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

$(a,b) + (a',b') := (a + a',\; b + b'), \qquad (a,b)\cdot(a',b') := (aa',\; bb'),$

and the distinguished elements

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

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 $uu' = 1_R = u'u$ and $vv' = 1_S = v'v$ then $(u,v)(u',v') = (1_R,1_S) = (u',v')(u,v)$; conversely, if $(u,v)(u',v') = (1_R,1_S) = (u',v')(u,v)$ then reading each component gives $uu' = 1_R = u'u$ and $vv' = 1_S = v'v$. 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 $a, a' \in R$, $(aa', 1_S) = (a,1_S)(a',1_S) = (a',1_S)(a,1_S) = (a'a, 1_S)$, so $aa' = a'a$; the same argument in the second component settles $S$.

Zero divisors. If $1_R \ne 0_R$ and $1_S \ne 0_S$, then $R \times S$ has zero divisors and is therefore never an integral domain (Zero divisor, and integral domain: a commutative ring with $1 \ne 0$ 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 $1 \ne 0$ 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 $1_S \ne 0_S$; 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.