Meaning of $IM$ where $I$ is an ideal of $R$ and $M$ is a left $R$ - module?

Question

I am doing exercise $13$ in section $10.2$ in Dummit & Fote. The relevant details are:

• $R$ is a commutative ring with identity

• $I$ is a nilpotent ideal of $R$

• $M$ is an $R$ - module

The exercise refers to $M/IM$. I can only interpret $IM$ as $\{i \cdot m| i \in I, m \in M\}$

So I guess my questions are:

$1)$ Is my interpretation of $IM$ correct?

$2)$ If yes, how do we show that $IM$ is a submodule?

I can show $IM$ closed under the ring action: $r \cdot (i \cdot m) = (ri) \cdot m = i_0 \cdot m$ with $i_0 \in I$. However, I don't know how to show that it's a subgroup. I know that I want $i_1 \cdot m_1 - i_2 \cdot m_2 = i_3 \cdot m_3$, but I'm stuck.

Thanks for the help.

Answer 1

No, your definition is not quite right. However, it is easily fixed: $IM=\{x_1m_1+\cdots+x_nm_n: x_1, \ldots, x_n\in I, m_1, \ldots, m_n\in M\}$. Using this definition, $IM$ is clearly a submodule of $M$.