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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?

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2 Answers 2

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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 $R_1\cap R_2$.

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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.
Now, the intersection of two subrings is a subring since all the axioms for a subring are satisfied.

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