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 $$R$$.