Do non-vanishing algebraic functions exist? A polynomial with no zeros in the complex plane must be constant.  A rational function with no zeros on the Riemann sphere must be constant.  Is there any kind of analogous statement for algebraic functions?
Specifically: does there exist a non-constant algebraic function $f(z)$ which does not vanish for any value of $z$?
To be sure: algebraic functions are in general multi-valued, so I mean any value of $z$ on any sheet of the Riemann surface associated to $f(z)$.
 A: Answering my own question: it's easy to see that the answer is no.
An algebraic function $f(x)$ is, by definition, one that satisfies an equation of the form
$$
a_n(x) f(x)^n + a_{n-1}(x) f(x)^{n-1} + \cdots + a_1(x) f(x) + a_0(x) = 0
$$
where the $a_i(x)$ are polynomials in $x$.  Without loss of generality, we can assume that $a_0$ does not vanish (otherwise we could factor out $f(x)$), and similarly we can assume that the $a_i(x)$ do not have any common roots (if they did, at some $x = x_0$, then we could factor out $(x - x_0$)).
If $a_0(x)$ has a root at some $x_0$ in the complex plane, it is clear that $f = 0$ is one of the roots of the above equation, so zero is a value of $f(x_0)$.  So, if $f(x)$ never attains the value zero, the polynomial $a_0(x)$ cannot have any zeros in the complex plane, so it must be constant.  We can normalize it to be $a_0(x) = 1$.
It remains to consider the possibility of zeros at infinite $x$.  Let $m$ be the highest power of $x$ in any of the polynomials $a_i(x)$ and define the polynomials
$$
b_i(x) = x^m a_i(1/x).
$$
Note that since at least one of the $a_i(x)$'s has a term proportional to $x^m$, it is not possible for all of the $b_i(x)$'s to vanish at $x=0$.
Now take $x \to 1/x$ in the first equation (recalling that $a_0(x) = 1$) and multiply through by $x^m$ to obtain
$$
b_n(x) f(1/x)^n + b_{n-1}(x) f(1/x)^{n-1} + \cdots + b_1(x) f(1/x) + x^m = 0.
$$
By setting $x=0$, we conclude that $f = 0$ is a root unless $m=0$.  But if $m=0$, then $f(x)$ is constant.
