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> For this homework exercise, we are asked to show that the ideal $I=(3,1+\sqrt{-5})$ is a flat $\mathbb{Z}[\sqrt{-5}]$-module. The hint is to show that $I$ becomes principal (and thus free as a module) when we invert $2$ **or** $3$, so that $I$ is locally flat.
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I'm having trouble understanding what happens to $I$ when we invert $2$ **or** $3$. I would say that if we invert $3$, then $3$ becomes a unit. How does this make $I$ principal? If we invert $2$, I don't see how this gets us anywhere.

Furthermore, I don't understand why it helps to prove that $I$ is locally flat. There is a lemma in our course notes saying that if $M$ is a finitely generated $R$-module in a local ring $R$, then $M$ is flat iff $M$ is free. So if $I$ is locally flat, why is it flat as well?

I hope my confusion is coming across. **Any** help relating to these questions is appreciated, and if anything is unclear, please tell me.