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When is length of a module equal to its dimension as a vector space?
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Suppose $(R,mathfrakm)$ is a local ring and $M$ is an $R$-module of finite length. What are some conditions, under which we have $$operatornamelength_R(M)=dim_kM,$$ for an "appropriate" field $k$ isomorphic to the residue field $R/mathfrakm$?
This ($leftarrow$ click) question is a very simple case, where $M$ is assumed to be an $R/mathfrakm$-module, that is, $M$ is assumed to be annihilated by $mathfrakm$.
Here ($leftarrow$ click) is another special case where the ring $R$ is assumed to be Artinian, which I think is not necessary.
In general, I think the condition $operatornamelength_R(M)=dim_kM$ holds if $R$ contains a field $k$ such that the composition $khookrightarrow Rtwoheadrightarrow R/mathfrakm$ is an isomorphism. For instance, every complete local ring of equal characteristic satisfies this condition.
The proof that I have in mind is by induction on $operatornamelength_R(M)$. First, the hypothesis "$R$ contains a field $k$ such that the composition $khookrightarrow Rtwoheadrightarrow R/mathfrakm$ is an isomorphism" shows that $dim_kR/mathfrakm=1$. So, clearly $operatornamelength_R(R/mathfrakm)=dim_k R/mathfrakm$ holds. Now suppose the statement holds for any $R$-module of length $<n$ and let $M$ be an $R$-module of length $n$. Then we have an exact sequence of $R$-modules $$0rightarrow M^primerightarrow Mrightarrow R/mathfrakmrightarrow 0.$$
As $operatornamelength_R(M^prime)=operatornamelength_R(M)-1=n-1$, it follows from the induction hypothesis that $$operatornamelength_R(M)=dim_k(M^prime)+1.$$ But viewing the above exact sequence as an exact sequence of $k$-vector spaces, we also have $$dim_k M=dim_kM^prime+1,$$ from which it follows that $operatornamelength_R(M)=dim_kM$.
Am I correct?
linear-algebra abstract-algebra commutative-algebra
edited Jul 25 at 15:04
Bernard
110k635103
asked Jul 25 at 15:01
Must
1828
add a comment |Â
up vote
0
down vote
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Suppose $(R,mathfrakm)$ is a local ring and $M$ is an $R$-module of finite length. What are some conditions, under which we have $$operatornamelength_R(M)=dim_kM,$$ for an "appropriate" field $k$ isomorphic to the residue field $R/mathfrakm$?
This ($leftarrow$ click) question is a very simple case, where $M$ is assumed to be an $R/mathfrakm$-module, that is, $M$ is assumed to be annihilated by $mathfrakm$.
Here ($leftarrow$ click) is another special case where the ring $R$ is assumed to be Artinian, which I think is not necessary.
In general, I think the condition $operatornamelength_R(M)=dim_kM$ holds if $R$ contains a field $k$ such that the composition $khookrightarrow Rtwoheadrightarrow R/mathfrakm$ is an isomorphism. For instance, every complete local ring of equal characteristic satisfies this condition.
The proof that I have in mind is by induction on $operatornamelength_R(M)$. First, the hypothesis "$R$ contains a field $k$ such that the composition $khookrightarrow Rtwoheadrightarrow R/mathfrakm$ is an isomorphism" shows that $dim_kR/mathfrakm=1$. So, clearly $operatornamelength_R(R/mathfrakm)=dim_k R/mathfrakm$ holds. Now suppose the statement holds for any $R$-module of length $<n$ and let $M$ be an $R$-module of length $n$. Then we have an exact sequence of $R$-modules $$0rightarrow M^primerightarrow Mrightarrow R/mathfrakmrightarrow 0.$$
As $operatornamelength_R(M^prime)=operatornamelength_R(M)-1=n-1$, it follows from the induction hypothesis that $$operatornamelength_R(M)=dim_k(M^prime)+1.$$ But viewing the above exact sequence as an exact sequence of $k$-vector spaces, we also have $$dim_k M=dim_kM^prime+1,$$ from which it follows that $operatornamelength_R(M)=dim_kM$.
Am I correct?
linear-algebra abstract-algebra commutative-algebra
edited Jul 25 at 15:04
Bernard
110k635103
asked Jul 25 at 15:01
Must
1828
You are right. For an $R$-module, even to talk about $dim_k M$, $k$ should be a subring of $R$ and then what you say is correct.
– Mohan
Jul 25 at 18:26
@Mohan Thank you!
– Must
Jul 25 at 18:37
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Suppose $(R,mathfrakm)$ is a local ring and $M$ is an $R$-module of finite length. What are some conditions, under which we have $$operatornamelength_R(M)=dim_kM,$$ for an "appropriate" field $k$ isomorphic to the residue field $R/mathfrakm$?
This ($leftarrow$ click) question is a very simple case, where $M$ is assumed to be an $R/mathfrakm$-module, that is, $M$ is assumed to be annihilated by $mathfrakm$.
Here ($leftarrow$ click) is another special case where the ring $R$ is assumed to be Artinian, which I think is not necessary.
In general, I think the condition $operatornamelength_R(M)=dim_kM$ holds if $R$ contains a field $k$ such that the composition $khookrightarrow Rtwoheadrightarrow R/mathfrakm$ is an isomorphism. For instance, every complete local ring of equal characteristic satisfies this condition.
The proof that I have in mind is by induction on $operatornamelength_R(M)$. First, the hypothesis "$R$ contains a field $k$ such that the composition $khookrightarrow Rtwoheadrightarrow R/mathfrakm$ is an isomorphism" shows that $dim_kR/mathfrakm=1$. So, clearly $operatornamelength_R(R/mathfrakm)=dim_k R/mathfrakm$ holds. Now suppose the statement holds for any $R$-module of length $<n$ and let $M$ be an $R$-module of length $n$. Then we have an exact sequence of $R$-modules $$0rightarrow M^primerightarrow Mrightarrow R/mathfrakmrightarrow 0.$$
As $operatornamelength_R(M^prime)=operatornamelength_R(M)-1=n-1$, it follows from the induction hypothesis that $$operatornamelength_R(M)=dim_k(M^prime)+1.$$ But viewing the above exact sequence as an exact sequence of $k$-vector spaces, we also have $$dim_k M=dim_kM^prime+1,$$ from which it follows that $operatornamelength_R(M)=dim_kM$.
Am I correct?
linear-algebra abstract-algebra commutative-algebra
edited Jul 25 at 15:04
Bernard
110k635103
asked Jul 25 at 15:01
Must
1828
Suppose $(R,mathfrakm)$ is a local ring and $M$ is an $R$-module of finite length. What are some conditions, under which we have $$operatornamelength_R(M)=dim_kM,$$ for an "appropriate" field $k$ isomorphic to the residue field $R/mathfrakm$?
This ($leftarrow$ click) question is a very simple case, where $M$ is assumed to be an $R/mathfrakm$-module, that is, $M$ is assumed to be annihilated by $mathfrakm$.
Here ($leftarrow$ click) is another special case where the ring $R$ is assumed to be Artinian, which I think is not necessary.
In general, I think the condition $operatornamelength_R(M)=dim_kM$ holds if $R$ contains a field $k$ such that the composition $khookrightarrow Rtwoheadrightarrow R/mathfrakm$ is an isomorphism. For instance, every complete local ring of equal characteristic satisfies this condition.
The proof that I have in mind is by induction on $operatornamelength_R(M)$. First, the hypothesis "$R$ contains a field $k$ such that the composition $khookrightarrow Rtwoheadrightarrow R/mathfrakm$ is an isomorphism" shows that $dim_kR/mathfrakm=1$. So, clearly $operatornamelength_R(R/mathfrakm)=dim_k R/mathfrakm$ holds. Now suppose the statement holds for any $R$-module of length $<n$ and let $M$ be an $R$-module of length $n$. Then we have an exact sequence of $R$-modules $$0rightarrow M^primerightarrow Mrightarrow R/mathfrakmrightarrow 0.$$
As $operatornamelength_R(M^prime)=operatornamelength_R(M)-1=n-1$, it follows from the induction hypothesis that $$operatornamelength_R(M)=dim_k(M^prime)+1.$$ But viewing the above exact sequence as an exact sequence of $k$-vector spaces, we also have $$dim_k M=dim_kM^prime+1,$$ from which it follows that $operatornamelength_R(M)=dim_kM$.
Am I correct?
linear-algebra abstract-algebra commutative-algebra
edited Jul 25 at 15:04
Bernard
110k635103
asked Jul 25 at 15:01
Must
1828
edited Jul 25 at 15:04
Bernard
110k635103
edited Jul 25 at 15:04
Bernard
110k635103
edited Jul 25 at 15:04
Bernard
110k635103
110k635103
asked Jul 25 at 15:01
Must
1828
asked Jul 25 at 15:01
Must
1828
asked Jul 25 at 15:01
Must
1828
1828
You are right. For an $R$-module, even to talk about $dim_k M$, $k$ should be a subring of $R$ and then what you say is correct.
– Mohan
Jul 25 at 18:26
@Mohan Thank you!
– Must
Jul 25 at 18:37
add a comment |Â
You are right. For an $R$-module, even to talk about $dim_k M$, $k$ should be a subring of $R$ and then what you say is correct.
– Mohan
Jul 25 at 18:26
@Mohan Thank you!
– Must
Jul 25 at 18:37
You are right. For an $R$-module, even to talk about $dim_k M$, $k$ should be a subring of $R$ and then what you say is correct.
– Mohan
Jul 25 at 18:26
You are right. For an $R$-module, even to talk about $dim_k M$, $k$ should be a subring of $R$ and then what you say is correct.
– Mohan
Jul 25 at 18:26
@Mohan Thank you!
– Must
Jul 25 at 18:37
@Mohan Thank you!
– Must
Jul 25 at 18:37
add a comment |Â
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You are right. For an $R$-module, even to talk about $dim_k M$, $k$ should be a subring of $R$ and then what you say is correct.
– Mohan
Jul 25 at 18:26
@Mohan Thank you!
– Must
Jul 25 at 18:37