Mixing
Isomorphism of Tor and tor functor
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In Theorem 6.32. pg 355 of Rotman's Hom. Alg., he proves that two different construction of Torsions conincide,
$$Tor_n^R(A,B) cong tor_n^R(A,B)$$
where
If $B$ is a left $R$-module and $T = - otimes_R B$, define left derived functor $$Tor_n^R(-, B) = L_nT.$$
If $A$ is a right $R$-module and $T=A otimes_R - $, define
$$tor_n^R(A,-) = L_nT.$$
It is a diagrammatic proof so I hope you may have a look at link. There is one part of proof which I cannot understand:
$W=Tor_1(K_i-1,V_j-1), X=Tor_1(K_i-1,V_j), ldots $
This isn't clear from the definition given. What we should have is that
$$Tor_1(K_i-1,V_j) cong ker d_1/im , d_2$$
where with $T=- otimes_RV_j$, $$ TP_i+2 xrightarrowd_2 TP_i+1 xrightarrowd_1 TP_i xrightarrow d_0 K_i-1 $$
why do the equalities hold? In fact even if it holds, shouldnt it be an isomorphism?
abstract-algebra proof-explanation homological-algebra
asked 2 days ago
Cyryl L.
1,7122721
add a comment |Â
up vote
1
down vote
favorite
In Theorem 6.32. pg 355 of Rotman's Hom. Alg., he proves that two different construction of Torsions conincide,
$$Tor_n^R(A,B) cong tor_n^R(A,B)$$
where
If $B$ is a left $R$-module and $T = - otimes_R B$, define left derived functor $$Tor_n^R(-, B) = L_nT.$$
If $A$ is a right $R$-module and $T=A otimes_R - $, define
$$tor_n^R(A,-) = L_nT.$$
It is a diagrammatic proof so I hope you may have a look at link. There is one part of proof which I cannot understand:
$W=Tor_1(K_i-1,V_j-1), X=Tor_1(K_i-1,V_j), ldots $
This isn't clear from the definition given. What we should have is that
$$Tor_1(K_i-1,V_j) cong ker d_1/im , d_2$$
where with $T=- otimes_RV_j$, $$ TP_i+2 xrightarrowd_2 TP_i+1 xrightarrowd_1 TP_i xrightarrow d_0 K_i-1 $$
why do the equalities hold? In fact even if it holds, shouldnt it be an isomorphism?
abstract-algebra proof-explanation homological-algebra
asked 2 days ago
Cyryl L.
1,7122721
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
In Theorem 6.32. pg 355 of Rotman's Hom. Alg., he proves that two different construction of Torsions conincide,
$$Tor_n^R(A,B) cong tor_n^R(A,B)$$
where
If $B$ is a left $R$-module and $T = - otimes_R B$, define left derived functor $$Tor_n^R(-, B) = L_nT.$$
If $A$ is a right $R$-module and $T=A otimes_R - $, define
$$tor_n^R(A,-) = L_nT.$$
It is a diagrammatic proof so I hope you may have a look at link. There is one part of proof which I cannot understand:
$W=Tor_1(K_i-1,V_j-1), X=Tor_1(K_i-1,V_j), ldots $
This isn't clear from the definition given. What we should have is that
$$Tor_1(K_i-1,V_j) cong ker d_1/im , d_2$$
where with $T=- otimes_RV_j$, $$ TP_i+2 xrightarrowd_2 TP_i+1 xrightarrowd_1 TP_i xrightarrow d_0 K_i-1 $$
why do the equalities hold? In fact even if it holds, shouldnt it be an isomorphism?
abstract-algebra proof-explanation homological-algebra
asked 2 days ago
Cyryl L.
1,7122721
In Theorem 6.32. pg 355 of Rotman's Hom. Alg., he proves that two different construction of Torsions conincide,
$$Tor_n^R(A,B) cong tor_n^R(A,B)$$
where
If $B$ is a left $R$-module and $T = - otimes_R B$, define left derived functor $$Tor_n^R(-, B) = L_nT.$$
If $A$ is a right $R$-module and $T=A otimes_R - $, define
$$tor_n^R(A,-) = L_nT.$$
It is a diagrammatic proof so I hope you may have a look at link. There is one part of proof which I cannot understand:
$W=Tor_1(K_i-1,V_j-1), X=Tor_1(K_i-1,V_j), ldots $
This isn't clear from the definition given. What we should have is that
$$Tor_1(K_i-1,V_j) cong ker d_1/im , d_2$$
where with $T=- otimes_RV_j$, $$ TP_i+2 xrightarrowd_2 TP_i+1 xrightarrowd_1 TP_i xrightarrow d_0 K_i-1 $$
why do the equalities hold? In fact even if it holds, shouldnt it be an isomorphism?
abstract-algebra proof-explanation homological-algebra
asked 2 days ago
Cyryl L.
1,7122721
asked 2 days ago
Cyryl L.
1,7122721
asked 2 days ago
Cyryl L.
1,7122721
asked 2 days ago
Cyryl L.
1,7122721
1,7122721
add a comment |Â
add a comment |Â
1 Answer
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Note that if you have a short exact sequence
$0to Kto P to Lto 0$ with $P$ projective and $M$ is a module (to the correct side) then the long exact sequence reads ($P$ projective!)
$$operatornameTor_1(P,M)=0to operatornameTor_1(L,M)to Kotimes Mto Potimes Mto Lotimes Mto 0$$
so that $operatornameTor_1(L,M)$ is the kernel of $Kotimes Mto Potimes M$. This is what Rotman is using throughout.
answered 2 days ago
Pedro Tamaroff♦
93.6k10143290
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Note that if you have a short exact sequence
$0to Kto P to Lto 0$ with $P$ projective and $M$ is a module (to the correct side) then the long exact sequence reads ($P$ projective!)
$$operatornameTor_1(P,M)=0to operatornameTor_1(L,M)to Kotimes Mto Potimes Mto Lotimes Mto 0$$
so that $operatornameTor_1(L,M)$ is the kernel of $Kotimes Mto Potimes M$. This is what Rotman is using throughout.
answered 2 days ago
Pedro Tamaroff♦
93.6k10143290
add a comment |Â
up vote
0
down vote
Note that if you have a short exact sequence
$0to Kto P to Lto 0$ with $P$ projective and $M$ is a module (to the correct side) then the long exact sequence reads ($P$ projective!)
$$operatornameTor_1(P,M)=0to operatornameTor_1(L,M)to Kotimes Mto Potimes Mto Lotimes Mto 0$$
so that $operatornameTor_1(L,M)$ is the kernel of $Kotimes Mto Potimes M$. This is what Rotman is using throughout.
answered 2 days ago
Pedro Tamaroff♦
93.6k10143290
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Note that if you have a short exact sequence
$0to Kto P to Lto 0$ with $P$ projective and $M$ is a module (to the correct side) then the long exact sequence reads ($P$ projective!)
$$operatornameTor_1(P,M)=0to operatornameTor_1(L,M)to Kotimes Mto Potimes Mto Lotimes Mto 0$$
so that $operatornameTor_1(L,M)$ is the kernel of $Kotimes Mto Potimes M$. This is what Rotman is using throughout.
answered 2 days ago
Pedro Tamaroff♦
93.6k10143290
Note that if you have a short exact sequence
$0to Kto P to Lto 0$ with $P$ projective and $M$ is a module (to the correct side) then the long exact sequence reads ($P$ projective!)
$$operatornameTor_1(P,M)=0to operatornameTor_1(L,M)to Kotimes Mto Potimes Mto Lotimes Mto 0$$
so that $operatornameTor_1(L,M)$ is the kernel of $Kotimes Mto Potimes M$. This is what Rotman is using throughout.
answered 2 days ago
Pedro Tamaroff♦
93.6k10143290
answered 2 days ago
Pedro Tamaroff♦
93.6k10143290
answered 2 days ago
Pedro Tamaroff♦
93.6k10143290
answered 2 days ago
Pedro Tamaroff♦
93.6k10143290
93.6k10143290
add a comment |Â
add a comment |Â
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