Irreducible primary ideal in Noetherian local ring (Sharp, Exercise 8.29)











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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?






commutative-algebra







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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?






    commutative-algebra







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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?






      commutative-algebra







      share|cite|improve this question















      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?






      commutative-algebra




      commutative-algebra






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