Algebraic function

The informal definition of an algebraic function provides a number of clues about their properties. To gain an intuitive understanding, it may be helpful to regard algebraic functions as functions which can be formed by the usual algebraic operations: addition, multiplication, division, and taking an nth root. This is something of an oversimplification; because of the fundamental theorem of Galois theory, algebraic functions need not be expressible by radicals.

First, note that any polynomial function ${\displaystyle y=p(x)}$ is an algebraic function, since it is simply the solution y to the equation

${\displaystyle y-p(x)=0.\,}$

More generally, any rational function ${\displaystyle y={\frac {p(x)}{q(x)}}}$ is algebraic, being the solution to

${\displaystyle q(x)y-p(x)=0.}$

Moreover, the nth root of any polynomial ${\textstyle y={\sqrt[{n}]{p(x)}}}$ is an algebraic function, solving the equation

${\displaystyle y^{n}-p(x)=0.}$

Surprisingly, the inverse function of an algebraic function is an algebraic function. For supposing that y is a solution to

${\displaystyle a_{n}(x)y^{n}+\cdots +a_{0}(x)=0,}$

for each value of x, then x is also a solution of this equation for each value of y. Indeed, interchanging the roles of x and y and gathering terms,

${\displaystyle b_{m}(y)x^{m}+b_{m-1}(y)x^{m-1}+\cdots +b_{0}(y)=0.}$

Writing x as a function of y gives the inverse function, also an algebraic function.

However, not every function has an inverse. For example, y = x² fails the horizontal line test: it fails to be one-to-one. The inverse is the algebraic "function" ${\displaystyle x=\pm {\sqrt {y}}}$. Another way to understand this, is that the set of branches of the polynomial equation defining our algebraic function is the graph of an algebraic curve.

From an algebraic perspective, complex numbers enter quite naturally into the study of algebraic functions. First of all, by the fundamental theorem of algebra, the complex numbers are an algebraically closed field. Hence any polynomial relation p(y, x) = 0 is guaranteed to have at least one solution (and in general a number of solutions not exceeding the degree of p in y) for y at each point x, provided we allow y to assume complex as well as real values. Thus, problems to do with the domain of an algebraic function can safely be minimized.

A graph of three branches of the algebraic function y, where y³ − xy + 1 = 0, over the domain 3/2^2/3 < x < 50.

Furthermore, even if one is ultimately interested in real algebraic functions, there may be no means to express the function in terms of addition, multiplication, division and taking nth roots without resorting to complex numbers (see casus irreducibilis). For example, consider the algebraic function determined by the equation

${\displaystyle y^{3}-xy+1=0.\,}$

Using the cubic formula, we get

${\displaystyle y=-{\frac {2x}{\sqrt[{3}]{-108+12{\sqrt {81-12x^{3}}}}}}+{\frac {\sqrt[{3}]{-108+12{\sqrt {81-12x^{3}}}}}{6}}.}$

For ${\displaystyle x\leq {\frac {3}{\sqrt[{3}]{4}}},}$ the square root is real and the cubic root is thus well defined, providing the unique real root. On the other hand, for ${\displaystyle x>{\frac {3}{\sqrt[{3}]{4}}},}$ the square root is not real, and one has to choose, for the square root, either non-real square root. Thus the cubic root has to be chosen among three non-real numbers. If the same choices are done in the two terms of the formula, the three choices for the cubic root provide the three branches shown, in the accompanying image.

It may be proven that there is no way to express this function in terms of nth roots using real numbers only, even though the resulting function is real-valued on the domain of the graph shown.

On a more significant theoretical level, using complex numbers allows one to use the powerful techniques of complex analysis to discuss algebraic functions. In particular, the argument principle can be used to show that any algebraic function is in fact an analytic function, at least in the multiple-valued sense.

Formally, let p(x, y) be a complex polynomial in the complex variables x and y. Suppose that x₀ ∈ C is such that the polynomial p(x₀, y) of y has n distinct zeros. We shall show that the algebraic function is analytic in a neighborhood of x₀. Choose a system of n non-overlapping discs Δ_i containing each of these zeros. Then by the argument principle

${\displaystyle {\frac {1}{2\pi i}}\oint _{\partial \Delta _{i}}{\frac {p_{y}(x_{0},y)}{p(x_{0},y)}}\,dy=1.}$

By continuity, this also holds for all x in a neighborhood of x₀. In particular, p(x, y) has only one root in Δ_i, given by the residue theorem:

${\displaystyle f_{i}(x)={\frac {1}{2\pi i}}\oint _{\partial \Delta _{i}}y{\frac {p_{y}(x,y)}{p(x,y)}}\,dy}$

which is an analytic function.

Note that the foregoing proof of analyticity derived an expression for a system of n different function elements f_i(x), provided that x is not a critical point of p(x, y). A critical point is a point where the number of distinct zeros is smaller than the degree of p, and this occurs only where the highest degree term of p or the discriminant vanish. Hence there are only finitely many such points c₁, ..., c_m.

A close analysis of the properties of the function elements f_i near the critical points can be used to show that the monodromy cover is ramified over the critical points (and possibly the point at infinity). Thus the holomorphic extension of the f_i has at worst algebraic poles and ordinary algebraic branchings over the critical points.

Note that, away from the critical points, we have

${\displaystyle p(x,y)=a_{n}(x)(y-f_{1}(x))(y-f_{2}(x))\cdots (y-f_{n}(x))}$

since the f_i are by definition the distinct zeros of p. The monodromy group acts by permuting the factors, and thus forms the monodromy representation of the Galois group of p. (The monodromy action on the universal covering space is related but different notion in the theory of Riemann surfaces.)