Subring: a subset containing $1_R$ and closed under addition, additive inverses and multiplication

Let $R$ be a ring (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides) with zero $0_R$ and identity $1_R$. A subset $S\subseteq R$ is a subring of $R$ when

• (T1) $1_R\in S$;
• (T2) $x,y\in S$ implies $x+y\in S$;
• (T3) $x\in S$ implies $-x\in S$;
• (T4) $x,y\in S$ implies $xy\in S$.

Why $S$ is then a ring, with the same zero and the same identity. From (T1) and (T3), $-1_R\in S$, and from (T2), $0_R=1_R+(-1_R)\in S$. So $S$ contains $0_R$, is closed under addition and closed under additive inverses, which are exactly conditions (S1), (S2) and (S3) of Subgroup for the abelian group $(R,+,0_R)$; hence $S$ is a subgroup of $(R,+,0_R)$, and Subgroup states that a subgroup with the restricted operation is itself a group, whose identity and whose inverses are those of the ambient group. So $(S,+,0_R)$ is an abelian group, commutativity being inherited.

By (T4) multiplication restricts to a binary operation on $S$ (Binary operation on a set; associativity, commutativity, and a subset closed under the operation), and that restriction is associative because it is associative on $R$. By (T1) the element $1_R$ lies in $S$ and satisfies $1_{R}x=x=x{1}_R$ there, so it is a two-sided identity for the restricted multiplication (Left identity, right identity, and two-sided identity for a binary operation); thus $(S,\cdot ,1_R)$ is a monoid. Both distributive laws hold in $S$ because they hold in $R$ and are equations between elements of $S$.

Therefore $(S,+,\cdot ,0_R,1_R)$ is a ring, and

$0_S=0_R,1_S=1_R,{-}_{S}x={-}_{R}x(x\in S)\mathrm{.}$

This last line is not decoration: it is what makes a statement proved in $S$ transportable to $R$ without translation, and it is used when a subfield of an ordered field is shown to inherit the order and when scalars are restricted along a subfield inclusion.

• (T1) is not redundant, and it is forced by the ring convention. Because Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides builds the identity into a ring, a subring must carry the same identity, and a subset closed under (T2), (T3) and (T4) need not contain it. The companion page exhibits the standard witness: the even integers are closed under addition, negation and multiplication and do not contain $1$. Such a subset is a non-unital ring, not a subring.

• A subring need not be an ideal and an ideal need not be a subring. Nothing on this page defines ideals; the point of the remark is only that closure under multiplication by arbitrary elements of $R$ is not asked for here.

• The economical test. Conditions (T2) and (T3) can be replaced by the single condition $x-y\in S$, exactly as One-step subgroup test: a nonempty $H\subseteq G$ is a subgroup iff $g{h}^{-1}\in H$ for all $g,h\in H$; the identity and the inverses of $H$ are then those of $G$ replaces the three subgroup conditions by one; that is Subring criterion: $S\subseteq R$ is a subring if and only if $1_R\in S$ and $a-b\in S$ and $ab\in S$ for all $a,b\in S$; and an intersection of subrings is a subring, which also shows that an intersection of subrings is a subring.