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Transcendental numbers are a well-known phenomenon: a number $x$ is transcendental if no polynomial with integer coefficients has $x$ as its root, or $p(x)\neq0$ for all polynomials $p$ with integer coefficients.

I was wondering if this concept could be extended to multivariate polynomials:

Is there a point $\vec{x}=(x_1, x_2, \dots, x_n)$ for any $n$ such that $p(\vec{x})\neq0$ for all multinomials $p$ with integer coefficients?

$\pi$ is transcendental, but $(\pi, \pi)$ is a root of $p(x,y)=x-y$, and $(\pi,0)$ is a root of $p(x,y)=x \cdot y$. However, I can't easily find a bivariate polynomial having $(\pi, e)$ as root.

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closed as too localized by Andrés E. Caicedo, Angelo, Qiaochu Yuan, Harry Gindi, Felipe Voloch Jan 17 '11 at 18:57

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The concept that you are looking for is called Algebraic independence. Try looking on wikipedia. – Daniel Loughran Jan 17 '11 at 16:46
What you write is not correct: You need to specify what coefficients you allow in your polynomials. If they are rational numbers, the question is trivial. – Andrés E. Caicedo Jan 17 '11 at 17:10
Algebraic independence is indeed exactly what I'm looking for, thanks. @Andres Caicedo: my question was indeed incorrect (I fixed it), I thought of that problem the moment I got out of reach of a computer to correct it >.< – Alex ten Brink Jan 17 '11 at 18:04