Mixing
Length of module $otimes A[X]$
$begingroup$
Let $M$ be an $A$-module of finite length, i.e. $ell_A(M)<+infty$. I would like to compute the length $$(*) : : : ell_{A[X]}(Motimes_AA[X]),$$ where $A[X]$ is the canonical polynomial ring on $A$. As it has been pointed out in the comments we have to assume that ($*$) is finite. I think that ($*$) is equal to $ell_A(M)$, but I do not know how to proceed. Any help?
abstract-algebra modules
edited Jan 19 at 16:47
user26857
39.3k124183
asked Jan 19 at 1:16
Vincenzo ZaccaroVincenzo Zaccaro
1,254720
$endgroup$
add a comment |
$begingroup$
Let $M$ be an $A$-module of finite length, i.e. $ell_A(M)<+infty$. I would like to compute the length $$(*) : : : ell_{A[X]}(Motimes_AA[X]),$$ where $A[X]$ is the canonical polynomial ring on $A$. As it has been pointed out in the comments we have to assume that ($*$) is finite. I think that ($*$) is equal to $ell_A(M)$, but I do not know how to proceed. Any help?
abstract-algebra modules
edited Jan 19 at 16:47
user26857
39.3k124183
asked Jan 19 at 1:16
Vincenzo ZaccaroVincenzo Zaccaro
1,254720
$endgroup$
$begingroup$
Have you tried this with $A$ a field, $M=A$?
$endgroup$
– Mohan
Jan 19 at 2:16
$begingroup$
Yes. I have to add the assumption $ell_{A[X]}(Motimes_AA[X])<+infty$.
$endgroup$
– Vincenzo Zaccaro
Jan 19 at 2:18
1
$begingroup$
The only time (*) is finite is when $M=0$, so there is not much to do.
$endgroup$
– Mohan
Jan 19 at 15:44
add a comment |
$begingroup$
Let $M$ be an $A$-module of finite length, i.e. $ell_A(M)<+infty$. I would like to compute the length $$(*) : : : ell_{A[X]}(Motimes_AA[X]),$$ where $A[X]$ is the canonical polynomial ring on $A$. As it has been pointed out in the comments we have to assume that ($*$) is finite. I think that ($*$) is equal to $ell_A(M)$, but I do not know how to proceed. Any help?
abstract-algebra modules
edited Jan 19 at 16:47
user26857
39.3k124183
asked Jan 19 at 1:16
Vincenzo ZaccaroVincenzo Zaccaro
1,254720
$endgroup$
Let $M$ be an $A$-module of finite length, i.e. $ell_A(M)<+infty$. I would like to compute the length $$(*) : : : ell_{A[X]}(Motimes_AA[X]),$$ where $A[X]$ is the canonical polynomial ring on $A$. As it has been pointed out in the comments we have to assume that ($*$) is finite. I think that ($*$) is equal to $ell_A(M)$, but I do not know how to proceed. Any help?
abstract-algebra modules
abstract-algebra modules
edited Jan 19 at 16:47
user26857
39.3k124183
asked Jan 19 at 1:16
Vincenzo ZaccaroVincenzo Zaccaro
1,254720
edited Jan 19 at 16:47
user26857
39.3k124183
asked Jan 19 at 1:16
Vincenzo ZaccaroVincenzo Zaccaro
1,254720
edited Jan 19 at 16:47
user26857
39.3k124183
edited Jan 19 at 16:47
user26857
39.3k124183
edited Jan 19 at 16:47
user26857
39.3k124183
39.3k124183
asked Jan 19 at 1:16
Vincenzo ZaccaroVincenzo Zaccaro
1,254720
asked Jan 19 at 1:16
Vincenzo ZaccaroVincenzo Zaccaro
1,254720
asked Jan 19 at 1:16
Vincenzo ZaccaroVincenzo Zaccaro
1,254720
1,254720
$begingroup$
Have you tried this with $A$ a field, $M=A$?
$endgroup$
– Mohan
Jan 19 at 2:16
$begingroup$
Yes. I have to add the assumption $ell_{A[X]}(Motimes_AA[X])<+infty$.
$endgroup$
– Vincenzo Zaccaro
Jan 19 at 2:18
1
$begingroup$
The only time (*) is finite is when $M=0$, so there is not much to do.
$endgroup$
– Mohan
Jan 19 at 15:44
add a comment |
$begingroup$
Have you tried this with $A$ a field, $M=A$?
$endgroup$
– Mohan
Jan 19 at 2:16
$begingroup$
Yes. I have to add the assumption $ell_{A[X]}(Motimes_AA[X])<+infty$.
$endgroup$
– Vincenzo Zaccaro
Jan 19 at 2:18
1
$begingroup$
The only time (*) is finite is when $M=0$, so there is not much to do.
$endgroup$
– Mohan
Jan 19 at 15:44
$begingroup$
Have you tried this with $A$ a field, $M=A$?
$endgroup$
– Mohan
Jan 19 at 2:16
$begingroup$
Have you tried this with $A$ a field, $M=A$?
$endgroup$
– Mohan
Jan 19 at 2:16
$begingroup$
Yes. I have to add the assumption $ell_{A[X]}(Motimes_AA[X])<+infty$.
$endgroup$
– Vincenzo Zaccaro
Jan 19 at 2:18
$begingroup$
Yes. I have to add the assumption $ell_{A[X]}(Motimes_AA[X])<+infty$.
$endgroup$
– Vincenzo Zaccaro
Jan 19 at 2:18
1
1
$begingroup$
The only time (*) is finite is when $M=0$, so there is not much to do.
$endgroup$
– Mohan
Jan 19 at 15:44
$begingroup$
The only time (*) is finite is when $M=0$, so there is not much to do.
$endgroup$
– Mohan
Jan 19 at 15:44
add a comment |
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$begingroup$
Have you tried this with $A$ a field, $M=A$?
$endgroup$
– Mohan
Jan 19 at 2:16
$begingroup$
Yes. I have to add the assumption $ell_{A[X]}(Motimes_AA[X])<+infty$.
$endgroup$
– Vincenzo Zaccaro
Jan 19 at 2:18
1
$begingroup$
The only time (*) is finite is when $M=0$, so there is not much to do.
$endgroup$
– Mohan
Jan 19 at 15:44