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Let $(R,M)$ be a Noetherian local ring, and let $Q$ be an $M$-primary ideal of $R$. Note that the $R$-module $(Q:M)/Q$ is annihilated by $M$ so it can be regarded as an $R/M$-vector space. Show that the following statement are equivalent

i) $Q$ is irreducible

ii) $\dim_{R/M} (Q:M)/Q=1$

iii) The set of all ideals of $R$ which strictly contain $Q$ admits $(Q:M)$ as smallest element.

I can see that $i) \Rightarrow ii)$ by the fact that there exists $n \in \mathbb{N}$ such that $M^n \subset Q$ but i can't indicate other cases. Can anyone help me?

user26857
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Desunkid
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1 Answers1

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I will reduce to the Artinian case, and assume that $Q=0$. We need to show that the following are equivalent.
i) $0$ is irreducible;
ii) $\dim_{(R/M)} 0:M = 1$;
iii) Every non-zero ideal of $R$ contains $0:M$.

You already showed that $i=>ii$.

($ii=>iii$:) If $Q'$ is a non-zero non-unit ideal, then $\dim_{(R/M)} Q' < \infty$. Thus, we may find an element $x \in Q'$ such that $\dim_{(R/M)} (x) = 1$. Therefore, $xM = 0$, so $(x) \subset 0:M$. Since both ideals have the same length (the $R/M$-vector space dimension), we have the equality.

($iii=>i$:) Suppose that $0$ is reducible, say $0 = P \cap P'$ for some nonzero ideals $P,P'$. By iii), both ideals contain $0:M$. Then $0 \neq 0:M \subset P \cap P' = 0$, and this is a contradiction.

I would like to point out that the conditions above are the defining property of $R$ (or $R/Q$ as in your post) being an Artinian Gorenstein ring.

Youngsu
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