Intersection of Submodules is Submodule/General Result

Theorem
Let $R$ be a ring.

Let $\struct {G, +_G}$ be an abelian group.

Let $M = \struct {G, +, \circ}_R$ be an $R$-module. Let $S$ be a set of submodules of $M$.

Then the intersection $\ds \bigcap S$ is a submodule of $M$.

Proof
From Intersection of Subgroups is Subgroup:General Result, it follows that $\ds \bigcap S$ is a subgroup of $M$.

As a subgroup is closed for its operation, it follows that for all $x,y \in \ds \bigcap S$, we have $x+y, y+x \in \ds \bigcap S$.

As $M$ is an $R$-module, and the addition $+$ on $\ds \bigcap S$ is the restriction of the addition on $M$, it follows that $x+y = y+x$.

Hence, $\ds \bigcap S$ is an abelian group.

Let $\lambda \in R$, and let $x \in \ds \bigcap S$.

Let $H$ be a submodule of $M$ such that $H \in S$.

By definition of intersection, it follows that $x \in H$.

As a submodule is closed for scalar product, it follows that $\lambda \circ x, x \circ \lambda \in H$.

As this is true for all submodules in the set $S$, it follows that $\lambda \circ x, x \circ \lambda \in \ds \bigcap S$.

Hence, $\ds \bigcap S$ is closed for scalar product.

Next, we show that $\ds \struct {\bigcap S, +, \circ}_R$ satisfy the left module axioms, and the right module axioms.

By definition of $R$-module, each of these axioms are satisfied by $\struct {G, +, \circ}_R$.

As the addition $+$ and the scalar multiplication $\circ$ on $\ds \bigcap S$ are the restrictions of the operations $+, \circ$ on $M = \struct {G, +, \circ}_R$, it follows that the axioms also are satisfied by $\ds \struct {\bigcap S, +, \circ}_R$.

For instance, the can be shown by:

Hence, $\ds \struct {\bigcap S, +, \circ}_R$ is a submodule of $M$.