What is an algebraic function?

I am doing first year university calculus, and we are learning about the different kinds of functions. According to wikipedia, an algebraic function is informally a function that satisfies a polynomial equation whose coefficients are themselves polynomials with rational coefficients.

I understand what this means, and I get the gist of what an algebraic function typically looks like, but I'm curious, what is the formal definition of an algebraic function?

-
I'd treat polynomials, rational functions, functions with rational exponents, and compositions thereof as algebraic... –  Ｊ. Ｍ. Sep 10 '11 at 19:01

Formally, an algebraic function in $n$ variables over the field $K$ is an element of the algebraic closure of the field of rational functions $K(x_1,\ldots,x_n)$.

Here's my best shot at (informally) explaining the terms in the formal definition:

• A ring is a structure in which we can add and multiply.

• A field is a ring in which we also can divide by any non-zero number.

• When $K$ is a field, the polynomial ring in $n$ variables, denoted by the expression $K[x_1,\ldots,x_n]$, is the collection of polynomials in the variables $x_1,\ldots,x_n$ whose coefficients are in $K$.

• The field of rational functions $K(x_1,\ldots,x_n)$ is defined to be $$K(x_1,\ldots,x_n)=\left\{\,\frac{f}{g}\,\Bigg|\,\,\,f,g\in K[x_1,\ldots,x_n], g\neq0\right\},$$ or in other words, the fractions of polynomials in $K[x_1,\ldots,x_n]$. The name "rational functions" makes sense because they are "ratios" of polynomials; in fact, the field $K(x_1,\ldots,x_n)$ stands in the same relation to the ring $K[x_1,\ldots,x_n]$ as does the field $\mathbb{Q}$ of rational numbers to the ring of integers $\mathbb{Z}$.

• When $L$ is a field, the algebraic closure of $L$ is the field $\overline{L}$ that consists of all the solutions to polynomials in $L[x]$.
$\text{ }$
For example, not every polynomial in $\mathbb{R}[x]$ (where $\mathbb{R}$ denotes the real numbers) has a solution in $\mathbb{R}$ - for example, $x^2+1=0$ has no solutions in $\mathbb{R}$. But the set of all roots of polynomials in $\mathbb{R}[x]$ forms a bigger field, containing $\mathbb{R}$ - namely $\mathbb{C}$, the complex numbers! So $\mathbb{C}=\overline{\mathbb{R}}$.

Let's consider the case of rational functions. Let $L=K(x_1,\ldots,x_n)$, and consider $L[\,t\,]$, where $t$ is a variable. The polynomial $t^2-x$ has no roots in $L$; the roots are $\sqrt{x}$ and $-\sqrt{x}$, but these don't live in $L$. They do, however, live in $\overline{L}$, which is the field of algebraic functions.

-
That the answer lies in and cites the same wikipedia article that the question cites... well, that's just strange! :) –  The Chaz 2.0 Sep 10 '11 at 19:05
Isn't the essential point that one can formalise the informal definition by adopting a greater level of mathematical sophistication, but without changing the essential idea. In my education, rational functions were "informally" defined in this way much earlier than the concepts of rings and fields, and were useful too. I could integrate rational functions (one variable) when I left school. I encountered the formality of rings and fields in my second year at university. The essential insight is "there are no surprises" ... –  Mark Bennet Sep 10 '11 at 19:27
@Zev This is really cool. I believe it is simply too much new material for me to grapple with at once, but I do get a vague sense of what is going on here. I will have ruminate on this and read some more. –  Matt Munson Sep 10 '11 at 19:38
Generally if $\rm\: T \supset R\:$ is ring extension then one says that $\rm\:t\in T\:$ is algebraic over $\rm\:R\:$ if $\rm\:t\:$ is the root of some polynomial $\rm\:0\ne f(x)\in R[x]\:.\:$ Some authors restrict the definition to the case of fields or domains. One calls the extension $\rm\:T\supset R\:$ algebraic if every element of $\rm\:t\:$ is algebraic over $\rm\:R\:.\:$ Elements of $\rm\:t\:$ that are not algebraic over $\rm\:R\:$ are called transcendental over $\rm\:R\:.\:$ For example, a simple extension $\rm\:R[t]\:$ is a polynomial ring $\rm\:R[t]\:\cong R[x]\:$ iff $\rm\:t\:$ is transcendental over $\rm\:R\:.\:$
Algebraic "functions" are elements algebraic over a "function" ring, i.e. a polynomial ring $\rm\:R[x_1,\:\ldots,x_n]\:$ in $\rm\:n \ge 1\:$ variables. Just as one does in the case of polynomial and rational "functions", one distinguishes between formal algebraic "functions", and functional ones, i.e. elements of an abstract ring vs. elements of a ring whose elements are (set-theoretic) functions.