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 ringR 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, thereforeb \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 condition 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, S being subset of R, associative and commutative laws of multiplication over addition holds in S. Thus S is a subring of R.