$ operatorname{Ext}_{k[x]}^n(k,k)$ for a field $k$ [closed]
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Consider the polynomial ring $k[x]$ for a field $k$ and the $k[x]$-module $k$, letting $x$ act trivially on $k$. What is $ operatorname{Ext}_{k[x]}^n(k,k)$ for $ngeq0$?
abstract-algebra general-topology algebraic-topology homological-algebra
edited Jan 19 at 14:18
Bernard
121k740116
asked Jan 19 at 13:40
CHwCCHwC
290111
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closed as off-topic by user26857, Cesareo, max_zorn, user91500, mrtaurho Jan 21 at 10:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user26857, Cesareo, max_zorn, user91500, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Consider the polynomial ring $k[x]$ for a field $k$ and the $k[x]$-module $k$, letting $x$ act trivially on $k$. What is $ operatorname{Ext}_{k[x]}^n(k,k)$ for $ngeq0$?
abstract-algebra general-topology algebraic-topology homological-algebra
edited Jan 19 at 14:18
Bernard
121k740116
asked Jan 19 at 13:40
CHwCCHwC
290111
$endgroup$
closed as off-topic by user26857, Cesareo, max_zorn, user91500, mrtaurho Jan 21 at 10:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user26857, Cesareo, max_zorn, user91500, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Consider the polynomial ring $k[x]$ for a field $k$ and the $k[x]$-module $k$, letting $x$ act trivially on $k$. What is $ operatorname{Ext}_{k[x]}^n(k,k)$ for $ngeq0$?
abstract-algebra general-topology algebraic-topology homological-algebra
edited Jan 19 at 14:18
Bernard
121k740116
asked Jan 19 at 13:40
CHwCCHwC
290111
$endgroup$
Consider the polynomial ring $k[x]$ for a field $k$ and the $k[x]$-module $k$, letting $x$ act trivially on $k$. What is $ operatorname{Ext}_{k[x]}^n(k,k)$ for $ngeq0$?
abstract-algebra general-topology algebraic-topology homological-algebra
abstract-algebra general-topology algebraic-topology homological-algebra
edited Jan 19 at 14:18
Bernard
121k740116
asked Jan 19 at 13:40
CHwCCHwC
290111
edited Jan 19 at 14:18
Bernard
121k740116
asked Jan 19 at 13:40
CHwCCHwC
290111
edited Jan 19 at 14:18
Bernard
121k740116
edited Jan 19 at 14:18
Bernard
121k740116
edited Jan 19 at 14:18
Bernard
121k740116
121k740116
asked Jan 19 at 13:40
CHwCCHwC
290111
asked Jan 19 at 13:40
CHwCCHwC
290111
asked Jan 19 at 13:40
CHwCCHwC
290111
290111
closed as off-topic by user26857, Cesareo, max_zorn, user91500, mrtaurho Jan 21 at 10:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user26857, Cesareo, max_zorn, user91500, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user26857, Cesareo, max_zorn, user91500, mrtaurho Jan 21 at 10:42
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user26857, Cesareo, max_zorn, user91500, mrtaurho
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
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votes
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A free resolution of $k$ is $k[x] overset{x}to k[x]$. Now apply $ operatorname{Hom}(-,k)$. This gives the complex $koverset{0}leftarrow k$. So $ operatorname{Ext}^0, operatorname{Ext}^1 = k$ and higher exts are zero (which we already knew because $k[x]$ has homological dimension 1).
edited Jan 19 at 14:19
Bernard
121k740116
answered Jan 19 at 13:48
BenBen
3,871616
$endgroup$
$begingroup$
(And the algebra structure is the only possible one, an exterior two dimensional algebra.)
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– Pedro Tamaroff♦
Jan 19 at 16:36
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
A free resolution of $k$ is $k[x] overset{x}to k[x]$. Now apply $ operatorname{Hom}(-,k)$. This gives the complex $koverset{0}leftarrow k$. So $ operatorname{Ext}^0, operatorname{Ext}^1 = k$ and higher exts are zero (which we already knew because $k[x]$ has homological dimension 1).
edited Jan 19 at 14:19
Bernard
121k740116
answered Jan 19 at 13:48
BenBen
3,871616
$endgroup$
$begingroup$
(And the algebra structure is the only possible one, an exterior two dimensional algebra.)
$endgroup$
– Pedro Tamaroff♦
Jan 19 at 16:36
add a comment |
$begingroup$
A free resolution of $k$ is $k[x] overset{x}to k[x]$. Now apply $ operatorname{Hom}(-,k)$. This gives the complex $koverset{0}leftarrow k$. So $ operatorname{Ext}^0, operatorname{Ext}^1 = k$ and higher exts are zero (which we already knew because $k[x]$ has homological dimension 1).
edited Jan 19 at 14:19
Bernard
121k740116
answered Jan 19 at 13:48
BenBen
3,871616
$endgroup$
$begingroup$
(And the algebra structure is the only possible one, an exterior two dimensional algebra.)
$endgroup$
– Pedro Tamaroff♦
Jan 19 at 16:36
add a comment |
$begingroup$
A free resolution of $k$ is $k[x] overset{x}to k[x]$. Now apply $ operatorname{Hom}(-,k)$. This gives the complex $koverset{0}leftarrow k$. So $ operatorname{Ext}^0, operatorname{Ext}^1 = k$ and higher exts are zero (which we already knew because $k[x]$ has homological dimension 1).
edited Jan 19 at 14:19
Bernard
121k740116
answered Jan 19 at 13:48
BenBen
3,871616
$endgroup$
A free resolution of $k$ is $k[x] overset{x}to k[x]$. Now apply $ operatorname{Hom}(-,k)$. This gives the complex $koverset{0}leftarrow k$. So $ operatorname{Ext}^0, operatorname{Ext}^1 = k$ and higher exts are zero (which we already knew because $k[x]$ has homological dimension 1).
edited Jan 19 at 14:19
Bernard
121k740116
answered Jan 19 at 13:48
BenBen
3,871616
edited Jan 19 at 14:19
Bernard
121k740116
edited Jan 19 at 14:19
Bernard
121k740116
edited Jan 19 at 14:19
Bernard
121k740116
121k740116
answered Jan 19 at 13:48
BenBen
3,871616
answered Jan 19 at 13:48
BenBen
3,871616
answered Jan 19 at 13:48
BenBen
3,871616
3,871616
$begingroup$
(And the algebra structure is the only possible one, an exterior two dimensional algebra.)
$endgroup$
– Pedro Tamaroff♦
Jan 19 at 16:36
add a comment |
$begingroup$
(And the algebra structure is the only possible one, an exterior two dimensional algebra.)
$endgroup$
– Pedro Tamaroff♦
Jan 19 at 16:36
$begingroup$
(And the algebra structure is the only possible one, an exterior two dimensional algebra.)
$endgroup$
– Pedro Tamaroff♦
Jan 19 at 16:36
$begingroup$
(And the algebra structure is the only possible one, an exterior two dimensional algebra.)
$endgroup$
– Pedro Tamaroff♦
Jan 19 at 16:36
add a comment |
