Mixing
Projective resolution of short exact sequence: definition in Jacobson
$begingroup$
Let $0rightarrow M'rightarrow M rightarrow M''rightarrow 0$ be a short-exact sequence of modules (over a ring $R$). By a projective resolution of this sequence, we mean (according to Jacobson volume II, p. 344):
(i) A projective resolution $cdots rightarrow C_1'rightarrow C_0'rightarrow M'rightarrow 0$ of $M'$;
(ii) A projective resolution $cdots rightarrow C_1rightarrow C_0rightarrow Mrightarrow 0$ of $M$;
(iii) A projective resolution $cdots rightarrow C_1''rightarrow C_0''rightarrow M''rightarrow 0$ of $M''$;
(iv) maps $i_n:C_n'rightarrow C_n$ and $p_n:C_nrightarrow C_n''$
such that
(I) for every $ige 0$, $0rightarrow C_i'rightarrow C_irightarrow C_i''rightarrow 0$ is a short exact sequence.
(II) The maps between $C_0',C_0,C_0''$ and $M',M,M''$ [which appear in (i)-(iv)] make two squares commutative.
Q. Is it not necessary in the definition that maps between $C_n',C_n,C_n''$ and $C_{n-1}',C_{n-1},C_{n-1}''$ [which appear in (i)-(iv)] make two squares commutative?
To look at question in simple way, think of three sequences in (i)-(iii) are written exactly one below the other; then it is given that the column of $M$'s is exact. Then in the definition of projective resolution of this exact sequence of $M$'s, we want
the column of $C_0$'s, column of $C_1$'s, etc. exact,
the column of $C_0$'s and $M$'s makes two squares commutative.
My question is, don't we require (in definition) the column of each $C_n$'s and $C_{n-1}$'s make two commutative squares?
As far the argument in Jacobson's book is considered, the author constructs the projective resolution of $M$ in such a way that the maps in that resolution (of $M$) make all small squares commutative.
homology-cohomology homological-algebra projective-module
asked Jan 27 at 6:32
BeginnerBeginner
3,98411226
$endgroup$
add a comment |
$begingroup$
Let $0rightarrow M'rightarrow M rightarrow M''rightarrow 0$ be a short-exact sequence of modules (over a ring $R$). By a projective resolution of this sequence, we mean (according to Jacobson volume II, p. 344):
(i) A projective resolution $cdots rightarrow C_1'rightarrow C_0'rightarrow M'rightarrow 0$ of $M'$;
(ii) A projective resolution $cdots rightarrow C_1rightarrow C_0rightarrow Mrightarrow 0$ of $M$;
(iii) A projective resolution $cdots rightarrow C_1''rightarrow C_0''rightarrow M''rightarrow 0$ of $M''$;
(iv) maps $i_n:C_n'rightarrow C_n$ and $p_n:C_nrightarrow C_n''$
such that
(I) for every $ige 0$, $0rightarrow C_i'rightarrow C_irightarrow C_i''rightarrow 0$ is a short exact sequence.
(II) The maps between $C_0',C_0,C_0''$ and $M',M,M''$ [which appear in (i)-(iv)] make two squares commutative.
Q. Is it not necessary in the definition that maps between $C_n',C_n,C_n''$ and $C_{n-1}',C_{n-1},C_{n-1}''$ [which appear in (i)-(iv)] make two squares commutative?
To look at question in simple way, think of three sequences in (i)-(iii) are written exactly one below the other; then it is given that the column of $M$'s is exact. Then in the definition of projective resolution of this exact sequence of $M$'s, we want
the column of $C_0$'s, column of $C_1$'s, etc. exact,
the column of $C_0$'s and $M$'s makes two squares commutative.
My question is, don't we require (in definition) the column of each $C_n$'s and $C_{n-1}$'s make two commutative squares?
As far the argument in Jacobson's book is considered, the author constructs the projective resolution of $M$ in such a way that the maps in that resolution (of $M$) make all small squares commutative.
homology-cohomology homological-algebra projective-module
asked Jan 27 at 6:32
BeginnerBeginner
3,98411226
$endgroup$
2
$begingroup$
Yes, every square should commute!
$endgroup$
– Lord Shark the Unknown
Jan 27 at 6:36
$begingroup$
In book it is mentioned that only two squares commute, so my doubt was whether commutativity of other squares follows from this?
$endgroup$
– Beginner
Jan 27 at 6:37
1
$begingroup$
If you can draw diagram and add it in post it will be easier to understand your question..
$endgroup$
– Praphulla Koushik
Jan 27 at 7:28
2
$begingroup$
In my edition of Jacobson, it is stated that a resolution of $0to M'to Mto M''to 0$ is a resolution $C',C,C''$ of $M',M,M''$ together with chain homomorphism $C'to C, Cto C''$ such that ... (your condition (I) and (II)). The fact that these are chain homomorphism means that every square must commute.
$endgroup$
– Roland
Jan 27 at 9:01
$begingroup$
@Roland: this is actually the thing I was missing from my reading. Thanks for pointing it. You may post it as answer and I will accept.
$endgroup$
– Beginner
Jan 28 at 8:46
add a comment |
$begingroup$
Let $0rightarrow M'rightarrow M rightarrow M''rightarrow 0$ be a short-exact sequence of modules (over a ring $R$). By a projective resolution of this sequence, we mean (according to Jacobson volume II, p. 344):
(i) A projective resolution $cdots rightarrow C_1'rightarrow C_0'rightarrow M'rightarrow 0$ of $M'$;
(ii) A projective resolution $cdots rightarrow C_1rightarrow C_0rightarrow Mrightarrow 0$ of $M$;
(iii) A projective resolution $cdots rightarrow C_1''rightarrow C_0''rightarrow M''rightarrow 0$ of $M''$;
(iv) maps $i_n:C_n'rightarrow C_n$ and $p_n:C_nrightarrow C_n''$
such that
(I) for every $ige 0$, $0rightarrow C_i'rightarrow C_irightarrow C_i''rightarrow 0$ is a short exact sequence.
(II) The maps between $C_0',C_0,C_0''$ and $M',M,M''$ [which appear in (i)-(iv)] make two squares commutative.
Q. Is it not necessary in the definition that maps between $C_n',C_n,C_n''$ and $C_{n-1}',C_{n-1},C_{n-1}''$ [which appear in (i)-(iv)] make two squares commutative?
To look at question in simple way, think of three sequences in (i)-(iii) are written exactly one below the other; then it is given that the column of $M$'s is exact. Then in the definition of projective resolution of this exact sequence of $M$'s, we want
the column of $C_0$'s, column of $C_1$'s, etc. exact,
the column of $C_0$'s and $M$'s makes two squares commutative.
My question is, don't we require (in definition) the column of each $C_n$'s and $C_{n-1}$'s make two commutative squares?
As far the argument in Jacobson's book is considered, the author constructs the projective resolution of $M$ in such a way that the maps in that resolution (of $M$) make all small squares commutative.
homology-cohomology homological-algebra projective-module
asked Jan 27 at 6:32
BeginnerBeginner
3,98411226
$endgroup$
Let $0rightarrow M'rightarrow M rightarrow M''rightarrow 0$ be a short-exact sequence of modules (over a ring $R$). By a projective resolution of this sequence, we mean (according to Jacobson volume II, p. 344):
(i) A projective resolution $cdots rightarrow C_1'rightarrow C_0'rightarrow M'rightarrow 0$ of $M'$;
(ii) A projective resolution $cdots rightarrow C_1rightarrow C_0rightarrow Mrightarrow 0$ of $M$;
(iii) A projective resolution $cdots rightarrow C_1''rightarrow C_0''rightarrow M''rightarrow 0$ of $M''$;
(iv) maps $i_n:C_n'rightarrow C_n$ and $p_n:C_nrightarrow C_n''$
such that
(I) for every $ige 0$, $0rightarrow C_i'rightarrow C_irightarrow C_i''rightarrow 0$ is a short exact sequence.
(II) The maps between $C_0',C_0,C_0''$ and $M',M,M''$ [which appear in (i)-(iv)] make two squares commutative.
Q. Is it not necessary in the definition that maps between $C_n',C_n,C_n''$ and $C_{n-1}',C_{n-1},C_{n-1}''$ [which appear in (i)-(iv)] make two squares commutative?
To look at question in simple way, think of three sequences in (i)-(iii) are written exactly one below the other; then it is given that the column of $M$'s is exact. Then in the definition of projective resolution of this exact sequence of $M$'s, we want
the column of $C_0$'s, column of $C_1$'s, etc. exact,
the column of $C_0$'s and $M$'s makes two squares commutative.
My question is, don't we require (in definition) the column of each $C_n$'s and $C_{n-1}$'s make two commutative squares?
As far the argument in Jacobson's book is considered, the author constructs the projective resolution of $M$ in such a way that the maps in that resolution (of $M$) make all small squares commutative.
homology-cohomology homological-algebra projective-module
homology-cohomology homological-algebra projective-module
asked Jan 27 at 6:32
BeginnerBeginner
3,98411226
asked Jan 27 at 6:32
BeginnerBeginner
3,98411226
asked Jan 27 at 6:32
BeginnerBeginner
3,98411226
asked Jan 27 at 6:32
BeginnerBeginner
3,98411226
asked Jan 27 at 6:32
BeginnerBeginner
3,98411226
3,98411226
2
$begingroup$
Yes, every square should commute!
$endgroup$
– Lord Shark the Unknown
Jan 27 at 6:36
$begingroup$
In book it is mentioned that only two squares commute, so my doubt was whether commutativity of other squares follows from this?
$endgroup$
– Beginner
Jan 27 at 6:37
1
$begingroup$
If you can draw diagram and add it in post it will be easier to understand your question..
$endgroup$
– Praphulla Koushik
Jan 27 at 7:28
2
$begingroup$
In my edition of Jacobson, it is stated that a resolution of $0to M'to Mto M''to 0$ is a resolution $C',C,C''$ of $M',M,M''$ together with chain homomorphism $C'to C, Cto C''$ such that ... (your condition (I) and (II)). The fact that these are chain homomorphism means that every square must commute.
$endgroup$
– Roland
Jan 27 at 9:01
$begingroup$
@Roland: this is actually the thing I was missing from my reading. Thanks for pointing it. You may post it as answer and I will accept.
$endgroup$
– Beginner
Jan 28 at 8:46
add a comment |
2
$begingroup$
Yes, every square should commute!
$endgroup$
– Lord Shark the Unknown
Jan 27 at 6:36
$begingroup$
In book it is mentioned that only two squares commute, so my doubt was whether commutativity of other squares follows from this?
$endgroup$
– Beginner
Jan 27 at 6:37
1
$begingroup$
If you can draw diagram and add it in post it will be easier to understand your question..
$endgroup$
– Praphulla Koushik
Jan 27 at 7:28
2
$begingroup$
In my edition of Jacobson, it is stated that a resolution of $0to M'to Mto M''to 0$ is a resolution $C',C,C''$ of $M',M,M''$ together with chain homomorphism $C'to C, Cto C''$ such that ... (your condition (I) and (II)). The fact that these are chain homomorphism means that every square must commute.
$endgroup$
– Roland
Jan 27 at 9:01
$begingroup$
@Roland: this is actually the thing I was missing from my reading. Thanks for pointing it. You may post it as answer and I will accept.
$endgroup$
– Beginner
Jan 28 at 8:46
2
2
$begingroup$
Yes, every square should commute!
$endgroup$
– Lord Shark the Unknown
Jan 27 at 6:36
$begingroup$
Yes, every square should commute!
$endgroup$
– Lord Shark the Unknown
Jan 27 at 6:36
$begingroup$
In book it is mentioned that only two squares commute, so my doubt was whether commutativity of other squares follows from this?
$endgroup$
– Beginner
Jan 27 at 6:37
$begingroup$
In book it is mentioned that only two squares commute, so my doubt was whether commutativity of other squares follows from this?
$endgroup$
– Beginner
Jan 27 at 6:37
1
1
$begingroup$
If you can draw diagram and add it in post it will be easier to understand your question..
$endgroup$
– Praphulla Koushik
Jan 27 at 7:28
$begingroup$
If you can draw diagram and add it in post it will be easier to understand your question..
$endgroup$
– Praphulla Koushik
Jan 27 at 7:28
2
2
$begingroup$
In my edition of Jacobson, it is stated that a resolution of $0to M'to Mto M''to 0$ is a resolution $C',C,C''$ of $M',M,M''$ together with chain homomorphism $C'to C, Cto C''$ such that ... (your condition (I) and (II)). The fact that these are chain homomorphism means that every square must commute.
$endgroup$
– Roland
Jan 27 at 9:01
$begingroup$
In my edition of Jacobson, it is stated that a resolution of $0to M'to Mto M''to 0$ is a resolution $C',C,C''$ of $M',M,M''$ together with chain homomorphism $C'to C, Cto C''$ such that ... (your condition (I) and (II)). The fact that these are chain homomorphism means that every square must commute.
$endgroup$
– Roland
Jan 27 at 9:01
$begingroup$
@Roland: this is actually the thing I was missing from my reading. Thanks for pointing it. You may post it as answer and I will accept.
$endgroup$
– Beginner
Jan 28 at 8:46
$begingroup$
@Roland: this is actually the thing I was missing from my reading. Thanks for pointing it. You may post it as answer and I will accept.
$endgroup$
– Beginner
Jan 28 at 8:46
add a comment |
1 Answer
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As requested, this answer is the same as a previous comment.
It is in fact stated that a resolution of $0to M′to Mto M''to 0$ is the data of resolutions $C′,C,C′′$ of $M′,M,M′′$ together with chain homomorphisms $C′to C,Cto C′′$ such that the conditions (I) and (II) in the OP hold. The fact that these are chain homomorphisms means that every square must commute.
answered Jan 28 at 9:39
RolandRoland
7,44911015
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add a comment |
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$begingroup$
As requested, this answer is the same as a previous comment.
It is in fact stated that a resolution of $0to M′to Mto M''to 0$ is the data of resolutions $C′,C,C′′$ of $M′,M,M′′$ together with chain homomorphisms $C′to C,Cto C′′$ such that the conditions (I) and (II) in the OP hold. The fact that these are chain homomorphisms means that every square must commute.
answered Jan 28 at 9:39
RolandRoland
7,44911015
$endgroup$
add a comment |
$begingroup$
As requested, this answer is the same as a previous comment.
It is in fact stated that a resolution of $0to M′to Mto M''to 0$ is the data of resolutions $C′,C,C′′$ of $M′,M,M′′$ together with chain homomorphisms $C′to C,Cto C′′$ such that the conditions (I) and (II) in the OP hold. The fact that these are chain homomorphisms means that every square must commute.
answered Jan 28 at 9:39
RolandRoland
7,44911015
$endgroup$
add a comment |
$begingroup$
As requested, this answer is the same as a previous comment.
It is in fact stated that a resolution of $0to M′to Mto M''to 0$ is the data of resolutions $C′,C,C′′$ of $M′,M,M′′$ together with chain homomorphisms $C′to C,Cto C′′$ such that the conditions (I) and (II) in the OP hold. The fact that these are chain homomorphisms means that every square must commute.
answered Jan 28 at 9:39
RolandRoland
7,44911015
$endgroup$
As requested, this answer is the same as a previous comment.
It is in fact stated that a resolution of $0to M′to Mto M''to 0$ is the data of resolutions $C′,C,C′′$ of $M′,M,M′′$ together with chain homomorphisms $C′to C,Cto C′′$ such that the conditions (I) and (II) in the OP hold. The fact that these are chain homomorphisms means that every square must commute.
answered Jan 28 at 9:39
RolandRoland
7,44911015
answered Jan 28 at 9:39
RolandRoland
7,44911015
answered Jan 28 at 9:39
RolandRoland
7,44911015
answered Jan 28 at 9:39
RolandRoland
7,44911015
7,44911015
add a comment |
add a comment |
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$begingroup$
Yes, every square should commute!
$endgroup$
– Lord Shark the Unknown
Jan 27 at 6:36
$begingroup$
In book it is mentioned that only two squares commute, so my doubt was whether commutativity of other squares follows from this?
$endgroup$
– Beginner
Jan 27 at 6:37
1
$begingroup$
If you can draw diagram and add it in post it will be easier to understand your question..
$endgroup$
– Praphulla Koushik
Jan 27 at 7:28
2
$begingroup$
In my edition of Jacobson, it is stated that a resolution of $0to M'to Mto M''to 0$ is a resolution $C',C,C''$ of $M',M,M''$ together with chain homomorphism $C'to C, Cto C''$ such that ... (your condition (I) and (II)). The fact that these are chain homomorphism means that every square must commute.
$endgroup$
– Roland
Jan 27 at 9:01
$begingroup$
@Roland: this is actually the thing I was missing from my reading. Thanks for pointing it. You may post it as answer and I will accept.
$endgroup$
– Beginner
Jan 28 at 8:46