IM is a submodule of M

Question

In the fisrt answer of this question where $R$ is a ring, $I$ is a left ideal of $R$ and $M$ an $R$-module; I don't know why can't I see that $IM$ is closed under addition. If we take two elements $x$ and $y$ of $IM$ they will be in this form

$$x=\sum_{i=1}^{n}a_ix_i\ , y=\sum_{i=1}^{m}b_iy_i$$ where $a_i,b_i\in I$ and $x_i,y_i\in M$.

My question is why $x+y\in IM$ ?? Thank you for your time

Answer 1

For $i=1, \dots, n+m$ call $$c_i = \left\{ \begin{matrix} a_i & \mbox{ if } &i \leq n \\ b_{i-n} & \mbox{ if } &i \geq n+1 \end{matrix} \right. $$ and analogously $$w_i = \left\{ \begin{matrix} x_i & \mbox{ if } &i \leq n \\ y_{i-n} & \mbox{ if } &i \geq n+1 \end{matrix} \right.$$

Then $$x+y = \sum_{i=1}^{n+m} c_iw_i \in IM$$

Answer 2

Let $z_1=x_1,\dots,z_n=x_n,z_{n+1}=y_1,\dots,z_{n+m}=y_m$, and let $c_1,\dots,c_n,c_{n+1},\dots,c_{n+m}$ be defined similarly based on $a_i,b_j$. Then:$$x+y = \sum c_iz_i$$