How do I understand the module structure on Yoneda Ext?
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Suppose $R$ is a (commutative) ring, and $M$ and $N$ are (finitely generated) $R$-modules. Then I know each $mathrm{Ext}_R^i(M, N)$ has the structure of an $R$-module. On the other hand, via Yoneda's description of Ext, each $varepsilon in mathrm{Ext}_R^i(M, N)$ corresponds to an equivalence class of exact sequences starting with $N$ and ending with $M$. My question is this: suppose $r in R$ and $varepsilon in mathrm{Ext}_R^i(M, N)$. How can I understand $r varepsilon$ in terms of $varepsilon$? To what extension does $r varepsilon$ correspond?
homological-algebra
asked Nov 19 at 22:31
Eric Canton
311110
add a comment |
up vote
6
down vote
favorite
Suppose $R$ is a (commutative) ring, and $M$ and $N$ are (finitely generated) $R$-modules. Then I know each $mathrm{Ext}_R^i(M, N)$ has the structure of an $R$-module. On the other hand, via Yoneda's description of Ext, each $varepsilon in mathrm{Ext}_R^i(M, N)$ corresponds to an equivalence class of exact sequences starting with $N$ and ending with $M$. My question is this: suppose $r in R$ and $varepsilon in mathrm{Ext}_R^i(M, N)$. How can I understand $r varepsilon$ in terms of $varepsilon$? To what extension does $r varepsilon$ correspond?
homological-algebra
asked Nov 19 at 22:31
Eric Canton
311110
add a comment |
up vote
6
down vote
favorite
up vote
6
down vote
favorite
Suppose $R$ is a (commutative) ring, and $M$ and $N$ are (finitely generated) $R$-modules. Then I know each $mathrm{Ext}_R^i(M, N)$ has the structure of an $R$-module. On the other hand, via Yoneda's description of Ext, each $varepsilon in mathrm{Ext}_R^i(M, N)$ corresponds to an equivalence class of exact sequences starting with $N$ and ending with $M$. My question is this: suppose $r in R$ and $varepsilon in mathrm{Ext}_R^i(M, N)$. How can I understand $r varepsilon$ in terms of $varepsilon$? To what extension does $r varepsilon$ correspond?
homological-algebra
asked Nov 19 at 22:31
Eric Canton
311110
Suppose $R$ is a (commutative) ring, and $M$ and $N$ are (finitely generated) $R$-modules. Then I know each $mathrm{Ext}_R^i(M, N)$ has the structure of an $R$-module. On the other hand, via Yoneda's description of Ext, each $varepsilon in mathrm{Ext}_R^i(M, N)$ corresponds to an equivalence class of exact sequences starting with $N$ and ending with $M$. My question is this: suppose $r in R$ and $varepsilon in mathrm{Ext}_R^i(M, N)$. How can I understand $r varepsilon$ in terms of $varepsilon$? To what extension does $r varepsilon$ correspond?
homological-algebra
homological-algebra
asked Nov 19 at 22:31
Eric Canton
311110
asked Nov 19 at 22:31
Eric Canton
311110
asked Nov 19 at 22:31
Eric Canton
311110
asked Nov 19 at 22:31
Eric Canton
311110
asked Nov 19 at 22:31
Eric Canton
311110
311110
add a comment |
add a comment |
1 Answer
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up vote
3
down vote
accepted
we can just see the case $i=1$:
In general,if $M$ is $A-B$ bimodule(i.e.left $A$ module and right $B$ module and $(am)b=a(mb)$),N is $A-C$ bimodule,then $Ext^1(M,N)$ is a $B-C$ bimodule.the structure of left $B$ and right $C$ module as follows:
if $delta:0rightarrow Nxrightarrow f Xxrightarrow g Mrightarrow 0$ is a short exact sequence in left $A$-modules.$forall bin B$,there is a left $A$-module homomorphism $varphi_b:Mrightarrow M$ by senting $m$ to $mb$.take pullback with $varphi_b$ and $g$,we get an element in $Ext^1(N,M)$ this is $bcdot delta$.
similarly,if $forall cin C$,take pushout with natural right multiplication $psi_c:Nrightarrow N$ and $f$,we get an element in $Ext^1(N,M)$ this is $deltacdot c$.
it is easy to Check this: structre of left $B$ module and right $C$ module has associativity.
As follows:
Then $varphi_a=alpha^-$ is using the unique map induced by kernel.
Now we consider the commutative case,we only need to check $rcdot varepsilon=varepsiloncdot r$,suppose $varepsilon:0rightarrow Nxrightarrow f Xxrightarrow g Mrightarrow 0$ is
a short exact sequence in $R-Mod$.Since $R$ is commutative ring,we have pullback diagram:
so $varepsilon= rcdotvarepsilon$,similarly $varepsilon =varepsiloncdot r$.we are done.
for $i>1$,the same.
answered Nov 20 at 1:48
Sky
1,228212
Thanks for your comment, this is very helpful. It is still somewhat unclear to me what happens for higher Exts, though. Is there some way to extend what you have written? I'm happy to go read on my own, if there is some source that describes how to take this structure on Ext^1 and deduce the structure on Ext^i.
– Eric Canton
Nov 20 at 19:33
@EricCanton GTM4
– Sky
Nov 21 at 0:01
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
we can just see the case $i=1$:
In general,if $M$ is $A-B$ bimodule(i.e.left $A$ module and right $B$ module and $(am)b=a(mb)$),N is $A-C$ bimodule,then $Ext^1(M,N)$ is a $B-C$ bimodule.the structure of left $B$ and right $C$ module as follows:
if $delta:0rightarrow Nxrightarrow f Xxrightarrow g Mrightarrow 0$ is a short exact sequence in left $A$-modules.$forall bin B$,there is a left $A$-module homomorphism $varphi_b:Mrightarrow M$ by senting $m$ to $mb$.take pullback with $varphi_b$ and $g$,we get an element in $Ext^1(N,M)$ this is $bcdot delta$.
similarly,if $forall cin C$,take pushout with natural right multiplication $psi_c:Nrightarrow N$ and $f$,we get an element in $Ext^1(N,M)$ this is $deltacdot c$.
it is easy to Check this: structre of left $B$ module and right $C$ module has associativity.
As follows:
Then $varphi_a=alpha^-$ is using the unique map induced by kernel.
Now we consider the commutative case,we only need to check $rcdot varepsilon=varepsiloncdot r$,suppose $varepsilon:0rightarrow Nxrightarrow f Xxrightarrow g Mrightarrow 0$ is
a short exact sequence in $R-Mod$.Since $R$ is commutative ring,we have pullback diagram:
so $varepsilon= rcdotvarepsilon$,similarly $varepsilon =varepsiloncdot r$.we are done.
for $i>1$,the same.
answered Nov 20 at 1:48
Sky
1,228212
Thanks for your comment, this is very helpful. It is still somewhat unclear to me what happens for higher Exts, though. Is there some way to extend what you have written? I'm happy to go read on my own, if there is some source that describes how to take this structure on Ext^1 and deduce the structure on Ext^i.
– Eric Canton
Nov 20 at 19:33
@EricCanton GTM4
– Sky
Nov 21 at 0:01
add a comment |
up vote
3
down vote
accepted
we can just see the case $i=1$:
In general,if $M$ is $A-B$ bimodule(i.e.left $A$ module and right $B$ module and $(am)b=a(mb)$),N is $A-C$ bimodule,then $Ext^1(M,N)$ is a $B-C$ bimodule.the structure of left $B$ and right $C$ module as follows:
if $delta:0rightarrow Nxrightarrow f Xxrightarrow g Mrightarrow 0$ is a short exact sequence in left $A$-modules.$forall bin B$,there is a left $A$-module homomorphism $varphi_b:Mrightarrow M$ by senting $m$ to $mb$.take pullback with $varphi_b$ and $g$,we get an element in $Ext^1(N,M)$ this is $bcdot delta$.
similarly,if $forall cin C$,take pushout with natural right multiplication $psi_c:Nrightarrow N$ and $f$,we get an element in $Ext^1(N,M)$ this is $deltacdot c$.
it is easy to Check this: structre of left $B$ module and right $C$ module has associativity.
As follows:
Then $varphi_a=alpha^-$ is using the unique map induced by kernel.
Now we consider the commutative case,we only need to check $rcdot varepsilon=varepsiloncdot r$,suppose $varepsilon:0rightarrow Nxrightarrow f Xxrightarrow g Mrightarrow 0$ is
a short exact sequence in $R-Mod$.Since $R$ is commutative ring,we have pullback diagram:
so $varepsilon= rcdotvarepsilon$,similarly $varepsilon =varepsiloncdot r$.we are done.
for $i>1$,the same.
answered Nov 20 at 1:48
Sky
1,228212
Thanks for your comment, this is very helpful. It is still somewhat unclear to me what happens for higher Exts, though. Is there some way to extend what you have written? I'm happy to go read on my own, if there is some source that describes how to take this structure on Ext^1 and deduce the structure on Ext^i.
– Eric Canton
Nov 20 at 19:33
@EricCanton GTM4
– Sky
Nov 21 at 0:01
add a comment |
up vote
3
down vote
accepted
up vote
3
down vote
accepted
we can just see the case $i=1$:
In general,if $M$ is $A-B$ bimodule(i.e.left $A$ module and right $B$ module and $(am)b=a(mb)$),N is $A-C$ bimodule,then $Ext^1(M,N)$ is a $B-C$ bimodule.the structure of left $B$ and right $C$ module as follows:
if $delta:0rightarrow Nxrightarrow f Xxrightarrow g Mrightarrow 0$ is a short exact sequence in left $A$-modules.$forall bin B$,there is a left $A$-module homomorphism $varphi_b:Mrightarrow M$ by senting $m$ to $mb$.take pullback with $varphi_b$ and $g$,we get an element in $Ext^1(N,M)$ this is $bcdot delta$.
similarly,if $forall cin C$,take pushout with natural right multiplication $psi_c:Nrightarrow N$ and $f$,we get an element in $Ext^1(N,M)$ this is $deltacdot c$.
it is easy to Check this: structre of left $B$ module and right $C$ module has associativity.
As follows:
Then $varphi_a=alpha^-$ is using the unique map induced by kernel.
Now we consider the commutative case,we only need to check $rcdot varepsilon=varepsiloncdot r$,suppose $varepsilon:0rightarrow Nxrightarrow f Xxrightarrow g Mrightarrow 0$ is
a short exact sequence in $R-Mod$.Since $R$ is commutative ring,we have pullback diagram:
so $varepsilon= rcdotvarepsilon$,similarly $varepsilon =varepsiloncdot r$.we are done.
for $i>1$,the same.
answered Nov 20 at 1:48
Sky
1,228212
we can just see the case $i=1$:
In general,if $M$ is $A-B$ bimodule(i.e.left $A$ module and right $B$ module and $(am)b=a(mb)$),N is $A-C$ bimodule,then $Ext^1(M,N)$ is a $B-C$ bimodule.the structure of left $B$ and right $C$ module as follows:
if $delta:0rightarrow Nxrightarrow f Xxrightarrow g Mrightarrow 0$ is a short exact sequence in left $A$-modules.$forall bin B$,there is a left $A$-module homomorphism $varphi_b:Mrightarrow M$ by senting $m$ to $mb$.take pullback with $varphi_b$ and $g$,we get an element in $Ext^1(N,M)$ this is $bcdot delta$.
similarly,if $forall cin C$,take pushout with natural right multiplication $psi_c:Nrightarrow N$ and $f$,we get an element in $Ext^1(N,M)$ this is $deltacdot c$.
it is easy to Check this: structre of left $B$ module and right $C$ module has associativity.
As follows:
Then $varphi_a=alpha^-$ is using the unique map induced by kernel.
Now we consider the commutative case,we only need to check $rcdot varepsilon=varepsiloncdot r$,suppose $varepsilon:0rightarrow Nxrightarrow f Xxrightarrow g Mrightarrow 0$ is
a short exact sequence in $R-Mod$.Since $R$ is commutative ring,we have pullback diagram:
so $varepsilon= rcdotvarepsilon$,similarly $varepsilon =varepsiloncdot r$.we are done.
for $i>1$,the same.
answered Nov 20 at 1:48
Sky
1,228212
edited Nov 21 at 1:10
answered Nov 20 at 1:48
Sky
1,228212
answered Nov 20 at 1:48
Sky
1,228212
answered Nov 20 at 1:48
Sky
1,228212
1,228212
Thanks for your comment, this is very helpful. It is still somewhat unclear to me what happens for higher Exts, though. Is there some way to extend what you have written? I'm happy to go read on my own, if there is some source that describes how to take this structure on Ext^1 and deduce the structure on Ext^i.
– Eric Canton
Nov 20 at 19:33
@EricCanton GTM4
– Sky
Nov 21 at 0:01
add a comment |
Thanks for your comment, this is very helpful. It is still somewhat unclear to me what happens for higher Exts, though. Is there some way to extend what you have written? I'm happy to go read on my own, if there is some source that describes how to take this structure on Ext^1 and deduce the structure on Ext^i.
– Eric Canton
Nov 20 at 19:33
@EricCanton GTM4
– Sky
Nov 21 at 0:01
Thanks for your comment, this is very helpful. It is still somewhat unclear to me what happens for higher Exts, though. Is there some way to extend what you have written? I'm happy to go read on my own, if there is some source that describes how to take this structure on Ext^1 and deduce the structure on Ext^i.
– Eric Canton
Nov 20 at 19:33
Thanks for your comment, this is very helpful. It is still somewhat unclear to me what happens for higher Exts, though. Is there some way to extend what you have written? I'm happy to go read on my own, if there is some source that describes how to take this structure on Ext^1 and deduce the structure on Ext^i.
– Eric Canton
Nov 20 at 19:33
@EricCanton GTM4
– Sky
Nov 21 at 0:01
@EricCanton GTM4
– Sky
Nov 21 at 0:01
add a comment |
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