Property of unfaithful module over PID

Question

Proposition

Let $R$ be a PID and $M$ a torsion module over $R$. Suppose $M$ is not faithful, with $\operatorname{Ann(M)}=Ra$ and $a=u{p_1}^{\alpha_1}...{p_n}^{\alpha_n}$ an irreducible factorization of $a$. Then $$M=\bigoplus M[p_i]$$ (here $M[p_i]=Ker(f:M \to M, m \rightarrow {p_i}^{\alpha_i}x$).

I got stuck trying to show $\sum=\bigoplus$, and I don't know what to do to prove $M=\sum M[p_i]$. I'll write what I did up to now.

-$\sum=\bigoplus$:

Let $x_i \in M[p_i] : \sum_{i=1}^n x_i=0$. I define $g_i=\prod_{j \neq i}{p_j}^{\alpha_j}$, then $\prod_{j \neq i}{p_j}^{\alpha_j}x_i=g_i(\sum_{i=1}^n x_i=0)=0$.

I want to show $x_i=0$. I know that $x_i \in M[p_i] \cap M[\prod_{j \neq i}{p_j}^{\alpha_j}]$ but I couldn't deduce from here that $x_i=0$.

Any suggestions to prove this proposition and help me where I got stuck would be appreciated.

Answer 1

For the time being assume $n = 2$. Since $R$ is a PID, the ideals $(p_1^{\alpha_1})$ and $(p_2^{\alpha_2})$ are comaximals. So there exists $r_1, r_2 \in R$ such that $r_1p_1^{\alpha_1} + r_2p_2^{\alpha_2} = 1$. Thus, for any $m \in M$, we can write $m = r_1p_1^{\alpha_1}m + r_2p_2^{\alpha_2}m$ and so we have, $p_1^{\alpha_1}m \in M[p_2], p_2^{\alpha_2}m \in M[p_1]$. This proves that $M = M[p_1] + M[p_2]$. Now let $m \in M$ be such that $m \in M[p_1]\bigcap M[p_2]$. Then $p_1^{\alpha_1}m = 0$ and $p_2^{\alpha_2}m = 0$. So, $m = r_1p_1^{\alpha_1}m + r_2p_2^{\alpha_2}m = 0$. This shows that $M = M[p_1] \oplus M[p_2]$.

Now for the general case, note that the ideals $(p_1^{\alpha_1})$ and $(\Pi_{i \neq 1}p_i^{\alpha_i})$ are comaximals. We can use inductions.

Answer 2

You can use unique decomposition into prime factors (up to units, of course).

Since the $p_i^{\alpha_i}$ have gcd equal to $1_R$ (up to a unit), the same is true for the $g_i$. So the ideal generated by the $g_i$ contains $1_R$. So we can write $$ 1_R = \sum_{i=1}^nh_ig_i$$ for some suitable $h_i \in R$. Put $$ q_i := h_ig_i.$$ Then we have (or it's easy to see) that $$ 1_R = \sum q_i $$ and $$ q_iq_j \equiv 0 \mod a $$ and thus $$ q_i^2-q_i = q_i(1_R-q_i) = q_i\sum_{j\neq i}q_j \equiv 0 \mod a$$ and thus $$ q_i^2 \equiv q_i \mod a $$ All this shows that the $q_i$ are a complete set of orthogonal idempotents mod $a$. Moreover, $q_i$ an $p_i$ are relatively prime, since otherwise $p_i$ would divide $1_R$, which is not true. This also implies $$ q_i\not\equiv 0 \mod a. $$ From all this, we get $$ M = 1_RM = \sum q_iM = \bigoplus q_iM, $$ where the directness of the sum follows from the orthogonality of the $q_i$ mod $a$. Since $q_i \not\equiv 0 \mod a$, we have $q_iM \neq 0$. Also, $p_i^{\alpha_i}q_iM = 0$ since $a$ divides $p_i^{\alpha_i}q_i$ and $p_i^{\alpha_i}q_jM \neq 0$ for $i \neq j$ since $a$ doesn't divide $p_i^{\alpha_i}q_j$ since $p_j$ doesn't divide $p_i^{\alpha_i}q_j$ (see above). This shows $$ M[p_i] = q_iM. $$ So we have the desired decomposition of $M$.