Before enunciating and trying to prove the theorem, here are some assumptions and definitions:
Definitions:
- $\text{ann}(m) = \{ r \in R \mid rm = 0 \text{ for } m \in M \} $
- $\text{ann}(M) = \{ r \in R \mid rM = \{0\} \} $
- $M(p_i) = \{ x \in M \mid p^kx = 0 \}$ for some $k \in \mathbb{N}^{+}$.
Assumptions:
Let $R$ be a ring that is a PID, $M\neq 0$ a $R$ torsion module that is finitely generated. Suppose that $\{m_1,\cdots,m_n\}$ generates $M$. Now define $d = r_1 \cdots r_n$ where $r_i \in \text{ann}(m_i)$ for $i \in \{1, \cdots , n\}$. Since $R$ is a PID it's a UFD. From that it follows that $d=p_1^{k_1}\cdots p_m^{k_m}$ for $k_i \in \mathbb{N}^+$ and $p_i \in R$ primes.
The theorem states that:
If $\text{ann}(M) = Rd$ then $M(p_i) \neq 0$ for $i \in \{1, \cdots , m \}$.
And here is the attempt to prove it:
It was previously proved that with those assumptions it is true that $$ M = M(p_1) \oplus \cdots \oplus M(p_m) $$ Now, aiming for a contradiction, suppose that for a $j \in \{1, \cdots , m \}$ it's true that $M(p_j) = \{0\}$. Therefore $\frac{d}{p_j^{k_j}} \in \text{ann}(M)$, and from that it follows that $d \mid \frac{d}{p_j^{k_j}}$ which implies that $\exists \alpha \in R$ such that $\alpha d = \frac{d}{p_j^{k_j}}$, hence:
$$ \alpha d = \alpha \cdot p_1^{k_1}\cdots p_m^{k_m} = \frac{p_1^{k_1}\cdots p_m^{k_m}}{p_j^{k_j}} \implies \alpha p_j^{k_j} = 1 \implies p_j^{k_j} \in R^X \implies p_j \in R^X $$ And that is a contraction since $p_j \in R$ is a prime, and by definition, $p_j$ can't be a unity in $R$.
Can someone please check if my proof is correct? Thanks!