# Subrings

Let $R$ be a ring. A non –empty subset $S$ of the set $R$ is said to be a subring of $R$ if $S$ is closed under addition and multiplication in $R$ and $S$ itself is a ring for those operations.
If $R$ is any ring, then $\left\{ 0 \right\}$ and $R$ are always subrings of $R$. These are said to be improper subrings. The subrings of $R$ other than these two, if any, are said to be proper subrings of $R$.

Theorem: The necessary and sufficient condition for a non-empty subset $S$ of a ring$R$ to be a subring of $R$ are
(i) $a,b \in S \Rightarrow a - b \in S$
(ii) $a,b \in S \Rightarrow ab \in S$

Proof: To prove that the conditions are necessary let us suppose that $S$ is a subring of $R$. Obviously $S$ is a group with respect to addition, therefore$b \in S \Rightarrow - b \in S$.
Since $S$ is closed under addition,

Also $S$ is closed with respect to multiplication,

Now to prove that the condition are sufficient suppose $S$ is a non-empty subset of $R$ for which the conditions (i) and (ii) are satisfied.
From condition (i)

Hence additive identity is in $S$. Now
$0 \in S,\,a \in S \Rightarrow 0 - a \in S \Rightarrow - a \in S$
i.e. each element of $S$ possesses additive inverse.
Let $a,b \in S$ then $- b \in S$ and then from condition (i)

Thus $S$ is closed under addition, $S$ being subset of $R$, associative and commutative laws of multiplication over addition holds in $S$. Thus $S$ is a subring of $R$.