The intersection of a nonempty collection of subrings is a subring

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


Let R be a ring and let A be a nonempty set of subrings of R.

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

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