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In theorem 6.4 of Steven Roman's Advanced Linear Algebra, he shows:

If $M$ is a module over a PID $R$ and $u_1, \dots, u_n\in M$ have coprime orders, then the sum $$\langle\!\langle u_1\rangle\!\rangle + \cdots + \langle\!\langle u_n\rangle\!\rangle $$ is direct.

His argument is: suppose $v_i\in \langle\!\langle u_i\rangle\!\rangle$ and $$v_1 + \cdots +v_n = 0$$ Then the order of $\langle\!\langle v_1 + \cdots + v_n\rangle\!\rangle$ is $1$, but this is impossible unless all $v_i$ are zero.

My question is: why is this only possible if all $v_i$ are $0$? Presumably this follows from the fact that the orders of the $u_i$ are coprime, but I can't figure out how.

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  • $\begingroup$ What do you know about the Chinese Remainder Theorem? $\endgroup$ Commented May 8 at 16:37
  • $\begingroup$ @SammyBlack I know the basics. I used it to show that $v_1 + \cdots v_n = c(u_1 + \cdots u_n)$ for some $c\in R$, from which it follows that the product of the orders of the $u_i$ divide $c$. But I don't know if this helps me. $\endgroup$
    – user236343
    Commented May 8 at 16:43

1 Answer 1

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Let the order of $u_i$ be $m_i$. Let $M_i = m_1\cdots m_n/m_i$.

Then $M_i(v_1+\cdots +v_n)= M_iv_i$ has the same order as $v_i$ (because it has order $\mathrm{order}(v_i)/\gcd(M_i,\mathrm{order}(v_i))$). But $M_i(v_1+\cdots +v_n)=0$, so it has order $1$. Thus, $v_i$ has order $1$, hence $v_i=0$.

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