Maximum number of linearly independent elements in a module
The rank of an $R$-module is the maximum number of $R$-linearly independent elements in the module.
If $R$ (commutative with identity) is an integral domain and $M$ is a finitely generated $R$-module then $operatorname{rank}(M)=operatorname{rank}(N)+operatorname{rank}(M/N)$, where $N$ is an $R$-submodule of $M$.
If it was given $M/N$ is a free $R$-module then we can write $M$ as a direct sum of $N$ and $M/N$. Here we can only say that $M/N$ is finitely generated as $M$ is finitely generated. How do I go further?
ring-theory commutative-algebra modules
edited Dec 29 '18 at 15:25
egreg
179k1485202
asked Dec 29 '18 at 10:11
user362405user362405
1236
add a comment |
The rank of an $R$-module is the maximum number of $R$-linearly independent elements in the module.
If $R$ (commutative with identity) is an integral domain and $M$ is a finitely generated $R$-module then $operatorname{rank}(M)=operatorname{rank}(N)+operatorname{rank}(M/N)$, where $N$ is an $R$-submodule of $M$.
If it was given $M/N$ is a free $R$-module then we can write $M$ as a direct sum of $N$ and $M/N$. Here we can only say that $M/N$ is finitely generated as $M$ is finitely generated. How do I go further?
ring-theory commutative-algebra modules
edited Dec 29 '18 at 15:25
egreg
179k1485202
asked Dec 29 '18 at 10:11
user362405user362405
1236
@Student7 submodule need not be finitely generated even when the module is finitely generated
– user362405
Dec 29 '18 at 12:38
Ah sorry, thought of a noetherian ring.
– Student7
Dec 29 '18 at 12:43
Reduce the problem to when $R$ is a field, by going to the fraction field.
– Mohan
Dec 29 '18 at 14:16
@Mohan can you elaborate. I am not well versed with the techniques
– user362405
Dec 29 '18 at 14:23
add a comment |
The rank of an $R$-module is the maximum number of $R$-linearly independent elements in the module.
If $R$ (commutative with identity) is an integral domain and $M$ is a finitely generated $R$-module then $operatorname{rank}(M)=operatorname{rank}(N)+operatorname{rank}(M/N)$, where $N$ is an $R$-submodule of $M$.
If it was given $M/N$ is a free $R$-module then we can write $M$ as a direct sum of $N$ and $M/N$. Here we can only say that $M/N$ is finitely generated as $M$ is finitely generated. How do I go further?
ring-theory commutative-algebra modules
edited Dec 29 '18 at 15:25
egreg
179k1485202
asked Dec 29 '18 at 10:11
user362405user362405
1236
The rank of an $R$-module is the maximum number of $R$-linearly independent elements in the module.
If $R$ (commutative with identity) is an integral domain and $M$ is a finitely generated $R$-module then $operatorname{rank}(M)=operatorname{rank}(N)+operatorname{rank}(M/N)$, where $N$ is an $R$-submodule of $M$.
If it was given $M/N$ is a free $R$-module then we can write $M$ as a direct sum of $N$ and $M/N$. Here we can only say that $M/N$ is finitely generated as $M$ is finitely generated. How do I go further?
ring-theory commutative-algebra modules
ring-theory commutative-algebra modules
edited Dec 29 '18 at 15:25
egreg
179k1485202
asked Dec 29 '18 at 10:11
user362405user362405
1236
edited Dec 29 '18 at 15:25
egreg
179k1485202
asked Dec 29 '18 at 10:11
user362405user362405
1236
edited Dec 29 '18 at 15:25
egreg
179k1485202
edited Dec 29 '18 at 15:25
egreg
179k1485202
edited Dec 29 '18 at 15:25
egreg
179k1485202
179k1485202
asked Dec 29 '18 at 10:11
user362405user362405
1236
asked Dec 29 '18 at 10:11
user362405user362405
1236
asked Dec 29 '18 at 10:11
user362405user362405
1236
1236
@Student7 submodule need not be finitely generated even when the module is finitely generated
– user362405
Dec 29 '18 at 12:38
Ah sorry, thought of a noetherian ring.
– Student7
Dec 29 '18 at 12:43
Reduce the problem to when $R$ is a field, by going to the fraction field.
– Mohan
Dec 29 '18 at 14:16
@Mohan can you elaborate. I am not well versed with the techniques
– user362405
Dec 29 '18 at 14:23
add a comment |
@Student7 submodule need not be finitely generated even when the module is finitely generated
– user362405
Dec 29 '18 at 12:38
Ah sorry, thought of a noetherian ring.
– Student7
Dec 29 '18 at 12:43
Reduce the problem to when $R$ is a field, by going to the fraction field.
– Mohan
Dec 29 '18 at 14:16
@Mohan can you elaborate. I am not well versed with the techniques
– user362405
Dec 29 '18 at 14:23
@Student7 submodule need not be finitely generated even when the module is finitely generated
– user362405
Dec 29 '18 at 12:38
@Student7 submodule need not be finitely generated even when the module is finitely generated
– user362405
Dec 29 '18 at 12:38
Ah sorry, thought of a noetherian ring.
– Student7
Dec 29 '18 at 12:43
Ah sorry, thought of a noetherian ring.
– Student7
Dec 29 '18 at 12:43
Reduce the problem to when $R$ is a field, by going to the fraction field.
– Mohan
Dec 29 '18 at 14:16
Reduce the problem to when $R$ is a field, by going to the fraction field.
– Mohan
Dec 29 '18 at 14:16
@Mohan can you elaborate. I am not well versed with the techniques
– user362405
Dec 29 '18 at 14:23
@Mohan can you elaborate. I am not well versed with the techniques
– user362405
Dec 29 '18 at 14:23
add a comment |
1 Answer
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Just an elaboration. Let $R$ be a domain and $M$ any module. Then rank of $M$ is equal to the dimension of the vector space over $K$, the fraction field of $R$ of $Motimes_R K$ (if you are unfamiliar with tensor products, you can also use localization). So, rank of $M$ is rank of $Motimes K$ which is clearly the sum of the ranks of $Notimes K$ and $M/Notimes K$ and the rest should be clear.
answered Dec 29 '18 at 15:17
MohanMohan
11.6k1817
add a comment |
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1 Answer
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oldest
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Just an elaboration. Let $R$ be a domain and $M$ any module. Then rank of $M$ is equal to the dimension of the vector space over $K$, the fraction field of $R$ of $Motimes_R K$ (if you are unfamiliar with tensor products, you can also use localization). So, rank of $M$ is rank of $Motimes K$ which is clearly the sum of the ranks of $Notimes K$ and $M/Notimes K$ and the rest should be clear.
answered Dec 29 '18 at 15:17
MohanMohan
11.6k1817
add a comment |
Just an elaboration. Let $R$ be a domain and $M$ any module. Then rank of $M$ is equal to the dimension of the vector space over $K$, the fraction field of $R$ of $Motimes_R K$ (if you are unfamiliar with tensor products, you can also use localization). So, rank of $M$ is rank of $Motimes K$ which is clearly the sum of the ranks of $Notimes K$ and $M/Notimes K$ and the rest should be clear.
answered Dec 29 '18 at 15:17
MohanMohan
11.6k1817
add a comment |
Just an elaboration. Let $R$ be a domain and $M$ any module. Then rank of $M$ is equal to the dimension of the vector space over $K$, the fraction field of $R$ of $Motimes_R K$ (if you are unfamiliar with tensor products, you can also use localization). So, rank of $M$ is rank of $Motimes K$ which is clearly the sum of the ranks of $Notimes K$ and $M/Notimes K$ and the rest should be clear.
answered Dec 29 '18 at 15:17
MohanMohan
11.6k1817
Just an elaboration. Let $R$ be a domain and $M$ any module. Then rank of $M$ is equal to the dimension of the vector space over $K$, the fraction field of $R$ of $Motimes_R K$ (if you are unfamiliar with tensor products, you can also use localization). So, rank of $M$ is rank of $Motimes K$ which is clearly the sum of the ranks of $Notimes K$ and $M/Notimes K$ and the rest should be clear.
answered Dec 29 '18 at 15:17
MohanMohan
11.6k1817
answered Dec 29 '18 at 15:17
MohanMohan
11.6k1817
answered Dec 29 '18 at 15:17
MohanMohan
11.6k1817
answered Dec 29 '18 at 15:17
MohanMohan
11.6k1817
11.6k1817
add a comment |
add a comment |
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@Student7 submodule need not be finitely generated even when the module is finitely generated
– user362405
Dec 29 '18 at 12:38
Ah sorry, thought of a noetherian ring.
– Student7
Dec 29 '18 at 12:43
Reduce the problem to when $R$ is a field, by going to the fraction field.
– Mohan
Dec 29 '18 at 14:16
@Mohan can you elaborate. I am not well versed with the techniques
– user362405
Dec 29 '18 at 14:23