Question

It can be proved that the union of two subspaces of a vector space is a vector space iff one contains the other, suppose this.
Does it also happen for R-submodules of an R-module?

Answer 1

Yes it also happens for R-modulw of an R-module.

Proof:

Let M be a left R-module over a ring R, with submodules N and K. To that if N ∪ K is a submodule of M, then either N ⊆ K or K ⊆ N.

If N ⊆ K we are done. If not, there exists x ∈ N with x ∈ K. We will show that K ⊆ N. Let y ∈ K. Since N ∪ K is a submodule, either x + y ∈ K or x + y ∈ N. The first cannot happen, since x+y ∈ K implies x = (x+y)−y ∈ K, a contradiction. Therefore we must have x + y ∈ N, so y = (x + y) − x ∈ N.

Conversely if N ⊆ K, then  N ∪ K = K and it is submodule of M. Similerly if  K ⊆ N, then  N ∪ K = N and it is also a submodule of M.

Hence if N and K are submodule of an R-module M then  N ∪ K is submodule of M if and only if N ⊆ K and K ⊆ N.