$IM$ is a submodule of M? I would like to know if $IM$ is a submodule of $M$, I'm trying to use the submodule criterion which is: 

$G$ is a submodule of $M$ iff $G\neq \varnothing$ and whenever
  $g,g'\in G$ and $r,r'\in R$, then $rg+r'g'\in G$.

I'm having problems to prove this, I need some help.
 A: Let $R$ be a ring, $I$ a left ideal of $R$ and $M$ a left $R$-module.
In general, the set
$$
S=\{\,ax:a\in I, x\in M\,\}
$$
is not even closed under addition. Thus the notation $IM$ denotes the set of expressions
$$
\sum_{k=1}^n a_k x_k
$$
where $n$ is an arbitrary natural number and $a_k\in I$, $x_k\in M$ $(k=1,\dots,n)$.
Thus, it's obvious from the definition that $IM$ is closed under addition. The only thing to show is that it's closed under multiplication by scalars; but, if $r\in R$,
$$
r\Bigl(\sum_{k=1}^n a_k x_k\Bigr)=
\sum_{k=1}^n (ra_k) x_k
$$
which belongs to $IM$ because $ra_k\in I$ by definition of left ideal.
For an example when that set $S$ is not closed under addition, you can take the polynomial ring $R=\mathbb{Z}[x]$ and $I=M=(2,x)$ (the ideal with the generators $2$ and $x$). Then $2\cdot 2$ and $x\cdot x$ belong to the set $S$, but $4+x^2$ doesn't, because it can't be expressed as a product $fg$ with $f,g\in I$.
A: Presumably $I$ is a left ideal of $R$. Then $IM$ is the additive subgroup of $M$ generated by elements of the form $am$ where $a \in I$ and $m \in M$. Therefore you only need to show that this subgroup is closed under scalar multiplication by $R$.
A: I was also confused by this, and before I've read that $IM$ actually means the submodule generated by set $\{ am | a \in I, m \in M \}$, not the set itself, I proved the following:
For a commutative ring $R$ a and any ideal $I$ of $R$ (here we can forget about left or right) it holds, that $IM$ (as the set itself) forms a submodule for any $R$-module $M$ iff $I$ is principal.
The proof is trivial in one direction. In the other take ideal $I = (a, b)$ and $R$-module $R[x]$ and study the set $(a, b) R[x]$. You will soon find out that this set is not closed under addition (take the element $ax + b$). It can be proved that if $ax + b \in (a, b) R[x]$, then $(a, b)$ can be generated by one element.
