Intersection of Subrings

Theorem:

The intersection of two subrings is a subring.

Proof:

Let $${S_1}$$ and $${S_2}$$ be two subrings of ring $$R$$.

Since $$0 \in {S_1}$$ and $$0 \in {S_2}$$ at least $$0 \in {S_1} \cap {S_2}$$. Therefore $${S_1} \cap {S_2}$$ is non-empty.

Let $$a,b \in {S_1} \cap {S_2}$$, then
$$a \in {S_1} \cap {S_2} \Rightarrow a \in {S_1}$$ and $$a \in {S_2}$$

and
$$b \in {S_1} \cap {S_2} \Rightarrow b \in {S_1}$$ and $$b \in {S_2}$$

But $${S_1}$$ and $${S_2}$$ are subrings of $$R$$, therefore
$$a,b \in {S_1} \Rightarrow a – b \in {S_1}$$ and $$ab \in {S_1}$$

and
$$a,b \in {S_2} \Rightarrow a – b \in {S_2}$$ and $$ab \in {S_2}$$

Consequently, $$a,b \in {S_1} \cap {S_2} \Rightarrow a – b \in {S_1} \cap {S_2}$$ and $$ab \in {S_1} \cap {S_2}$$.

Hence, $${S_1} \cap {S_2}$$ is a subring of $$R$$.