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

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

with distinguished elements the constant functions

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

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 $X\mathrm{\ne }\mathrm{\varnothing }$ 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 $X=\mathrm{\varnothing }$ 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 $X\mathrm{\ne }\mathrm{\varnothing }$ and is stated with it.

Zero divisors. Suppose $R$ is not the one-element ring, that is $1_R\mathrm{\ne }{0}_R$ (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 $x_1\mathrm{\ne }{x}_2$. Define $f,g\in R^X$ by

$f(x):=\{\begin{array}{ll}{1}_R& x=x_1\\ 0_R& x\mathrm{\ne }{x}_1\end{array}g(x):=\{\begin{array}{ll}{0}_R& x=x_1\\ 1_R& x\mathrm{\ne }{x}_1\mathrm{.}\end{array}$

Then $f\mathrm{\ne }{0}_{R^X}$ because $f(x_1)=1_R\mathrm{\ne }{0}_R$, and $g\mathrm{\ne }{0}_{R^X}$ because $g(x_2)=1_R\mathrm{\ne }{0}_R$; 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 $1\mathrm{\ne }0$ 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. $R^{\mathrm{\varnothing }}$ has exactly one element and is the one-element ring, in which $1=0$; that is why the commutativity statement above carries the hypothesis $X\mathrm{\ne }\mathrm{\varnothing }$ in the direction where it is needed, and why the zero-divisor statement asks for two distinct points.