Motivation: The intersection of two free submodules is rarely free. For example, the intersection of two principal ideals in a commutative ring need not be principal once more. The correct example (as such a claim is true in any UFD) is to consider a ring of integers that fails to have unique factorization. For example, take $\mathbb{Z}[\sqrt{-5}]$ and then the intersection of $(6)$ and $(2\cdot(1+\sqrt{-5}))$ is not principal since there can be no least common multiple (in part due to lack of unique factorization).
So the idea is to weaken the condition of being free. In this case, I say an $R$-module $M$ is reflexive if the obvious map $M \to M^{\vee\vee} = \operatorname{Hom}_R(\operatorname{Hom}_R(M,R),R)$ is an isomorphism, i.e. it is isomorphic to its own double dual.
Question: Assuming no conditions on the ring $R$ except that $R$ is commutative with identity, and $L$ is an $R$-module, is the intersection of two reflexive submodules $M_1,M_2\subseteq L$ necessarily reflexive? If this fails, what are the weakest set of conditions in which this can hold?
With added assumptions on the ring $R$ and $M$, e.g. $M$ is finitely generated over $R$ and $R$ is an integrally closed Noetherian domain, I believe one can easily show this using one of the many equivalent conditions of reflexivity.