# 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 $R$.