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 $$R$$.
Saif
September 7 @ 9:57 am
Great work, because first you prove the set Sl intersection S2 is non-empty