> [!NOTE] **Theorem** (Subring Test) > Let $(R,+,\times)$ be a [[Rings|ring]]. A subset $S$ of $R$ is a [[Subring|subring]] iff it satisfies the following conditions > > $(1)$ Additive (& multiplicative) identity(ies): $0\in S$ (& $1\in S$). > > $(2)$ Closure under addition: $\forall a,b\in S: a+b \in S.$ > > $(3)$ Additive inverses: $\forall a : -a \in S$ > > $(4)$ Closure under multiplication: $\forall a,b \in S: a\times b \in S.$ > **Proof.** ($\implies$) Suppose that the subset $S$ is a subring of $R.$ Then, by definition of a ring, $(S,+)$ is a subgroup of $(R,+)$. So by [[Two-Step Subgroup Test|two step subgroup test]], we have that $(1)$-$(3)$ are true in the case that $R$ is ring without a unity. From the definition of a subring that the multiplicative identity of $R$ is in $S$ hence $(1)$ is true if $R$ is ring with unity. Also from the definition of a subring, if $a,b \in S$ then $ab \in S$, so $(4)$ is true. ($\impliedby$) Conversely, suppose $S$ is a subset of $R$ and the above conditions are true. By [[Two-Step Subgroup Test|two-step subgroup test]], $(S,+)$ is a subgroup of $(R,+)$ since $(1)-(3)$ hold. Commutativity of addition holds in $S$ since it holds in $R$. Hence $(S, +)$ is an abelian group. If $R$ is a ring with unity, then $(1)$ ensures $1\in S.$ Condition $(4)$ ensures closure under multiplication. Associativity of Multiplication & Distributivity hold in $S$ since they hold in $R$.