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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?

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    I'd treat polynomials, rational functions, functions with rational exponents, and compositions thereof as algebraic...2011-09-10

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According to Wikipedia,

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.

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    @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.2011-09-10
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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.