I read the statement as $IM = M \implies M = 0$, when $I$ is in the jacobson radical.

1. Essentially, it tells us that if a module $M$ "lives by the $I$", then it also "dies by the $I$".
2. Alternatively, we factor the equation as $M(I - 1) = 0$. Since our ideal $I$is a member of the jacobson radical, $(1 - I)$ is "morally" a unit and thus $M = 0$. This is of course completely bogus, but cute nontheless.
3. We can think of a graded ring, say $R[x]$ acting on some graded module $M$ (say, a subideal, $M = (x^2)$). When we compute $IM$, this will bump up the grading of $M$. If $IM = M$, then $M$ could not have had non-trivial elements in the first place, since the vector of, say, "non-zero elements in each grade" which used to look like $(v_0, v_1, v_2, \dots)$ will now look like $(0, v_0, v_1, \dots)$. Equating the two, we get $v_0 = 0, v_1 = v_0 = 0, v_2 = v_1 = 0$ and so on, collapsing the entire ring.