# Proving that the intersection of two subrings of R is also a subring of R

If $$R_1$$ and $$R_2$$ are both subrings of $$R$$ , how to prove that $$R_1 \cap R_2$$ is also a subring of $$R$$.

here is my attempt

(1)Since $$R_1$$ is a subring of $$R$$ then it must contain zero (identity with respect to addition) same thing for $$R_2$$ and so $$R_1 \cap R_2$$ must have at least zero so it is nonempty.

(2)Now how to prove that if $$x,y \in R_1 \cap R_2$$ then $$x-y \in R_1 \cap R_2$$ and $$xy \in R_1 \cap R_2$$ how can we guarantee that?

If $x, y\in R_1\cap R_2$, then $x,y\in R_1$ and $x,y \in R_2$. Now look at $x-y$ and $xy$ again and see if you can't figure out why they are also in $R_1\cap R_2$.
Let $R_1$ be a subring in $R$ and $R_2$ be a subring in $R$.
Allow $x, y$ belong in $R_1\bigcap R_2$. So $x \in R_1$ and $y \in R_2$, also $y \in R_1$ and $x \in R_2$ Thus $x-y \in R_1$ and $x-y \in R_2$ This means $x-y$ belongs to $R_1 \bigcap R_2$.
Since $R_1$ and $R_2$ are rings they posses an identity each so they cannot be empty.
• Your answer might be easier to read if you used MathJax to format your formulas. For example, $a_1+b^2=c$ produces $a_1+b^2=c$. See this guide for details: meta.math.stackexchange.com/q/5020/166535 Commented Mar 17, 2015 at 16:00