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Consider the set of (complex) algebraic numbers $\mathbb{A}$, defined as $\mathbb{A} = \{x \in \mathbb{C} \mid p(x) = 0 \text{ for some $p \in \mathbb{Z}[x]$}\}.$ Suppose we now wanted to generalize this to multivariate polynomials via $$ \mathbb{A}_n = \{(x_1, \dots, x_n) \in \mathbb{C}^n \mid p(x_1,\dots,x_n) = 0 \text{ for some $p \in \mathbb{Z}[x_1, \dots, x_n]$} \}.$$

Right away I noticed that $\mathbb{A}_n \neq \mathbb{A}^n$ for $n \neq 1$, as all pairs of the form $(z,z,0,\dots,0)$ (for complex $z$) are in $\mathbb{A}_n$ since $(z,z,0,\dots,0)$ is a root of the polynomial with integer coefficients $p(x_1,\dots,x_n) = x_1 - x_2.$ In particular when $n \neq 2$ we have $(\pi, \pi) \in \mathbb{A}_2$ but $(\pi, \pi) \not\in \mathbb{A}^2$.

Is this construction meaningful in any way? I know that $\mathbb{A}^n \subseteq \mathbb{A}_n$. Does it hold for $n > 1$ that $\mathbb{A}_n = \mathbb{C}^n$? My guess is no, but I’m not entirely sure how I’d go about proving this (I imagine that a counterexample would be nice, but proving that a set of $n$ numbers satisfies no polynomial with integer coefficients seems a lot harder than proving that any one number doesn’t). What other information, useful or otherwise, can be gathered from the structure of $\mathbb{A}_n$ for $n > 1$?

Rodrigo de Azevedo
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Robert
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  • Regarding whether $\mathbb A_n = \mathbb C^n$; note that if $(z,k_1,\ldots,k_{n-1}) \in \mathbb A^n$ and $k_1,\ldots, k_{n-1}\in \mathbb Z$ then $z \in \mathbb A$. Ineed, if $p \in \mathbb Z[X_1,\ldots,X_n]$ is such that $p(z,k_1,\ldots,k_{n-1}) = 0$, then $z$ vanishes at $p(-,k_1,\ldots,k_{n-1}) \in \mathbb Z[X]$.

    In particular, $(\pi,0,\ldots,0) \not \in \mathbb A^n$.

     – qualcuno Dec 17 '21 at 05:19
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    Note that it is unknown whether there is a nonzero integer-coefficient polynomial $p(x,y)$ such that $p(\pi,e)=0$. – Gerry Myerson Dec 17 '21 at 05:36
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    But it is known that the transcendence degree of the complex numbers over the rationals is infinite, so for any $n$ there exist complex numbers $\alpha_1,\dots,\alpha_n$ for which no nonzero integer-coefficient polynomial $p(x_1,\dots,x_n)$ exists satisfying $p(\alpha_1,\dots,\alpha_n)=0$. – Gerry Myerson Dec 17 '21 at 05:38

1 Answers1

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(I assume you want to require $p\neq 0$ in your definition of $\mathbb{A}_n$ to avoid trivialities.)

This is kind of tautological, but perhaps a more enlightening way of describing your $\mathbb{A}_n$ is that it is the set of $n$-tuples of complex numbers that are not algebraically independent over $\mathbb{Q}$. This makes it immediately clear that $\mathbb{A}_n$ is not all of $\mathbb{C}^n$. Explicitly, you can choose the coordinates one at a time to get a point in the complement: first pick $x_1\in\mathbb{C}$ that is transcendental over $\mathbb{Q}$, then pick $x_2\in\mathbb{C}$ that is transcendental over $\mathbb{Q}(x_1)$, then pick $x_3\in\mathbb{C}$ that is transcendental over $\mathbb{Q}(x_1,x_2)$, and so on. It is always possible to keep choosing more transcendental elements like this since the algebraic closure of $\mathbb{Q}(x_1,\dots,x_k)$ is always countable. (This also illustrates that $\mathbb{A}_n$ is quite "small": for instance, it has Lebesgue measure $0$, since these countable sets on each coordinate have measure $0$. See this answer of mine for some related ideas which also give an alternate way of proving $\mathbb{A}_n$ is not all of $\mathbb{C}^n$.)

Eric Wofsey
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  • Thank you! I have a hard time understanding why the algebraic closure of $\mathbb{Q}(x_1, \dots, x_k)$ is countable for $k > 1.$ In particular, wouldn’t the tuple $(z, z, 0, \dots, 0)$ (for $z \in \mathbb{C}$) always be in $\mathbb{A}_n?$ As a result, we would have a surjection $\phi: \mathbb{A}_n \to \mathbb{C}$ by $\phi(z,z,0,\dots,0) = z$ and $\phi(x_1,\dots,x_k) = 0$ otherwise, right? Would this then imply that $\mathbb{A}_n$ is uncountable? If $\mathbb{A}_n$ is uncountable, why does it have Lebesgue measure $0$? Or is it possible for a set to have measure zero while being uncountable? – Robert Dec 18 '21 at 14:39
  • The algebraic closure of any countable field $K$ is countable: there are only countably many nonzero single-variable polynomials over $K$, and each one has only finitely many roots. – Eric Wofsey Dec 18 '21 at 15:03
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    $\mathbb{A}_n$ is uncountable for $n>1$ but always has measure $0$. For instance, $(x,y)\in\mathbb{A}_2$ iff either $x$ is algebraic over $\mathbb{Q}$ or $y$ is algebraic over $\mathbb{Q}(x,y)$. Each of these two cases has measure $0$ by Fubini's theorem (integrating the characteristic function of your set over $\mathbb{C}^2$): in the first case, there is a measure $0$ set of values of $x$ that are possible, and in the second case, for fixed $x$, there is a measure $0$ set of values of $y$ that are possible (so you get $0$ if you integrate with respect to $y$ first). – Eric Wofsey Dec 18 '21 at 15:11