# Generator of direct product of modules is direct product of each generator?

Let assume I have an R-module M over a ring $$R$$ with unit and I want to consider the direct product (or direct sum, since I understand that they are both the same when the product has finitely many factors) $$M \oplus M$$. Furthermore, let $$N_1 \oplus N_2 \subseteq M \oplus M$$ be a submodule in $$M \oplus M$$ which is generated by \begin{align*} \left\langle \begin{pmatrix} a_1 \\ b_1 \end{pmatrix}, \begin{pmatrix} a_2 \\ b_2\end{pmatrix} \right\rangle , a_i, b_i \in M \end{align*} and lets assume I want to understand the structure of this submodule. My question is: Can I instead study the generator \begin{align*} \left\langle a_1 , a_2 \right\rangle\oplus \left\langle b_1 , b_2 \right\rangle , \end{align*} and is it true that \begin{align*} N_1 \oplus N_2 = \left\langle a_1 , a_2 \right\rangle\oplus \left\langle b_1 , b_2 \right\rangle. \end{align*}

Sorry if this is a stupid question but I am not very familiar with these concepts. Thank you in advance!

EDIT: My intuition tells me that this is not possible. And if you know helpful literature regarding this topic and tell me some of them, it will be appreciated.

$$M=\mathbb{Z}$$ , $$N_1=0, N_2=M$$ . then $$0 \oplus \mathbb{Z} \subseteq \mathbb{Z} \oplus \mathbb{Z}$$ now take : \begin{align*} S= \left\langle \begin{pmatrix} 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 \\ 1\end{pmatrix} \right\rangle \end{align*} $$S$$ is isomorphic to $$0 \oplus \mathbb{Z}$$ . in fact $$0 \oplus \mathbb{Z}$$ is isomorphic to $$\mathbb{Z}$$ and $$\mathbb{Z}$$ is isomorphic to $$S=\left\langle (1 , 1) \right\rangle$$ via the map : $$x\rightarrow (x,x)$$ .
Now do we have : \begin{align*} 0 \oplus \mathbb{Z} = \left\langle 0 , 1 \right\rangle\oplus \left\langle 0 , 1 \right\rangle. \end{align*} The answer is obviously a no, since if we had this, then $$\mathbb{Z}=\mathbb{Z} \oplus \mathbb{Z}$$ which is not true.