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The intersection of a nonempty collection of subrings is a subring


Solution to Abstract Algebra by Dummit & Foote 3rd edition Chapter 7.1 Exercise 7.1.4

Prove that the intersection of any nonempty collection of subrings of a ring is also a subring.


Solution: Let $R$ be a ring and let $S_a$ be any nonempty collection set of subrings of $R$ indexed by $a\in A$.

Now $\bigcap_{a\in A} S_a \subseteq R$ is a subgroup, so it suffices to show that $\bigcap_{a\in A} S_a$ is closed under multiplication. To that end, let $x,y \in \bigcap_{a\in A} S_a$. Then $x,y \in S_a$ for all $a \in A$, and $xy \in S_a$ for all $a \in A$. Thus $$xy \in \bigcap_{a\in A}S_a,$$ and by definition $\bigcap_{a\in A} S_a \subseteq R$ is a subring.


Linearity

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