Hello I am teaching myself algebra and have come across this fact.
For coprime ideals $ I_1,\dots, I_n$ of $R$ and $R$-module $M$, we have $(I_1 I_2 \cdots I_n) M= I_1 M \cap \cdots\cap I_n M$.
One direction is trivial, but I am having considerable difficulty showing that $I_1 M \cap ... I_n M \in (I_1 I_2 .. I_n) M$.
Note that $(I_1...I_n)M= \{\sum i_1 ... i_n m:$ finite sums with $ i_j \in I_j, m \in M \}$. Coprime implies that for each $i,j$ not equal, there is an $k_i \in I_i, k_j \in I_j$ such that $k_i + k_j = 1$
I can't manage to show it for two of them hence can't do an induction. Would anybody be willing to help me? Thanks in advance.
If $n=2$, then there exist $k_1\in I_1,\,k_2\in I_2$ such that $k_1+k_2=1$. Hence any $x\in I_1M\cap I_2M$ is equal to $(k_1+k_2)x\in(I_1I_2)M$.
Suppose it holds for $n=m-1$. Then for $x\in I_1M\cap\cdots\cap I_mM$, by the induction hypothesis we have $x\in(I_1\cdots I_{m-1})M$. Also, we can multiply the equations $k_i+k_m=1$ together to find $l_1+l_2=1$ for $l_1\in I_1\cdots I_{m-1}$ and $l_2\in I_m$. Now $x=(l_1+l_2)x\in(I_1\cdots I_m)M$.
Hope this helps.