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Let $R$ be commutative ring with $1$. One says that an ideal $I$ is unmixed if $I$ has no embedded prime divisors (in other words, if the associated prime ideals of $R/I$ are the minimal prime ideals of $I$).

Let $I$ be an unmixed radical ideal of $R$. I want a hint to prove $(I:x)$ is unmixed ($x\in R\setminus I)$.
Thank you

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1 Answer 1

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$\newcommand{\Ass}{{\mathrm{Ass}}}$

I assume $R$ noetherian.

We write $I=Q_1 \cap \cdots \cap Q_r$ as the representation of $I$ as an intersection of primary ideals. Then $(I:x) = (Q_1:x) \cap \cdots \cap (Q_r:x)$ as this is a general property of intersection of ideals.

Now $(Q_i:x)$ is equal to $R$ if $x \in Q_i$ and $(Q_i:x) \subseteq \sqrt{Q_i} = P_i$ for $x \notin Q_i$. (Use the definition of primary ideal: $y z \in Q_i$ and $z \notin Q_i$ then $y^m \in Q_i$).

Additionally if $(Q_i:x) \neq (1)$ then $(Q_i:x)$ is $P_i$-primary: Say $z w \in (Q_i:x)$ and $z \notin (Q_i:x)$, so $z w x \in Q_i$ and $z x \notin Q_i$ so $w^n \in Q_i \subseteq (Q_i:x)$.

So we have a primary decomposition

$$(I:x) = (Q_{i_1}:x) \cap \cdots \cap (Q_{i_s}:x)$$

with $(Q_j:x) \neq (1)$ exactly when $j = i_\nu$ for a certain $\nu$.

So $\Ass (R/(I:x)) = \{P_{i_1},\ldots,P_{i_s}\}$

As $P_i$ is minimal over $I$ by assumption so $P_{i_\nu} \supseteq (I:x) $ is minimal over $(I:x) \supseteq I$ a fortiori.

This proves your proposition.

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  • $\begingroup$ thank you +1.what about non-noetherian case? have a counter-example? $\endgroup$
    – user 1
    Commented Apr 25, 2015 at 18:19

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