The ring $R^{X}$ of all functions from a set $X$ into a ring, with pointwise operations

Let $X$ be a set and $R$ a ring (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides). Write

$R^{X} \;:=\; \{\, f \;:\; f \text{ is a function } X \to R \,\}$

and define, for $f, g \in R^{X}$, the pointwise operations

$(f + g)(x) := f(x) + g(x), \qquad (f \cdot g)(x) := f(x)\,g(x) \qquad (x \in X),$

with distinguished elements the constant functions

$0_{R^{X}}(x) := 0_R, \qquad 1_{R^{X}}(x) := 1_R \qquad (x \in X).$

These data make $R^{X}$ a ring. Each rule assigns to a pair of functions $X \to R$ another function $X \to R$, so each is a binary operation on $R^{X}$ (Binary operation on a set; associativity, commutativity, and a subset closed under the operation). Two elements of $R^{X}$ are equal exactly when they agree at every $x \in X$, so every ring axiom, being an equation between elements of $R^{X}$, holds as soon as it holds at each point; and at each point it is the corresponding axiom of $R$. In particular $-f$ is the function $x \mapsto -f(x)$, the additive group of $R^{X}$ is abelian, multiplication is associative with two-sided identity $1_{R^{X}}$, and both distributive laws hold.

Commutativity. If $R$ is commutative (Commutative ring) then so is $R^{X}$, pointwise. If $X \ne \varnothing$ the converse holds: fixing $x_0 \in X$ and taking constant functions $f \equiv a$, $g \equiv b$ gives $ab = ba$ from $(fg)(x_0) = (gf)(x_0)$. If $X = \varnothing$ then $R^{X}$ has exactly one element, the empty function, so $1_{R^{X}} = 0_{R^{X}}$ and $R^{X}$ is commutative whatever $R$ is; the converse therefore needs the hypothesis $X \ne \varnothing$ and is stated with it.

Zero divisors. Suppose $R$ is not the one-element ring, that is $1_R \ne 0_R$ (In any ring $0 \cdot a = a \cdot 0 = 0$, $(-a)b = a(-b) = -(ab)$, $(-a)(-b) = ab$, $(-1)a = -a$ and $a(b - c) = ab - ac$), and suppose $X$ has two distinct elements $x_1 \ne x_2$. Define $f, g \in R^{X}$ by

$f(x) := \begin{cases} 1_R & x = x_1 \\ 0_R & x \ne x_1\end{cases} \qquad\qquad g(x) := \begin{cases} 0_R & x = x_1 \\ 1_R & x \ne x_1.\end{cases}$

Then $f \ne 0_{R^{X}}$ because $f(x_1) = 1_R \ne 0_R$, and $g \ne 0_{R^{X}}$ because $g(x_2) = 1_R \ne 0_R$; and $(fg)(x) = 0_R$ for every $x$, since one of the two factors is $0_R$ at each point and $0_R y = 0_R = y 0_R$ (In any ring $0 \cdot a = a \cdot 0 = 0$, $(-a)b = a(-b) = -(ab)$, $(-a)(-b) = ab$, $(-1)a = -a$ and $a(b - c) = ab - ac$). So $f$ and $g$ are zero divisors (Zero divisor, and integral domain: a commutative ring with $1 \ne 0$ and no zero divisors) and $R^{X}$ is not an integral domain.

• The same underlying set carries other structures. $R^{X}$ is the set of all functions $X \to R$, with no continuity, boundedness or finiteness condition. When $R$ is a field, this same set with this same addition carries a second structure of interest, obtained by replacing the pointwise product with a scalar multiplication. The addition of that second structure is this same pointwise rule $(f+g)(x) = f(x) + g(x)$; what differs is the second operation, which multiplies a function by a scalar rather than two functions together, so the two are not special cases of one another.

• This is where non-domains become plentiful. Products of two nonzero rings (The product ring $R \times S$ with componentwise operations, its identity $(1_R, 1_S)$ and its units $R^{\times} \times S^{\times}$) and function rings on a set with at least two points are the two standard sources of zero divisors, and neither needs any arithmetic beyond $0 \cdot y = 0$.

• The empty index set is a genuine case, not an edge case to be waved away. $R^{\varnothing}$ has exactly one element and is the one-element ring, in which $1 = 0$; that is why the commutativity statement above carries the hypothesis $X \ne \varnothing$ in the direction where it is needed, and why the zero-divisor statement asks for two distinct points.