# 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 conditions 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,
$\begin{gathered} a \in S,\,b \in S \Rightarrow a \in S,\, – b \in S \Rightarrow a + \left( { – b} \right) \in S \\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow a – b \in S \\ \end{gathered}$

Also $S$ is closed with respect to multiplication,
$a \in S,\,b \in S \Rightarrow ab \in S$

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

From condition (i)
$a \in S \Rightarrow a – a \in S \Rightarrow 0 \in S$

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)
$0 \in S,\, – b \in S\, \Rightarrow a – \left( { – b} \right) \in S\, \Rightarrow \left( {a + b} \right) \in S$

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