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Let $A$ be a p.i.d. and $M$ an $A$-module. Let $x,y\in M$ and $\text{Ann}(x)=(m)$ and $\text{Ann}(y)=(n)$. We want to show there is a $z\in M$ such that $\text{Ann}(z)=(m)\cap(n)$. Since $A$ is a u.f.d. we can build $m_1,n_1$ such that $m_1n_1=\text{lcm}(m,n)$ and let $z=\frac{m}{m}_1x+\frac{n}{n_1}y$. I now want to show that $\text{Ann}(z)=(m_1n_1)$ but I don't know how to do this. What I tried is to assume $s\in\text{Ann}(z)$. We have to prove that $m|s$ and $n|s$. We have $\frac{sm}{m_1}x+\frac{sn}{n_1}y=0$. If both terms had to be zero the rest would be easy but I'm not sure whether that's even true.

Navid Rashidian
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    I recommend to name the variables differently. $m,n$ should be elements of $M$. Just saying this because I was quite confused when reading your approach for the first time. (But of course it's not necessary to rename these, from a formal point of view.) – Martin Brandenburg Dec 29 '24 at 13:27
  • By the way, where did you find this statement? It doesn't look correct to me (without further assumptions on $M$). – Martin Brandenburg Dec 29 '24 at 15:47
  • @MartinBrandenburg What do you recommend I put instead of them? – Navid Rashidian Dec 30 '24 at 13:19
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    @MartinBrandenburg My professor told me about this. I don't know where he found it himself. The special case for Abelian groups is well-known (e.g. see here). We actually completed the proof yesterday and I will post an answer to my question and then you could verify it. – Navid Rashidian Dec 30 '24 at 13:20

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