# nLab polynomial

### Context

#### Higher algebra

higher algebra

universal algebra

# Contents

## Definition

Let $R$ be a commutative ring.

###### Example

The set of polynomials in one variable with coefficients in $R$ is the set $R[\mathbb{N}]$ of all formal linear combinations on elements $n \in \mathbb{N}$, thought of as powers $x^n$ of the variable $x$

$a_n z^n + \cdots + a_1 z + a_0 \,,$

where $n$ is an arbitrary natural number and $a_0, \dots, a_n \in R$, modulo the equivalence relation generated by equations of the form

$0 z^{n+1} + a_n z^n + \cdots + a_1 z + a_0 = a_n z^n + \cdots + a_1 z + a_0$

(so that we ignore coefficients of zero).

This set is equipped with the structure of a ring itself, in fact a commutative algebra over $R$, denoted $R[z]$ and called the polynomial ring or ring of polynomials given by the unique bilinear map

$R[z] \cdot R[z] \to R[z]$

which on monomials is given by

$z^k \cdot z^l = z^{k+l} \,.$

## Properties

###### Proposition

The polynomial ring $R[z]$ is the free $R$-algebra on one generator (the variable $z$).

###### Proof

By the definition of free objects one needs to check that ring homomorphisms

$f : R[z] \to K$

to another ring K are in natural bijection with functions of sets

$\bar f : * \to K$

from the singleton to the set underlying $K$. Take $\bar f \coloneqq f(z)$. Using $R$-linearity, this is directly seen to yield the desired bijection.

###### Remark

Similarly, the set of polynomials in any give set of variables with coefficients in $R$ is the free commutative $R$-algebra on that set of generators; see symmetric power and symmetric algebra.

###### Remark

The field of fractions of $R[z]$ is the field $R(z)$ of rational functions.

Revised on April 26, 2014 17:14:06 by Todd Trimble (67.81.95.215)