# 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

## References

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 $\mathbb{R}^2$):

Last revised on June 5, 2022 at 20:55:08. See the history of this page for a list of all contributions to it.