Zero set of nested radicals My question deals with a function on $\mathbb{R}^n$ that consists of nested radicals and polynomial functions. I'm not even sure how to properly formulate this question, i.e. precisely what class of functions I'm trying to specify.
Consider the function
$$
f(x_1,x_2,x_3)=\sqrt{x_1^2+x_3^3+x_2^4\sqrt{x_3x_1^7\sqrt{x_1+x_3}+x_2}}.
$$
Denote the vanishing set of $f$ as $V(f)=\{(x_1,x_2,x_3)\in\mathbb{R}^3\, |\, f(x_1,x_2,x_3)=0\}$. Through symbolic manipulation, I can a find polynomial $g(x_1,x_2,x_3)$ such that the vanishing set of $g$ denoted $V(g)$ contains $V(f)$.
\begin{align*}
\sqrt{x_1^2+x_3^3+x_2^4\sqrt{x_3x_1^7\sqrt{x_1+x_3}+x_2}}&=0\\
x_1^2+x_3^3+x_2^4\sqrt{x_3x_1^7\sqrt{x_1+x_3}+x_2}&=0\\
x_1^2+x_3^3&=-x_2^4\sqrt{x_3x_1^7\sqrt{x_1+x_3}+x_2}\\
(x_1^2+x_3^3)^2&=x_2^8(x_3x_1^7\sqrt{x_1+x_3}+x_2)\\
(x_1^2+x_3^3)^2-x_2^9&=x_2^8x_3x_1^7\sqrt{x_1+x_3}\\
((x_1^2+x_3^3)^2-x_2^9)^2&=x_2^{16}x_3^2x_1^{14}(x_1+x_3)\\
g&=((x_1^2+x_3^3)^2-x_2^9)^2-x_2^{16}x_3^2x_1^{14}(x_1+x_3)
\end{align*}
For any such function $f$ on $\mathbb{R}^n$ is it true that there exists a polynomial $g\in \mathbb{R}[x_1,...,x_n]$ such that $V(g)\supset V(f)$? What if we replace $\mathbb{R}$ with $\mathbb{C}$?
 A: Yes, this is true when working over $\Bbb R$, and we can explain via something called semi-algebraic geometry. A nice introduction is given in these notes of Coste.
A semi-algebraic set in $\Bbb R^n$ can be written as $A=\bigcup_{i=1}^p\{x\in\Bbb R^n\mid f_i=0,g_i>0\}$ for $f_i,g_i$ polynomials in $\Bbb R[x_1,\cdots,x_n]$, so if we can write the zero set of your function as a semi-algebraic set, the claim follows immediately: throw away the $g_i$, and then take the product of the $f_i$. To show that the zero set of your function is semi-algebraic, we'll need to introduce some basic properties of semialgebraic sets and maps.
First, semi-algebraic sets are closed under finite intersections, unions, complements, and coordinate projections. We call a function $\Bbb R^n\to\Bbb R^m$ semi-algebraic if its graph is a semi-algebraic subset of $\Bbb R^n\times\Bbb R^m$ (we can also extend this to a function from a semi-algebraic set in $\Bbb R^n$ to a semi-algebraic set in $\Bbb R^m$ in the obvious way). Compositions of semi-algebraic functions are again semi-algebraic, and level sets of semi-algebraic functions are again semi-algebraic.
This has an obvious application to your problem: if we can show that the function $\sqrt{-}$ is semi-algebraic, then your function obtained from nested radicals will be semi-algebraic, and thus the zero locus you're interested in will be too. But observing that the function $\sqrt{x}$ is defined on a semialgebraic set $x\geq 0$ and has semi-algebraic graph given by $\{(x,y)\in\Bbb R\times\Bbb R\mid y^2-x=0, y\geq 0\}$ immediately gives our conclusion. (I'm being a bit glib about the domain of definition of your function, and you are too, so I hope you forgive me - rest assured that it can be described by a collection of semi-algebraic expressions being non-negative and thus its a semialgebraic set.)
As far as extending to $\Bbb C$, you have some things to take care of before you go there. Square roots (any roots) in $\Bbb C$ are multi-valued, and you need to make sense of what you mean by that before talking about that extension of the problem.
