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Let $A$ be Krull ring and $\left\{R_l\right\}_{l \in L}$ a defining family of DVRs. Let $a \in A, a \neq 0$. Then $a R_l \neq R_l$ only for a finite number of indices $l$ (otherwise the image of $a$ under infinitely many valuations would be non-zero, contradiction). Let $R_1,\dots, R_t$ be the DVRs such that $a R_i \neq R_i$. Define $\mathfrak q_i = aR_i\cap A$ and $\mathfrak p_i=rad(R_i) \cap A$. Then $\mathfrak q_i$ is primary belonging to $\mathfrak p_i$.

Question: How can we see that $aA=\mathfrak q_1\cap \cdots \cap \mathfrak q_t$?

Remark: Certainly, $aA \subseteq \mathfrak q_1 \cap \cdots \cap \mathfrak q_t$, since $aA \subset \mathfrak q_i$ for every $i$. How about the other direction?

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Prove that $$aR_1\cap\cdots\cap aR_t\cap A=aA.$$ "$\supseteq$" is clear.

"$\subseteq$" it's easy: take $x\in aR_1\cap\cdots\cap aR_t\cap A$. Then $a^{-1}x\in R_1\cap\cdots\cap R_t$. Since $aR_l=R_l$ for all $l\neq 1,\dots,t$ it follows that $a^{-1}\in R_l$, so $a^{-1}x\in R_l$, and that's all.

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  • $\begingroup$ good job, +1 :) $\endgroup$ – Manos May 31 '13 at 15:21

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