The product ring $R\times S$ with componentwise operations, its identity $(1_R,1_S)$ and its units $R^{\times }\times S^{\times }$

Let $R$ and $S$ be rings (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides). The product ring $R\times S$ is the cartesian product of the underlying sets with the componentwise operations

$(a,b)+(a^{\mathrm{\prime }},b^{\mathrm{\prime }}):=(a+a^{\mathrm{\prime }},b+b^{\mathrm{\prime }}),(a,b)\cdot (a^{\mathrm{\prime }},b^{\mathrm{\prime }}):=(a{a}^{\mathrm{\prime }},b{b}^{\mathrm{\prime }}),$

and the distinguished elements

$0_{R\times S}:=(0_R,0_S),1_{R\times S}:=(1_R,1_S)\mathrm{.}$

These data make $R\times S$ a ring. Both rules take a pair of elements of $R\times S$ to an element of $R\times S$, so each is a binary operation. Every ring axiom is an equation between elements of $R\times S$, and two such elements are equal exactly when their components are; so each axiom holds componentwise from the corresponding axiom in $R$ and in $S$. Explicitly: $(R\times S,+,(0_R,0_S))$ is an abelian group with $-(a,b)=(-a,-b)$; multiplication is associative with two-sided identity $(1_R,1_S)$; and both distributive laws hold.

Units. An element $(u,v)$ is a unit of $R\times S$ if and only if $u$ is a unit of $R$ and $v$ is a unit of $S$, in which case $(u,v{)}^{-1}=(u^{-1},v^{-1})$. Indeed, if $u{u}^{\mathrm{\prime }}=1_R=u^{\mathrm{\prime }}u$ and $v{v}^{\mathrm{\prime }}=1_S=v^{\mathrm{\prime }}v$ then $(u,v)(u^{\mathrm{\prime }},v^{\mathrm{\prime }})=(1_R,1_S)=(u^{\mathrm{\prime }},v^{\mathrm{\prime }})(u,v)$; conversely, if $(u,v)(u^{\mathrm{\prime }},v^{\mathrm{\prime }})=(1_R,1_S)=(u^{\mathrm{\prime }},v^{\mathrm{\prime }})(u,v)$ then reading each component gives $u{u}^{\mathrm{\prime }}=1_R=u^{\mathrm{\prime }}u$ and $v{v}^{\mathrm{\prime }}=1_S=v^{\mathrm{\prime }}v$. So

$(R\times S{)}^{\times }=R^{\times }\times S^{\times },$

and by The units of a ring are the invertible elements of its multiplicative monoid, and $R^{\times }$ is a group under multiplication; $0\in R^{\times }$ only in the zero ring that set is a group under multiplication (Left inverse, right inverse, and invertible element of a monoid).

Commutativity. $R\times S$ is commutative (Commutative ring) if and only if both $R$ and $S$ are. If both are, the componentwise product is commutative. Conversely, if $R\times S$ is commutative then for $a,a^{\mathrm{\prime }}\in R$, $(a{a}^{\mathrm{\prime }},1_S)=(a,1_S)(a^{\mathrm{\prime }},1_S)=(a^{\mathrm{\prime }},1_S)(a,1_S)=(a^{\mathrm{\prime }}a,1_S)$, so $a{a}^{\mathrm{\prime }}=a^{\mathrm{\prime }}a$; the same argument in the second component settles $S$.

Zero divisors. If $1_R\mathrm{\ne }{0}_R$ and $1_S\mathrm{\ne }{0}_S$, then $R\times S$ has zero divisors and is therefore never an integral domain (Zero divisor, and integral domain: a commutative ring with $1\mathrm{\ne }0$ and no zero divisors), whatever $R$ and $S$ are. Indeed $(1_R,0_S)$ and $(0_R,1_S)$ are both nonzero, and

$(1_R,0_S)\cdot (0_R,1_S)=(1_R{0}_R,0_S{1}_S)=(0_R,0_S),$

using $x\cdot 0=0=0\cdot x$ (In any ring $0\cdot a=a\cdot 0=0$, $(-a)b=a(-b)=-(ab)$, $(-a)(-b)=ab$, $(-1)a=-a$ and $a(b-c)=ab-ac$).

• The product of two domains is never a domain, by the zero-divisor computation above, since a domain has $1\mathrm{\ne }0$ by definition. This is the cleanest source of zero divisors available at this point, and the companion page records the instance $\mathbb{Z}\times \mathbb{Z}$.

• The two projections are ring homomorphisms; the standard injections need not be, and fail when the other factor is nonzero. The maps $(a,b)\mapsto a$ and $(a,b)\mapsto b$ satisfy (RH1), (RH2) and (RH3) of Ring homomorphism: additive, multiplicative, and required to send $1$ to $1$. The map $a\mapsto (a,0_S)$ satisfies (RH1) and (RH2) but sends $1_R$ to $(1_R,0_S)$, which is not the identity of $R\times S$ when $1_S\mathrm{\ne }{0}_S$; the companion page uses exactly that map to show (RH3) is not redundant.

• Only the product of two rings is defined here, which is all this page and its companion use. Nothing below needs an indexed family.