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If $A$ is a Dedekind domain and $a,b$ are ideals, then why does $aA_p⊂bA_p$ for every prime ideal $p$ imply that $a\subset b$?

I read it in Milne's notes but it alludes to DVR's and I'm not familiar with that concept.

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Let $x$ lie in the first ideal. Consider the set $I$ of elements $a$ of $A$ such that $ax$ is in the second ideal. Check that $I$ is an ideal and that the hypothesis implies that $I$ is not contained in any nonzero prime ideal of $A$. Hence $I=A$, and so $1$ is in $I$.

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Factor $a$ and $b$ into its finite product of prime ideals. If say $$b=(p_1) \cdots (p_n),$$ $b_{p_i} = (p_i)_p$, so you have that $a$ also have $$(p_1) \cdots (p_n)$$ in its factorization. Thus, it must be contained in $b$.

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