Mixing
Length of tensor product of finite length modules is finite
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Let $R$ be a commutative ring. If $M$ and $N$ are finite length $R$-modules, then $Motimes_R N$ has finite length, and $l(Motimes_R N) le l(M)l(N)$.
I know the question has been posted previously here, but only half of it was answered and I do not understand the argument about the inequality.
I have tried to explicitly construct a composition series and use exact sequences, but in using the latter I can't see what modules to use in the sequences.
abstract-algebra commutative-algebra modules tensor-products
edited Apr 13 '17 at 12:19
Community♦
1
asked Apr 14 '15 at 11:04
Rhys Evans
18510
add a comment |
up vote
5
down vote
favorite
Let $R$ be a commutative ring. If $M$ and $N$ are finite length $R$-modules, then $Motimes_R N$ has finite length, and $l(Motimes_R N) le l(M)l(N)$.
I know the question has been posted previously here, but only half of it was answered and I do not understand the argument about the inequality.
I have tried to explicitly construct a composition series and use exact sequences, but in using the latter I can't see what modules to use in the sequences.
abstract-algebra commutative-algebra modules tensor-products
edited Apr 13 '17 at 12:19
Community♦
1
asked Apr 14 '15 at 11:04
Rhys Evans
18510
add a comment |
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Let $R$ be a commutative ring. If $M$ and $N$ are finite length $R$-modules, then $Motimes_R N$ has finite length, and $l(Motimes_R N) le l(M)l(N)$.
I know the question has been posted previously here, but only half of it was answered and I do not understand the argument about the inequality.
I have tried to explicitly construct a composition series and use exact sequences, but in using the latter I can't see what modules to use in the sequences.
abstract-algebra commutative-algebra modules tensor-products
edited Apr 13 '17 at 12:19
Community♦
1
asked Apr 14 '15 at 11:04
Rhys Evans
18510
Let $R$ be a commutative ring. If $M$ and $N$ are finite length $R$-modules, then $Motimes_R N$ has finite length, and $l(Motimes_R N) le l(M)l(N)$.
I know the question has been posted previously here, but only half of it was answered and I do not understand the argument about the inequality.
I have tried to explicitly construct a composition series and use exact sequences, but in using the latter I can't see what modules to use in the sequences.
abstract-algebra commutative-algebra modules tensor-products
abstract-algebra commutative-algebra modules tensor-products
edited Apr 13 '17 at 12:19
Community♦
1
asked Apr 14 '15 at 11:04
Rhys Evans
18510
edited Apr 13 '17 at 12:19
Community♦
1
asked Apr 14 '15 at 11:04
Rhys Evans
18510
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
1
asked Apr 14 '15 at 11:04
Rhys Evans
18510
asked Apr 14 '15 at 11:04
Rhys Evans
18510
asked Apr 14 '15 at 11:04
Rhys Evans
18510
18510
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
Let ${0}=N_0<N_1<cdots<N_n=N$ be a composition series for $N$. In particular, $N_i/N_{i-1}$ is a cyclic module, so $N$ can be generated by (at most) $n$ elements. Then there is a surjection $R^nto N$, and tensorizing by $M$ we get $Motimes_R R^nto Motimes_RNto 0$. But $Motimes_R R^nsimeq M^n$, so $Motimes_R R^n$ has finite length which is greater or equal than the length of $Motimes_RN$. Since the length of $M^n$ is $ell(M)ell(N)$, we are done.
answered Apr 14 '15 at 20:13
user26857
39.1k123882
I thought that this approach would struggle at the "generated by (at most) n elements" part, but I don't know why. Thanks.
– Rhys Evans
Apr 15 '15 at 7:14
1
@user26857: Nice solution.
– user371231
5 hours ago
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
Let ${0}=N_0<N_1<cdots<N_n=N$ be a composition series for $N$. In particular, $N_i/N_{i-1}$ is a cyclic module, so $N$ can be generated by (at most) $n$ elements. Then there is a surjection $R^nto N$, and tensorizing by $M$ we get $Motimes_R R^nto Motimes_RNto 0$. But $Motimes_R R^nsimeq M^n$, so $Motimes_R R^n$ has finite length which is greater or equal than the length of $Motimes_RN$. Since the length of $M^n$ is $ell(M)ell(N)$, we are done.
answered Apr 14 '15 at 20:13
user26857
39.1k123882
I thought that this approach would struggle at the "generated by (at most) n elements" part, but I don't know why. Thanks.
– Rhys Evans
Apr 15 '15 at 7:14
1
@user26857: Nice solution.
– user371231
5 hours ago
add a comment |
up vote
3
down vote
accepted
Let ${0}=N_0<N_1<cdots<N_n=N$ be a composition series for $N$. In particular, $N_i/N_{i-1}$ is a cyclic module, so $N$ can be generated by (at most) $n$ elements. Then there is a surjection $R^nto N$, and tensorizing by $M$ we get $Motimes_R R^nto Motimes_RNto 0$. But $Motimes_R R^nsimeq M^n$, so $Motimes_R R^n$ has finite length which is greater or equal than the length of $Motimes_RN$. Since the length of $M^n$ is $ell(M)ell(N)$, we are done.
answered Apr 14 '15 at 20:13
user26857
39.1k123882
I thought that this approach would struggle at the "generated by (at most) n elements" part, but I don't know why. Thanks.
– Rhys Evans
Apr 15 '15 at 7:14
1
@user26857: Nice solution.
– user371231
5 hours ago
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
Let ${0}=N_0<N_1<cdots<N_n=N$ be a composition series for $N$. In particular, $N_i/N_{i-1}$ is a cyclic module, so $N$ can be generated by (at most) $n$ elements. Then there is a surjection $R^nto N$, and tensorizing by $M$ we get $Motimes_R R^nto Motimes_RNto 0$. But $Motimes_R R^nsimeq M^n$, so $Motimes_R R^n$ has finite length which is greater or equal than the length of $Motimes_RN$. Since the length of $M^n$ is $ell(M)ell(N)$, we are done.
answered Apr 14 '15 at 20:13
user26857
39.1k123882
Let ${0}=N_0<N_1<cdots<N_n=N$ be a composition series for $N$. In particular, $N_i/N_{i-1}$ is a cyclic module, so $N$ can be generated by (at most) $n$ elements. Then there is a surjection $R^nto N$, and tensorizing by $M$ we get $Motimes_R R^nto Motimes_RNto 0$. But $Motimes_R R^nsimeq M^n$, so $Motimes_R R^n$ has finite length which is greater or equal than the length of $Motimes_RN$. Since the length of $M^n$ is $ell(M)ell(N)$, we are done.
answered Apr 14 '15 at 20:13
user26857
39.1k123882
edited 4 hours ago
answered Apr 14 '15 at 20:13
user26857
39.1k123882
answered Apr 14 '15 at 20:13
user26857
39.1k123882
answered Apr 14 '15 at 20:13
user26857
39.1k123882
39.1k123882
I thought that this approach would struggle at the "generated by (at most) n elements" part, but I don't know why. Thanks.
– Rhys Evans
Apr 15 '15 at 7:14
1
@user26857: Nice solution.
– user371231
5 hours ago
add a comment |
I thought that this approach would struggle at the "generated by (at most) n elements" part, but I don't know why. Thanks.
– Rhys Evans
Apr 15 '15 at 7:14
1
@user26857: Nice solution.
– user371231
5 hours ago
I thought that this approach would struggle at the "generated by (at most) n elements" part, but I don't know why. Thanks.
– Rhys Evans
Apr 15 '15 at 7:14
I thought that this approach would struggle at the "generated by (at most) n elements" part, but I don't know why. Thanks.
– Rhys Evans
Apr 15 '15 at 7:14
1
1
@user26857: Nice solution.
– user371231
5 hours ago
@user26857: Nice solution.
– user371231
5 hours ago
add a comment |
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