Polynomial ring

In algebra, the polynomial ring over a commutative ring is a ring which formalises the polynomials of elementary algebra.

Construction of the polynomial ring

Let R be a ring. Consider the R-module of sequences

\left(a_{0},a_{1},\ldots ,a_{n},\ldots \right)\,

which have only finitely many non-zero terms, under pointwise addition

(a+b)_{n}=a_{n}+b_{n}.\,

We define the degree of a non-zero sequence (a_n) as the the largest integer d such that a_d is non-zero.

We define "convolution" of sequences by

(a\star b)_{n}=\sum _{i=0}^{n}a_{i}b_{n-i}.\,

Convolution is a commutative, associative operation on sequences which is distributive over addition.

Let X denote the sequence

X=(0,1,0,\ldots ).\,

We have

X^{2}=X\star X=(0,0,1,0,\ldots )\,

X^{3}=X\star X\star X=(0,0,0,1,0,\ldots )\,

and so on, so that

\left(a_{0},a_{1},\ldots ,a_{n},\ldots \right)=\sum _{n}a_{n}X^{n},\,

which makes sense as a finite sum since only finitely many of the a_n are non-zero.

The ring defined in this way is denoted $R[X]$ .

If R is commutative then so is R[X].
If R is an integral domain then so is R[X].
- In this case the degree function satisfies $\deg(fg)=\deg(f)+\deg(g)$ .
If R is a unique factorisation domain then so is R[X].
If R is a Noetherian domain then so is R[X].
If R is a field, then R[X] is a Euclidean domain.

If $f:A\rightarrow B$ is a ring homomorphism then there is a homomorphism, also denoted by f, from $A[X]\rightarrow B[X]$ which extends f. Any homomorphism on A[X] is determined by its restriction to A and its value at X.