nLab rational function




As partial functions on a field

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

There is a function

i:k(x){AOb(Part(k))|Hom(A,k)}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)Part(k), where the function algebra kkk \to k is the endomorphism kk-algebra on the improper subset kk. The function ii is defined as follows: given two polynomials p,qk[x]p, q \in k[x] where q0q \neq 0, for all x{ak|q(a)0}x \in \{a \in k | q(a) \neq 0\}

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

where for xkx \in k and yky \in k


is the division of xx by yy in kk, and for rk(x)r \in k(x) and sk(x)s \in k(x),


is the division of rr by ss in k(x)k(x). A rational function ff is an element of the image of ii, fim(i)f \in \im(i).

As functions on the projective line

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

D k i k [p/q]i 1(k)\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][p/q] on 1(k)\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” 1/01/0).


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)=xx1f(x) = \frac{x}{x} \neq 1, due to the fact that f(x)f(x) is undefined at 00, and thus has a different domain than the constant function 11, 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 3\mathbb{R}^3 and to the configuration space of points in 2\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.