# Submodules of a finitely generated module is contained in a maximal ideal

I need to prove the following fact: Let R be a unital ring, M is a finitely generated R-module. Submodules of a finitely generated $R$-module is contained in a maximal ideal. I found the following post that deals with the same problem.

Maximal submodule in a finitely generated module over a ring

However, I don't quite understand the proof given:

1. Why we can be sure that the submodules of the the $R$-module forms a chain? My interpretation of chain is that I have $N_1 \subseteq N_2 \subseteq .....$. What happens if some of the submodules is not finitely generated? Am I misunderstanding the definition of chain?

2. Am I correct to say that the maximal submodule is just the union of all the submodules of the $R$-module?

Thanks

As for the first call $X$ the collection of proper submodules (i.e. not containing $M$) that contain a given submodule $M'$. What we want to show is that $X$ is inductive, i.e. that every chain has an upper bound. Notice that we're not asserting that $X$ is a chain, but that every chain in $X$ has an upper bound. Of course for every chain the union of the elements of the chain is an upper bound (here is where you use the f.g. hypothesis, since if the union were the whole of $M$ then we can generate by finitely many elements, so these must have been included in the union of finitely many elements in the chain, a contradiction). Zorn's lemma now provides us with a maximal element, which in our context is precisely a maximal submodule.