Problem: Let $I$ be a left ideal of a ring $R$ and $M$ a left $R$-module. Show that $IM = \{\sum_1^n x_im_i \mid n \geq 1 x\in I, m \in M\}$ is a submodule of $M$.
Proof Attempt: It is evident that $IM \subseteq M$, because $I \subseteq R$ and $M$ is closed under the action of $R$, any $xm \in M$ and since $M$ is an abelian group under addition, finite sums of those elements are in $M$. Note that $0_M \in IM$ evidently. Let $a,b \in IM$ and, without loss of generality, write $a = \sum_1^nx_im_i$ and $b = \sum_1^n y_im_i$. This is valid because we can add $0$ coefficient terms to fill in the missing summands. Then \begin{align*} a-b = \sum_1^nx_im_i - \sum_1^n y_im_i & = \sum_1^n (x_i-y_i)m_i, \end{align*} because $x_i,y_i \in I$ it follows that $x_i - y_i \in I$ as an ideal and so $a-b \in IM$. Therefore $IM \leq M$, and it remains only to show that $IM$ is closed under the action of $R$. Let $r \in R$ and $a = \sum_1^n x_im_i$, then \begin{align*} r(\sum_1^n x_im_i) & = \sum_1^n r(x_im_i) \\ & = \sum_1^n (rx_i)m_i \\ & = \sum_1^n x_i'm_i. \end{align*} therefore $ra \in IM$ and so $IM$ is a submodule of $M$.
Is this valid? In particular the approach that we can make the sums have the same length and same $m_i$ without loss of generality by adding in $0$ coefficient terms to the "shorter" of the two sums $a,b$? Intuitively I think it would work, but is there anywhere I'm forgetting an issue. One of the reasons I ask is the answer here does something I didn't think of at all. Thanks in advance for the clarification.