Mixing
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
edited yesterday
user26857
39.1k123882
asked yesterday
Desunkid
16310
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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
edited yesterday
user26857
39.1k123882
asked yesterday
Desunkid
16310
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
edited yesterday
user26857
39.1k123882
asked yesterday
Desunkid
16310
edited yesterday
user26857
39.1k123882
asked yesterday
Desunkid
16310
edited yesterday
user26857
39.1k123882
edited yesterday
user26857
39.1k123882
edited yesterday
user26857
39.1k123882
39.1k123882
asked yesterday
Desunkid
16310
asked yesterday
Desunkid
16310
asked yesterday
Desunkid
16310
16310
add a comment |
add a comment |
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