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Say, you have a function from $C^n \to C$ which is built by adding and composing polynomials and square root functions (e.g., $f(x_1,x_2,x_3)=\sqrt{x_1^2+1}+\sqrt{x_2}+2\sqrt{x_3}$). Is there an easy argument why the set of roots of such functions has measure 0 (provided that this is not the zero-function)? The argument is clear for polynomials, but what if we have these square roots? Any reference or pointers to related results are appreciated!

(a suitable squaring of the resulting equation leads to a polynomial system for this particular example, but this argument seems not to work in general...)

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    $\begingroup$ The square roots induce an algebraic function from a branched cover $X$ of $\mathbb{C}^n$ to $\mathbb{C}$. If the function is not identically zero, then the zero locus is Zariski closed in $X$ but not the whole space. Now consider the image in $\mathbb{C}^n$. $\endgroup$
    – S. Carnahan ♦
    Jun 29, 2011 at 19:40
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    $\begingroup$ Alternatively Consider the set of conjugates $f_1,\ldots,f_n$ of $f$ obtained by changing the signs of the square roots, and look at the zeros of the polynomial $\prod_if_i$. $\endgroup$
    – George Lowther
    Jun 29, 2011 at 20:34
  • $\begingroup$ Both answers are really nice! $\endgroup$
    – nullset
    Jun 29, 2011 at 21:17
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    $\begingroup$ I don't think that "adding and composing polynomials and square root functions" builds functions $\mathbb{C}^n\longrightarrow \mathbb{C}$ a priori. How do you deal with branch choices for the roots, for example? $\endgroup$
    – Ramsey
    Jun 29, 2011 at 22:26
  • $\begingroup$ I can't think of any examples to show that the multivariable version of the question is more interesting than the single-variable version. If you fix all the variables but one, then the complex plane will, I think, split up into domains on which the resulting function is holomorphic, and therefore either constant or having isolated zeros. As Jacques Carette observes, both can occur. If you somehow add a condition to rule out constant parts, then you'll get measure zero. Or am I missing something? $\endgroup$
    – gowers
    Jun 30, 2011 at 18:32

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Why is this not a counter-example: $\sqrt{x^2}+x$. This is not identically $0$ (it is $2$ at $x=1$), but it is $0$ for all $x$ such that $\Im(x)<0$, which is most certainly not measure $0$. I believe that both Scott and George implicitly assumed irreducibility, which is an unwarranted assumption.

Note that for this case, it turns out that the polynomial defined by the product of the conjugates indeed is the zero-function, even though the function itself is not identically $0$.

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    $\begingroup$ This is a good observation. Similarly, the definition of "algebraic function" at the popular level (such as Wikipedia) must be viewed as flawed. $\endgroup$
    – Gerald Edgar
    Jun 30, 2011 at 0:37

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