Duality on the symmetric algebra
2
If we have a commutative ring $A$, and a module over $A$, $M$, is it true that :
$$rm{S}^{bullet}_{A}(M^{ast})simeqrm{S}^{bullet}_{A}(M)^{ast}?$$
linear-algebra abstract-algebra commutative-algebra
asked Nov 28 '18 at 18:17
J. Ove
111
|
show 1 more comment
2
If we have a commutative ring $A$, and a module over $A$, $M$, is it true that :
$$rm{S}^{bullet}_{A}(M^{ast})simeqrm{S}^{bullet}_{A}(M)^{ast}?$$
linear-algebra abstract-algebra commutative-algebra
asked Nov 28 '18 at 18:17
J. Ove
111
What sort of dual are you taking on the outside? If you just take the dual "naively", then this will fail with any finite dimensional vector space.
– Tobias Kildetoft
Nov 28 '18 at 18:21
I´m taking the naive dual, i.e, $rm{Hom}_{A}(M,A)$. Why does this construction fail?
– J. Ove
Nov 28 '18 at 18:23
Because one side will have countable dimension while the other will have uncountable dimension.
– Tobias Kildetoft
Nov 28 '18 at 18:24
That´s totally true, I see now... thank you Tobias.
– J. Ove
Nov 28 '18 at 18:26
3
Note that this does have an easy fix, which is to take the graded dual instead. Then at least I think you should get an isomorphism when $A$ is a field and $M$ is finite dimensional.
– Tobias Kildetoft
Nov 28 '18 at 18:27
|
show 1 more comment
2
2
2
If we have a commutative ring $A$, and a module over $A$, $M$, is it true that :
$$rm{S}^{bullet}_{A}(M^{ast})simeqrm{S}^{bullet}_{A}(M)^{ast}?$$
linear-algebra abstract-algebra commutative-algebra
asked Nov 28 '18 at 18:17
J. Ove
111
If we have a commutative ring $A$, and a module over $A$, $M$, is it true that :
$$rm{S}^{bullet}_{A}(M^{ast})simeqrm{S}^{bullet}_{A}(M)^{ast}?$$
linear-algebra abstract-algebra commutative-algebra
linear-algebra abstract-algebra commutative-algebra
asked Nov 28 '18 at 18:17
J. Ove
111
asked Nov 28 '18 at 18:17
J. Ove
111
asked Nov 28 '18 at 18:17
J. Ove
111
asked Nov 28 '18 at 18:17
J. Ove
111
asked Nov 28 '18 at 18:17
J. Ove
111
111
What sort of dual are you taking on the outside? If you just take the dual "naively", then this will fail with any finite dimensional vector space.
– Tobias Kildetoft
Nov 28 '18 at 18:21
I´m taking the naive dual, i.e, $rm{Hom}_{A}(M,A)$. Why does this construction fail?
– J. Ove
Nov 28 '18 at 18:23
Because one side will have countable dimension while the other will have uncountable dimension.
– Tobias Kildetoft
Nov 28 '18 at 18:24
That´s totally true, I see now... thank you Tobias.
– J. Ove
Nov 28 '18 at 18:26
3
Note that this does have an easy fix, which is to take the graded dual instead. Then at least I think you should get an isomorphism when $A$ is a field and $M$ is finite dimensional.
– Tobias Kildetoft
Nov 28 '18 at 18:27
|
show 1 more comment
What sort of dual are you taking on the outside? If you just take the dual "naively", then this will fail with any finite dimensional vector space.
– Tobias Kildetoft
Nov 28 '18 at 18:21
I´m taking the naive dual, i.e, $rm{Hom}_{A}(M,A)$. Why does this construction fail?
– J. Ove
Nov 28 '18 at 18:23
Because one side will have countable dimension while the other will have uncountable dimension.
– Tobias Kildetoft
Nov 28 '18 at 18:24
That´s totally true, I see now... thank you Tobias.
– J. Ove
Nov 28 '18 at 18:26
3
Note that this does have an easy fix, which is to take the graded dual instead. Then at least I think you should get an isomorphism when $A$ is a field and $M$ is finite dimensional.
– Tobias Kildetoft
Nov 28 '18 at 18:27
What sort of dual are you taking on the outside? If you just take the dual "naively", then this will fail with any finite dimensional vector space.
– Tobias Kildetoft
Nov 28 '18 at 18:21
What sort of dual are you taking on the outside? If you just take the dual "naively", then this will fail with any finite dimensional vector space.
– Tobias Kildetoft
Nov 28 '18 at 18:21
I´m taking the naive dual, i.e, $rm{Hom}_{A}(M,A)$. Why does this construction fail?
– J. Ove
Nov 28 '18 at 18:23
I´m taking the naive dual, i.e, $rm{Hom}_{A}(M,A)$. Why does this construction fail?
– J. Ove
Nov 28 '18 at 18:23
Because one side will have countable dimension while the other will have uncountable dimension.
– Tobias Kildetoft
Nov 28 '18 at 18:24
Because one side will have countable dimension while the other will have uncountable dimension.
– Tobias Kildetoft
Nov 28 '18 at 18:24
That´s totally true, I see now... thank you Tobias.
– J. Ove
Nov 28 '18 at 18:26
That´s totally true, I see now... thank you Tobias.
– J. Ove
Nov 28 '18 at 18:26
3
3
Note that this does have an easy fix, which is to take the graded dual instead. Then at least I think you should get an isomorphism when $A$ is a field and $M$ is finite dimensional.
– Tobias Kildetoft
Nov 28 '18 at 18:27
Note that this does have an easy fix, which is to take the graded dual instead. Then at least I think you should get an isomorphism when $A$ is a field and $M$ is finite dimensional.
– Tobias Kildetoft
Nov 28 '18 at 18:27
|
show 1 more comment
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What sort of dual are you taking on the outside? If you just take the dual "naively", then this will fail with any finite dimensional vector space.
– Tobias Kildetoft
Nov 28 '18 at 18:21
I´m taking the naive dual, i.e, $rm{Hom}_{A}(M,A)$. Why does this construction fail?
– J. Ove
Nov 28 '18 at 18:23
Because one side will have countable dimension while the other will have uncountable dimension.
– Tobias Kildetoft
Nov 28 '18 at 18:24
That´s totally true, I see now... thank you Tobias.
– J. Ove
Nov 28 '18 at 18:26
3
Note that this does have an easy fix, which is to take the graded dual instead. Then at least I think you should get an isomorphism when $A$ is a field and $M$ is finite dimensional.
– Tobias Kildetoft
Nov 28 '18 at 18:27