An exercise with Zariski topology

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?

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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

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.

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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