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

$0_S = 0_R, \qquad 1_S = 1_R, \qquad -_S x = -_R x \quad (x \in S).$

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 $gh^{-1} \in H$ 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.