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How do I check whether a give function is 'algebraic' or not? I have a function $m(z) = 2\pi i z^n$ where $z \in \mathbb{C} \backslash \mathbb{R}$ and $n \in \mathbb{Z}$. I can write this as $w - 2\pi i z^n = 0$ where $w = m(z)$ and it is a polynomial. I think this is an algebraic function? Is my understanding correct?

What is the general approach to find whether a given function is algebraic or not?

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What's your definition of algebraic function? I would say that it's any polynomial function, or at a stretch, any function which satisfies a polynomial equation... – Zhen Lin Nov 1 '11 at 21:42
@ZhenLin Thats correct, its a function which satisfies a polynomial equation whose coefficients are polynomial too. So this implies that the roots have to be polynomial themselves too? Is that correct? – sauravrt Nov 1 '11 at 21:46
As $m(z)$ is a polynomial function, it is algebraic. Do you mean $2\pi i z^{1/n}$ ? – user18119 Nov 1 '11 at 21:51
The two things that puzzle me are 1. why you exclude the reals from the domain of the function, and 2. what kind of thing $n$ is supposed to be: positive integer? integer? real? complex? matrix? category of local sheaves over a quantum topos? – Gerry Myerson Nov 1 '11 at 21:59
@QiL No. Its just $z^n$. Thanks. – sauravrt Nov 1 '11 at 22:00
up vote 3 down vote accepted

If $n$ is a non-negative integer then $m$ is a polynomial, a fortiori, algebraic. As you say, it is a zero of the equation $w-2\pi iz^n=0$, an equation with polynomial coefficients. If $n$ is a negative integer then it is a zero of $z^{-n}w-2\pi i=0$ which again is an equation with polynomial coefficients, so in this case, too, $m$ is algebraic.

I don't think there is a general approach to telling whether a function is algebraic, any more than there is a general approach for telling whether a number is algebraic. Indeed, nobody knows whether $\gamma$ (Euler's constant) is rational, so I suppose nobody knows whether the function $z^{\gamma}$ is algebraic.

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If $F(z)=z^\gamma$ is an algebraic function (over $\mathbb Q$), then $F'(z)$ is also an algebraic function (over $\mathbb Q$), and thus $F'(1)$ is an algebraic number (over $\mathbb Q$). Therefore, $z^\gamma$ algebraic implies $\gamma$ algebraic. Equivalently, if $\gamma$ is not algebraic, then $z^\gamma$ is not algebraic.

In some cases, one knows a Puiseux series expansions of a function $F(z)$; such an expansion is unique and is of the type $$F(z) = \sum_{k\geq k_0} f_k \times \big((z-z_0)^{r} \big)^k \,.$$ What is more, the theory of Newton polygon for algebraic functions implies that the exponent $r$ in this Puiseux expansion is then a rational number. This gives a nice way to test if $F(z)$ is algebraic or not!

Therefore, one can say that $z^\gamma$ is an algebraic function if and only if $\gamma$ is a rational number (the same reasoning proves that $z^{\sqrt 2}$ is not an algebraic function). [It just requires a little lemma showing that $z^\gamma$ cannot be rewritten as an infinite sum of $z^{rk}$, which follows easily by considering the limit when $z\rightarrow 0$.]

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