Submodule of a torsion module

Question

Let $R$ be a principal ideal domain. Let $M$ be a torsion module over $R$ annihilated by $a \in R$, i.e, $am = 0$ for all $m\in M$. Let $a= u {p_1}^{e_1} {p_2}^{e_2} \cdots {p_k}^{e_k} $ be the prime decomposition of $a$. Define $N_i$ to be the submodule of $M$ consisting of elements annihilated by some power of $p_i$. Dummit & Foote says (in p465) that $N_i = \{ x \in M | {p_i}^{e_i} x=0 \}$. However, I can't see why this holds. Why we cannot have an element $x$ in $M$ such that ${p_i}^{e_i} x \ne 0$ for some $e_i$ while ${p_i}^{m} x =0$ for some $m> e_i$?

Answer 1

Let $m \in N_1$, and write $p_1^{l} m = 0$ for some $l \geq1$. Let $$b = up_2^{e_2} \cdots p_n^{e_n}$$ Since $p_1^{l}, b$ are relatively prime, there exist $x, y \in R$ such that $xb + yp_1^l = 1$.

Multiply both sides of this equation by $p_1^{e_1}$ to get $xa + yp_1^{l+e_1} = p_1^{e_1}$, which implies $$p_1^{e_1} m = (xa)m + (yp_1^{l+e_1})m = 0$$

Answer 2

This is because, using the Chinese remainder theorem: \begin{align*}\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\ann{Ann} \ann_M a&\simeq\Hom_R(R/(a),M)\simeq\Hom_R\Bigl(\bigoplus_{i=1}^k R/(p_k^{e_k}),M\Bigr)\\ &\simeq\bigoplus_{i=1}^k\Hom\bigl(R/(p_k^{e_k}),M\bigr)\simeq\bigoplus_{i=1}^k\ann_M\bigl(R/(p_l^{e_k})\bigr) \end{align*}