# Intersection of Subrings is Subring

Jump to navigation Jump to search

## Theorem

Let $\struct {R, +, \circ}$ be a ring.

Let $\mathbb L$ be a non-empty set of subrings of $R$.

Then the intersection $\displaystyle \bigcap \mathbb L$ of the members of $\mathbb L$ is itself a subring of $R$.

## Proof

Let $\displaystyle L = \bigcap \mathbb L$.

By Intersection of Subgroups is Subgroup, $\struct {L, +}$ is a subgroup of $\struct {R, +}$.

By the One-Step Subgroup Test:

$\forall x, y \in \struct {L, +}: x + \paren {-y} \in L$

By Intersection of Subsemigroups, $\struct {L, \circ}$ a subsemigroup of $\struct {R, \circ}$.

So by definition of subsemigroup:

$\forall x, y \in \struct {L, \circ}: x \circ y \in L$

By the Subring Test it follows that $\struct {L, +, \circ}$ is a subring of $\struct {R, +, \circ}$.

$\blacksquare$