Most definitions I've seen [1] [2] are agnostic about what kind of polynomials the function must satisfy. PlanetMath, for example, says

A function of one variable is said to be algebraic if it satisfies a polynomial equation whose coefficients are polynomials in the same variable.

However, some references [3] [4] are more strict and require that the polynomial have integer coefficients. Wolfram Mathworld has

An algebraic function is a function $f(x)$ which satisfies $p(x,f(x))=0$, where $p(x,y)$ is a polynomial in $x$ and $y$ with integer coefficients.

The only difference (I think) is that allowing arbitrary real coefficients enables the use of transcendental constants. Since adding these constants doesn't add much complexity (it's the functions that are the interesting part), I'm inclined to allow them. What's the consensus here? Can an algebraic function contain transcendental constants?

  • $\begingroup$ Certainly your reference [1] requires integer coefficients. Read the paragraph starting "In more precise terms ..." I didn't check the other references. I've always understood that integer coefficients are required. $\endgroup$ – saulspatz Dec 8 '18 at 14:39
  • $\begingroup$ You're right. I'll reshuffle the references $\endgroup$ – SCappella Dec 8 '18 at 14:40

An algebraic function is defined as algebraic function over a field. This field is called the field of constants of the algebraic function. An algebraic function of one variable $x$ over a field $K$ is a function $f$ with $y=f(x)$ and $f\colon x\mapsto y$, where $y$ satisfies an equation $F(y,x)=0$ where $F$ is an irreducible polynomial in $y$ and $x$ with coefficients in the field $K$. Take e.g. the definition from Encyclopedia of Mathematics: Algebraic function.

If only integer coefficients or only rational coefficients are allowed in $F$, $f$ is an algebraic function over the rational numbers.

Clearly, one can define algebraic functions over a field of constants which contains transcendental numbers, e.g. over the real numbers.

An algebraic function $f$ with $f\colon x\mapsto f(x)$ means, the function value $f(x)$ is generated by connecting the function argument, $x$, only with it self and with constants of the field of constants and only by algebraic operations.


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