Definition. Let $R$ be a commutative ring. A function $f: R\to R$ is called a polynomial function of $R$ , if there are some elements $a_0,\,a_1,\,\ldots,\,a_m$ of $R$ such that $$f(x) \;=\; a_0\!+\!a_1x\!+\cdots+\!a_mx^m \,\,\,\, \forall x\in R.$$

Remark. The coefficients $a_i$ in a polynomial function need not be unique; e.g. if $R = \{0,\,1\}$ is the ring (and field) of two elements, then the polynomials $X$ and $X^2$ both may be used for the same polynomial function. However, if we stipulate that $R$ is an infinite integral domain, the coefficients are guaranteed to be unique.

The set of all polynomial functions of $R$ , being a subset of the set $R^R$ of all functions from $R$ to $R$ , is here denoted by $R/^R$ .

Theorem 1   If $R$ is a commutative ring, then the set $R/^R$ of all polynomial functions of $R$ , equipped with the operations

(1)

is a commutative ring.

Proof. It's straightforward to show that the function set $R^R$ forms a commutative ring when equipped with the operations ``$+$ '' and ``$\cdot$ '' defined as (1). We show now that $R/^R$ forms a subring of $R^R$ . Let $f$ and $g$ be any two polynomial functions given by $$f(x) \;=\; a_0\!+\!a_1x\!+\cdots+\!a_mx^m, \,\,\, g(x) \;=\; b_0\!+\!b_1x\!+\cdots+\!b_nx^n.$$ Then we can give $f\!+\!g$ by $$(f\!+\!g)(x) \;=\; \sum_{i=0}^k(a_i\!+\!b_i)x^i$$ where $k = \max\{m,\,n\}$ and $a_i = 0$ (resp. $b_i = 0$ ) for $i > m$ (resp. $i > n$ ). This means that $f\!+\!g \in R/^R$ . Secondly, the equation $$(f\!\cdot\!g)(x) \;=\; a_0b_0+(a_0b_1\!+\!a_1b_0)x+(a_0b_2\!+\!a_1b_1\!+\!a_2b_0)x^2\!+\cdots+\!a_mb_nx^{m+n}$$ signifies that $f\!\cdot\!g \in R/^R$ . Because also the function $-\!f$ given by $$(-\!f)(x) \;=\; -\!a_0\!-\!a_1x\!-\cdots-\!a_mx^m$$ and satisfying $-\!f\!+\!f = 0:x\mapsto 0$ belongs to $R/^R$ , the subset $R/^R$ is a subring of $R^R$ .