Problem of Module on a PID

Question

Suppose that $R$ is a $\mathrm{PID}$. Suppose that $a$ and $b$ are nonzero elements of $R$ which are relatively prime. Let $M$ be an $R$ module so that $abM=\{0\}.$ Show that

1. $aM=M_b$ and $bM=M_a$
2. $M=M_a\oplus M_b$

Where $M_a=\{x\in M|ax=0\}$.

Answer 1

Here is a hint for both questions:

Since $a$ and $b$ relatively prime, $1= ak+bl$ for some $k,l\in R$.

Now obviously $bM\subset M_a$.

To get the other direction, let $x\in M_a$, and write $x=1\cdot x = (ak+bl)x$. I will let you proceed from there.

For the second part, again from the decomposition of $1$ and from the previous part we know that $M = aM+bM = M_b+M_a$. To show that the sum is direct, find what $M_a\cap M_b$ has to be.