Let RR be a commutative ring.


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

a nz n++a 1z+a 0, a_n z^n + \cdots + a_1 z + a_0 \,,

where nn is an arbitrary natural number and a 0,,a nRa_0, \dots, a_n \in R, modulo the equivalence relation generated by equations of the form

0z n+1+a nz n++a 1z+a 0=a nz n++a 1z+a 0 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 RR, denoted R[z]R[z] and called the polynomial ring or ring of polynomials given by the unique bilinear map

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

which on monomials is given by

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



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


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

f:R[z]K f : R[z] \to K

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

f¯:*K \bar f : * \to K

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


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


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

Revised on December 10, 2014 15:10:13 by Urs Schreiber (