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If $a,b\in R$ are nonzero in Dedekind domain $R$, and $\mathfrak{p}$ a prime ideal. Suppose $m=sup\{k:a\in \mathfrak{p}^k\}$, $n=sup\{k:b\in \mathfrak{p}^k\}$. I want to show $m+n=sup \{k:ab\in \mathfrak{p}^k\}$.

I know that both $m,n$ are finite, and I know $\leq$. To show $\geq$, I only need to derive a contradiction out of $ab\in \mathfrak{p}^{m+n+1}$. How do I do this?

Jun Xu
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  • One approach: localize at $\mathfrak{p}$ to reduce to a DVR, so that every nonzero ideal is a power of the maximal ideal, and show that distinct powers of the maximal ideal are distinct – math54321 May 18 '23 at 20:13

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