# nLab rational function

Contents

### Context

#### Algebra

higher algebra

universal algebra

# Contents

## Definition

### As partial functions on a field

If $k$ is a field, given the polynomial ring $k[x]$, there is a canonical ring homomorphism $j:k[x] \to (k \to k)$ which sends constant polynomials in $k[x]$ to constant functions in $k \to k$, and the generator $x \in k[x]$ to the identity function $id_k$, where $k \to k$ is the function algebra on $k$.

There is a function

$i:k(x) \to \{A \in Ob(Part(k)) \vert Hom(A, k)\}$

from the field of rational expression to the set of all partial functions in the category of partial functions $Part(k)$, where the function algebra $k \to k$ is the endomorphism $k$-algebra on the improper subset $k$. The function $i$ is defined as follows: given two polynomials $p, q \in k[x]$ where $q \neq 0$, for all $x \in \{a \in k | q(a) \neq 0\}$

$i\left(\frac{p}{q}\right)(x) \coloneqq \frac{j(p)(x)}{j(q)(x)}$

where for $x \in k$ and $y \in k$

$\frac{x}{y}$

is the division of $x$ by $y$ in $k$, and for $r \in k(x)$ and $s \in k(x)$,

$\frac{r}{s}$

is the division of $r$ by $s$ in $k(x)$. A rational function $f$ is an element of the image of $i$, $f \in \im(i)$.

### As functions on the projective line

It is perhaps more illuminating to think of this partial function (with domain $D$) as coming from a (total) function $[p/q]: \mathbb{P}^1(k) \to \mathbb{P}^1(k)$ on the projective line, where we have inclusion functions $i: k \to \mathbb{P}^1(k)$ and the partial function is given by the pair of projection maps in the pullback

$\array{ D & \to & k \\ \downarrow & & \downarrow_\mathrlap{i} \\ k & \underset{[p/q] \circ i}{\to} & \mathbb{P}^1(k) }$

It should also be noticed that such endofunctions $[p/q]$ on $\mathbb{P}^1(k)$ are closed under composition (except that special provision must be made for the constant function valued at $\infty$, which corresponds to the “fraction” $1/0$).

## Properties

### Rational functions on a field do not form a field

This comes from the fact that the reciprocal function is not a multiplicative inverse of the identity function $f(x) = \frac{x}{x} \neq 1$, due to the fact that $f(x)$ is undefined at $0$, and thus has a different domain than the constant function $1$, which is the multiplicative unit.

### As continuous functions

rational functions are continuous on their domain of definition

On the homotopy type of spaces of rational maps from the Riemann sphere to itself (related to the moduli space of monopoles in $\mathbb{R}^3$ and to the configuration space of points in $\mathbb{R}^2$):