sum of generalized eigenspaces of $S$. I can give a third proof that makes this more explicit. – Qiaochu Yuan Aug 20 '11 at 20:12
+1 for the question and the answer!!! I tried to do the exercise... Now that I see Qiaochu's solution, I realize that what I was missing is the most obvious piece of the puzzle: some nonzero polynomial vanishes on $S\subset K^n$ iff the monomials are linearly dependent on $S$... – Pierre-Yves Gaillard Aug 21 '11 at 3:18

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.