Let $M,N$ be submodules of an $R$-module $L$. We know that $M+N = \langle \alpha_1, \ldots, \alpha_r \rangle$ and $M\cap N = \langle \beta_1, \ldots, \beta_s \rangle$. Is it possible that some generators of $M+N$ are not in $M$ or $N$? That is, is it possible for $(M\cup N) \cap \{\alpha_1, \ldots, \alpha_r\} \subsetneq \{\alpha_1, \ldots, \alpha_r\} $?
For context, I am trying to prove that under the given conditions, $M$ and $N$ are also finitely generated. My idea is to form a generating set of $M$ from the $\alpha_i \in M\setminus N$ (which may be empty) and the $\alpha_i \in M\cap N$ (which is non-empty) but it seems like this method may miss some of the $\alpha_i$ that are necessary to generate $M$ (i.e. the ones mentioned above). Am I working in the right direction? If not, what is a better approach?
Since $\alpha_i\in M+N$, we can write $\alpha_i=m_i+n_i$ for $m_i\in M$ and $n_i\in M$.
You can show that $\left\{m_1,\ldots,m_r,\beta_1,\ldots,\beta_s\right\}$ generates $M$ and that $\left\{n_1,\ldots,n_r,\beta_1,\ldots,\beta_s\right\}$ generates $N$ (exercise).
As for the first question: We can take $M=\mathbb{Z}\oplus 0$, $N=0\oplus\mathbb{Z}$. Then $M+N=\mathbb{Z}\oplus\mathbb{Z}$ is finitely generated, but we can take generators which do not belong to $M$ nor $N$, for example $\alpha_1=(1,1)$ and $\alpha_2=(1,2)$.