# Why is the sum of cyclic submodules of coprime orders direct? [duplicate]

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.

• What do you know about the Chinese Remainder Theorem? Commented May 8 at 16:37
• @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. Commented May 8 at 16:43

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 $$v_i=0$$.