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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?

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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](http://math.stackexchange.com/questions/50015/set-that-is-not-algebraic/50018#50018) 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?2011-08-20

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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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    @Pierre: yes, I fi$n$d a useful attitude to keep in mind for problems of this type is not to be tied too closely to the initial formulation of the problem. The problem is phrased as algebraic geometry but it seems a little off: algebraic geometry _per se_ doesn't look terribly useful for the problem. So one looks for a more natural-looking reformulation.2011-08-21