In $R$-Mod Category, example for $Bcong A oplus C nRightarrow 0 to A to B to C to0$ splits. The Next CEO of Stack OverflowA nonsplit short exact sequence of abelian groups with $B cong A oplus C$Example of a non-splitting exact sequence $0rightarrow Mrightarrow Moplus Nrightarrow Nrightarrow 0$, questionExact sequence of modulesShowing $Mcong M'oplus M''$ given an exact sequence$0to C'to Cto C''to0$ splits if $Ccong C'oplus C''$ as a chain complex?Proposition 3.2.7 of Bland on short exact sequencesExample of a non-splitting exact sequence $0rightarrow Mrightarrow Moplus Nrightarrow Nrightarrow 0$, questionProposition 3.2.7 of Bland on short exact sequences, part 2A question about split short exact sequence of modulesA short exact sequence with $M=M_1 oplus M_2$ that does not splitProof that $ 0 to Hom(L,D) oversetvarphi 'rightarrow Hom(N,D) oversetpsi 'rightarrow Hom(M,D)$ is an exact sequenceA strange notation for the image of a morphism under a functor
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Multiple labels for a single equation
In $R$-Mod Category, example for $Bcong A oplus C nRightarrow 0 to A to B to C to0$ splits.
The Next CEO of Stack OverflowA nonsplit short exact sequence of abelian groups with $B cong A oplus C$Example of a non-splitting exact sequence $0rightarrow Mrightarrow Moplus Nrightarrow Nrightarrow 0$, questionExact sequence of modulesShowing $Mcong M'oplus M''$ given an exact sequence$0to C'to Cto C''to0$ splits if $Ccong C'oplus C''$ as a chain complex?Proposition 3.2.7 of Bland on short exact sequencesExample of a non-splitting exact sequence $0rightarrow Mrightarrow Moplus Nrightarrow Nrightarrow 0$, questionProposition 3.2.7 of Bland on short exact sequences, part 2A question about split short exact sequence of modulesA short exact sequence with $M=M_1 oplus M_2$ that does not splitProof that $ 0 to Hom(L,D) oversetvarphi 'rightarrow Hom(N,D) oversetpsi 'rightarrow Hom(M,D)$ is an exact sequenceA strange notation for the image of a morphism under a functor
$begingroup$
From splitting lemma, we know in $R$-Mod Category, short exact sequence
$0 to A stackrelfrightarrow B stackrelgrightarrow C to0$ splits
if it satisfies one of the following equivalent conditions:
$(1) exists f_1intextHom(B,A)text s.t. f_1circ f=textId_A$.
$(2) exists g_1intextHom(C,B)text s.t. gcirc g_1=textId_C$.
$(3) textIm f=textKer g$ is direct summand of $B$.
$(4) exists text isomorphism h:Bto A oplus C text s.t. $
$h circ f text is natural injection and g circ h^-1 text is natural projection.$
And $0 to A stackrelfrightarrow B stackrelgrightarrow C to0$ splits $implies Bcong A oplus C$.
If we only have $Bcong A oplus C$, is there any example for
$0 to A to B to C to0$ doesn't split?
Related questions:
$(1)$ Example for infinitely generated modules.
$(2)$ Example for abelian groups.
As proved in answer below, it's ture for modules on commutative ring with finite length.
Special thanks for jgon, for his knowledge, time, patience and friendliness.
abstract-algebra modules
asked Dec 13 '18 at 14:30
AndrewsAndrews
1,2812422
$endgroup$
add a comment |
$begingroup$
From splitting lemma, we know in $R$-Mod Category, short exact sequence
$0 to A stackrelfrightarrow B stackrelgrightarrow C to0$ splits
if it satisfies one of the following equivalent conditions:
$(1) exists f_1intextHom(B,A)text s.t. f_1circ f=textId_A$.
$(2) exists g_1intextHom(C,B)text s.t. gcirc g_1=textId_C$.
$(3) textIm f=textKer g$ is direct summand of $B$.
$(4) exists text isomorphism h:Bto A oplus C text s.t. $
$h circ f text is natural injection and g circ h^-1 text is natural projection.$
And $0 to A stackrelfrightarrow B stackrelgrightarrow C to0$ splits $implies Bcong A oplus C$.
If we only have $Bcong A oplus C$, is there any example for
$0 to A to B to C to0$ doesn't split?
Related questions:
$(1)$ Example for infinitely generated modules.
$(2)$ Example for abelian groups.
As proved in answer below, it's ture for modules on commutative ring with finite length.
Special thanks for jgon, for his knowledge, time, patience and friendliness.
abstract-algebra modules
asked Dec 13 '18 at 14:30
AndrewsAndrews
1,2812422
$endgroup$
$begingroup$
You mean, there exists an isomorphism, but it just doesn't satisfy the injection/projection conditions?
$endgroup$
– rschwieb
Dec 13 '18 at 14:40
$begingroup$
Possible duplicate of A nonsplit short exact sequence of abelian groups with $B cong A oplus C$
$endgroup$
– yamete kudasai
Dec 13 '18 at 14:42
$begingroup$
What do you mean by "drop the condition"?
$endgroup$
– positrón0802
Dec 13 '18 at 14:57
1
$begingroup$
I'm confused. Are you looking for another answer? How is this not answered by your linked question (2) (after all, abelian groups are $BbbZ$-modules)? Your answer doesn't prove that short exact sequences of finitely generated modules split when the middle module is isomorphic to the direct sum. It proves that they split when the modules have finite length. Finite generation is equivalent to finite length only when $R$ is Artinian. Also the answer to the linked question gives an example of a sequence of f.g. modules over a noncommutative ring that doesn't split.
$endgroup$
– jgon
2 days ago
$begingroup$
@jgon Thank you for your edit and kindly explaination, and sorry for my mistake. The linked questions completely answered my question. The answer below surely proves they split when the modules have finite length. I could't open that link when I edit hours ago so I made a big mistake. Thank you for your time and patience.
$endgroup$
– Andrews
2 days ago
add a comment |
$begingroup$
From splitting lemma, we know in $R$-Mod Category, short exact sequence
$0 to A stackrelfrightarrow B stackrelgrightarrow C to0$ splits
if it satisfies one of the following equivalent conditions:
$(1) exists f_1intextHom(B,A)text s.t. f_1circ f=textId_A$.
$(2) exists g_1intextHom(C,B)text s.t. gcirc g_1=textId_C$.
$(3) textIm f=textKer g$ is direct summand of $B$.
$(4) exists text isomorphism h:Bto A oplus C text s.t. $
$h circ f text is natural injection and g circ h^-1 text is natural projection.$
And $0 to A stackrelfrightarrow B stackrelgrightarrow C to0$ splits $implies Bcong A oplus C$.
If we only have $Bcong A oplus C$, is there any example for
$0 to A to B to C to0$ doesn't split?
Related questions:
$(1)$ Example for infinitely generated modules.
$(2)$ Example for abelian groups.
As proved in answer below, it's ture for modules on commutative ring with finite length.
Special thanks for jgon, for his knowledge, time, patience and friendliness.
abstract-algebra modules
asked Dec 13 '18 at 14:30
AndrewsAndrews
1,2812422
$endgroup$
From splitting lemma, we know in $R$-Mod Category, short exact sequence
$0 to A stackrelfrightarrow B stackrelgrightarrow C to0$ splits
if it satisfies one of the following equivalent conditions:
$(1) exists f_1intextHom(B,A)text s.t. f_1circ f=textId_A$.
$(2) exists g_1intextHom(C,B)text s.t. gcirc g_1=textId_C$.
$(3) textIm f=textKer g$ is direct summand of $B$.
$(4) exists text isomorphism h:Bto A oplus C text s.t. $
$h circ f text is natural injection and g circ h^-1 text is natural projection.$
And $0 to A stackrelfrightarrow B stackrelgrightarrow C to0$ splits $implies Bcong A oplus C$.
If we only have $Bcong A oplus C$, is there any example for
$0 to A to B to C to0$ doesn't split?
Related questions:
$(1)$ Example for infinitely generated modules.
$(2)$ Example for abelian groups.
As proved in answer below, it's ture for modules on commutative ring with finite length.
Special thanks for jgon, for his knowledge, time, patience and friendliness.
abstract-algebra modules
abstract-algebra modules
asked Dec 13 '18 at 14:30
AndrewsAndrews
1,2812422
asked Dec 13 '18 at 14:30
AndrewsAndrews
1,2812422
edited 2 days ago
Andrews
asked Dec 13 '18 at 14:30
AndrewsAndrews
1,2812422
asked Dec 13 '18 at 14:30
AndrewsAndrews
1,2812422
asked Dec 13 '18 at 14:30
AndrewsAndrews
1,2812422
1,2812422
$begingroup$
You mean, there exists an isomorphism, but it just doesn't satisfy the injection/projection conditions?
$endgroup$
– rschwieb
Dec 13 '18 at 14:40
$begingroup$
Possible duplicate of A nonsplit short exact sequence of abelian groups with $B cong A oplus C$
$endgroup$
– yamete kudasai
Dec 13 '18 at 14:42
$begingroup$
What do you mean by "drop the condition"?
$endgroup$
– positrón0802
Dec 13 '18 at 14:57
1
$begingroup$
I'm confused. Are you looking for another answer? How is this not answered by your linked question (2) (after all, abelian groups are $BbbZ$-modules)? Your answer doesn't prove that short exact sequences of finitely generated modules split when the middle module is isomorphic to the direct sum. It proves that they split when the modules have finite length. Finite generation is equivalent to finite length only when $R$ is Artinian. Also the answer to the linked question gives an example of a sequence of f.g. modules over a noncommutative ring that doesn't split.
$endgroup$
– jgon
2 days ago
$begingroup$
@jgon Thank you for your edit and kindly explaination, and sorry for my mistake. The linked questions completely answered my question. The answer below surely proves they split when the modules have finite length. I could't open that link when I edit hours ago so I made a big mistake. Thank you for your time and patience.
$endgroup$
– Andrews
2 days ago
add a comment |
$begingroup$
You mean, there exists an isomorphism, but it just doesn't satisfy the injection/projection conditions?
$endgroup$
– rschwieb
Dec 13 '18 at 14:40
$begingroup$
Possible duplicate of A nonsplit short exact sequence of abelian groups with $B cong A oplus C$
$endgroup$
– yamete kudasai
Dec 13 '18 at 14:42
$begingroup$
What do you mean by "drop the condition"?
$endgroup$
– positrón0802
Dec 13 '18 at 14:57
1
$begingroup$
I'm confused. Are you looking for another answer? How is this not answered by your linked question (2) (after all, abelian groups are $BbbZ$-modules)? Your answer doesn't prove that short exact sequences of finitely generated modules split when the middle module is isomorphic to the direct sum. It proves that they split when the modules have finite length. Finite generation is equivalent to finite length only when $R$ is Artinian. Also the answer to the linked question gives an example of a sequence of f.g. modules over a noncommutative ring that doesn't split.
$endgroup$
– jgon
2 days ago
$begingroup$
@jgon Thank you for your edit and kindly explaination, and sorry for my mistake. The linked questions completely answered my question. The answer below surely proves they split when the modules have finite length. I could't open that link when I edit hours ago so I made a big mistake. Thank you for your time and patience.
$endgroup$
– Andrews
2 days ago
$begingroup$
You mean, there exists an isomorphism, but it just doesn't satisfy the injection/projection conditions?
$endgroup$
– rschwieb
Dec 13 '18 at 14:40
$begingroup$
You mean, there exists an isomorphism, but it just doesn't satisfy the injection/projection conditions?
$endgroup$
– rschwieb
Dec 13 '18 at 14:40
$begingroup$
Possible duplicate of A nonsplit short exact sequence of abelian groups with $B cong A oplus C$
$endgroup$
– yamete kudasai
Dec 13 '18 at 14:42
$begingroup$
Possible duplicate of A nonsplit short exact sequence of abelian groups with $B cong A oplus C$
$endgroup$
– yamete kudasai
Dec 13 '18 at 14:42
$begingroup$
What do you mean by "drop the condition"?
$endgroup$
– positrón0802
Dec 13 '18 at 14:57
$begingroup$
What do you mean by "drop the condition"?
$endgroup$
– positrón0802
Dec 13 '18 at 14:57
1
1
$begingroup$
I'm confused. Are you looking for another answer? How is this not answered by your linked question (2) (after all, abelian groups are $BbbZ$-modules)? Your answer doesn't prove that short exact sequences of finitely generated modules split when the middle module is isomorphic to the direct sum. It proves that they split when the modules have finite length. Finite generation is equivalent to finite length only when $R$ is Artinian. Also the answer to the linked question gives an example of a sequence of f.g. modules over a noncommutative ring that doesn't split.
$endgroup$
– jgon
2 days ago
$begingroup$
I'm confused. Are you looking for another answer? How is this not answered by your linked question (2) (after all, abelian groups are $BbbZ$-modules)? Your answer doesn't prove that short exact sequences of finitely generated modules split when the middle module is isomorphic to the direct sum. It proves that they split when the modules have finite length. Finite generation is equivalent to finite length only when $R$ is Artinian. Also the answer to the linked question gives an example of a sequence of f.g. modules over a noncommutative ring that doesn't split.
$endgroup$
– jgon
2 days ago
$begingroup$
@jgon Thank you for your edit and kindly explaination, and sorry for my mistake. The linked questions completely answered my question. The answer below surely proves they split when the modules have finite length. I could't open that link when I edit hours ago so I made a big mistake. Thank you for your time and patience.
$endgroup$
– Andrews
2 days ago
$begingroup$
@jgon Thank you for your edit and kindly explaination, and sorry for my mistake. The linked questions completely answered my question. The answer below surely proves they split when the modules have finite length. I could't open that link when I edit hours ago so I made a big mistake. Thank you for your time and patience.
$endgroup$
– Andrews
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Let $R$ be a commutative ring and let
$$beginequationlabeleq:ses0to Ato Bto Cto 0tag*endequation$$
be a short exact sequence of modules of finite length over $R$. We show that if $Bcong Aoplus C$, then $eqrefeq:ses$ splits.
Since $Bcong Aoplus C$, we have that $$newcommandHomoperatornameHomHom_R(C,B)congHom_R(C,A)oplusHom_R(C,C).$$
Hence $$ell(Hom_R(C,B))=ell(Hom_R(C,A))+ell(Hom_R(C,C))$$ where $ell$ denotes length.
Write $f:Bto C$ for the map in $eqrefeq:ses$. Apply $Hom_R(C,-)$ to the sequence $eqrefeq:ses$.
From the left exactness of $Hom_R(C,-)$, we obtain the exact sequence
$$0to Hom_R(C,A)to Hom_R(C,B)to Hom_R(C,C),$$
where the map $f_* : Hom_R(C,B)to Hom_R(C,C)$ is given by $gmapsto fcirc g$.
Let $N$ denote the cokernel of $f_*$, which evidently has finite length. It follows that
$$0to Hom_R(C,A)to Hom_R(C,B)to Hom_R(C,C)to N to 0$$
is exact, and hence that $ell(N)=0$.
Hence $f_*$ is surjective, which shows that the sequence $eqrefeq:ses$ splits.
Here ring $R$ is commutative to make Hom$_R(C, - )$ a $R$-module.
answered Dec 13 '18 at 23:50
AndrewsAndrews
1,2812422
$endgroup$
1
$begingroup$
The proof only needs that the hom spaces have finite length, so the result holds in any hom-finite abelian category. In particular, this includes coherent sheaves on a projective $k$-scheme, for a field $k$.
$endgroup$
– Andrew Hubery
2 days ago
add a comment |
Your Answer
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$begingroup$
Let $R$ be a commutative ring and let
$$beginequationlabeleq:ses0to Ato Bto Cto 0tag*endequation$$
be a short exact sequence of modules of finite length over $R$. We show that if $Bcong Aoplus C$, then $eqrefeq:ses$ splits.
Since $Bcong Aoplus C$, we have that $$newcommandHomoperatornameHomHom_R(C,B)congHom_R(C,A)oplusHom_R(C,C).$$
Hence $$ell(Hom_R(C,B))=ell(Hom_R(C,A))+ell(Hom_R(C,C))$$ where $ell$ denotes length.
Write $f:Bto C$ for the map in $eqrefeq:ses$. Apply $Hom_R(C,-)$ to the sequence $eqrefeq:ses$.
From the left exactness of $Hom_R(C,-)$, we obtain the exact sequence
$$0to Hom_R(C,A)to Hom_R(C,B)to Hom_R(C,C),$$
where the map $f_* : Hom_R(C,B)to Hom_R(C,C)$ is given by $gmapsto fcirc g$.
Let $N$ denote the cokernel of $f_*$, which evidently has finite length. It follows that
$$0to Hom_R(C,A)to Hom_R(C,B)to Hom_R(C,C)to N to 0$$
is exact, and hence that $ell(N)=0$.
Hence $f_*$ is surjective, which shows that the sequence $eqrefeq:ses$ splits.
Here ring $R$ is commutative to make Hom$_R(C, - )$ a $R$-module.
answered Dec 13 '18 at 23:50
AndrewsAndrews
1,2812422
$endgroup$
1
$begingroup$
The proof only needs that the hom spaces have finite length, so the result holds in any hom-finite abelian category. In particular, this includes coherent sheaves on a projective $k$-scheme, for a field $k$.
$endgroup$
– Andrew Hubery
2 days ago
add a comment |
$begingroup$
Let $R$ be a commutative ring and let
$$beginequationlabeleq:ses0to Ato Bto Cto 0tag*endequation$$
be a short exact sequence of modules of finite length over $R$. We show that if $Bcong Aoplus C$, then $eqrefeq:ses$ splits.
Since $Bcong Aoplus C$, we have that $$newcommandHomoperatornameHomHom_R(C,B)congHom_R(C,A)oplusHom_R(C,C).$$
Hence $$ell(Hom_R(C,B))=ell(Hom_R(C,A))+ell(Hom_R(C,C))$$ where $ell$ denotes length.
Write $f:Bto C$ for the map in $eqrefeq:ses$. Apply $Hom_R(C,-)$ to the sequence $eqrefeq:ses$.
From the left exactness of $Hom_R(C,-)$, we obtain the exact sequence
$$0to Hom_R(C,A)to Hom_R(C,B)to Hom_R(C,C),$$
where the map $f_* : Hom_R(C,B)to Hom_R(C,C)$ is given by $gmapsto fcirc g$.
Let $N$ denote the cokernel of $f_*$, which evidently has finite length. It follows that
$$0to Hom_R(C,A)to Hom_R(C,B)to Hom_R(C,C)to N to 0$$
is exact, and hence that $ell(N)=0$.
Hence $f_*$ is surjective, which shows that the sequence $eqrefeq:ses$ splits.
Here ring $R$ is commutative to make Hom$_R(C, - )$ a $R$-module.
answered Dec 13 '18 at 23:50
AndrewsAndrews
1,2812422
$endgroup$
1
$begingroup$
The proof only needs that the hom spaces have finite length, so the result holds in any hom-finite abelian category. In particular, this includes coherent sheaves on a projective $k$-scheme, for a field $k$.
$endgroup$
– Andrew Hubery
2 days ago
add a comment |
$begingroup$
Let $R$ be a commutative ring and let
$$beginequationlabeleq:ses0to Ato Bto Cto 0tag*endequation$$
be a short exact sequence of modules of finite length over $R$. We show that if $Bcong Aoplus C$, then $eqrefeq:ses$ splits.
Since $Bcong Aoplus C$, we have that $$newcommandHomoperatornameHomHom_R(C,B)congHom_R(C,A)oplusHom_R(C,C).$$
Hence $$ell(Hom_R(C,B))=ell(Hom_R(C,A))+ell(Hom_R(C,C))$$ where $ell$ denotes length.
Write $f:Bto C$ for the map in $eqrefeq:ses$. Apply $Hom_R(C,-)$ to the sequence $eqrefeq:ses$.
From the left exactness of $Hom_R(C,-)$, we obtain the exact sequence
$$0to Hom_R(C,A)to Hom_R(C,B)to Hom_R(C,C),$$
where the map $f_* : Hom_R(C,B)to Hom_R(C,C)$ is given by $gmapsto fcirc g$.
Let $N$ denote the cokernel of $f_*$, which evidently has finite length. It follows that
$$0to Hom_R(C,A)to Hom_R(C,B)to Hom_R(C,C)to N to 0$$
is exact, and hence that $ell(N)=0$.
Hence $f_*$ is surjective, which shows that the sequence $eqrefeq:ses$ splits.
Here ring $R$ is commutative to make Hom$_R(C, - )$ a $R$-module.
answered Dec 13 '18 at 23:50
AndrewsAndrews
1,2812422
$endgroup$
Let $R$ be a commutative ring and let
$$beginequationlabeleq:ses0to Ato Bto Cto 0tag*endequation$$
be a short exact sequence of modules of finite length over $R$. We show that if $Bcong Aoplus C$, then $eqrefeq:ses$ splits.
Since $Bcong Aoplus C$, we have that $$newcommandHomoperatornameHomHom_R(C,B)congHom_R(C,A)oplusHom_R(C,C).$$
Hence $$ell(Hom_R(C,B))=ell(Hom_R(C,A))+ell(Hom_R(C,C))$$ where $ell$ denotes length.
Write $f:Bto C$ for the map in $eqrefeq:ses$. Apply $Hom_R(C,-)$ to the sequence $eqrefeq:ses$.
From the left exactness of $Hom_R(C,-)$, we obtain the exact sequence
$$0to Hom_R(C,A)to Hom_R(C,B)to Hom_R(C,C),$$
where the map $f_* : Hom_R(C,B)to Hom_R(C,C)$ is given by $gmapsto fcirc g$.
Let $N$ denote the cokernel of $f_*$, which evidently has finite length. It follows that
$$0to Hom_R(C,A)to Hom_R(C,B)to Hom_R(C,C)to N to 0$$
is exact, and hence that $ell(N)=0$.
Hence $f_*$ is surjective, which shows that the sequence $eqrefeq:ses$ splits.
Here ring $R$ is commutative to make Hom$_R(C, - )$ a $R$-module.
answered Dec 13 '18 at 23:50
AndrewsAndrews
1,2812422
edited 2 days ago
answered Dec 13 '18 at 23:50
AndrewsAndrews
1,2812422
answered Dec 13 '18 at 23:50
AndrewsAndrews
1,2812422
answered Dec 13 '18 at 23:50
AndrewsAndrews
1,2812422
1,2812422
1
$begingroup$
The proof only needs that the hom spaces have finite length, so the result holds in any hom-finite abelian category. In particular, this includes coherent sheaves on a projective $k$-scheme, for a field $k$.
$endgroup$
– Andrew Hubery
2 days ago
add a comment |
1
$begingroup$
The proof only needs that the hom spaces have finite length, so the result holds in any hom-finite abelian category. In particular, this includes coherent sheaves on a projective $k$-scheme, for a field $k$.
$endgroup$
– Andrew Hubery
2 days ago
1
1
$begingroup$
The proof only needs that the hom spaces have finite length, so the result holds in any hom-finite abelian category. In particular, this includes coherent sheaves on a projective $k$-scheme, for a field $k$.
$endgroup$
– Andrew Hubery
2 days ago
$begingroup$
The proof only needs that the hom spaces have finite length, so the result holds in any hom-finite abelian category. In particular, this includes coherent sheaves on a projective $k$-scheme, for a field $k$.
$endgroup$
– Andrew Hubery
2 days ago
add a comment |
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$begingroup$
You mean, there exists an isomorphism, but it just doesn't satisfy the injection/projection conditions?
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– rschwieb
Dec 13 '18 at 14:40
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Possible duplicate of A nonsplit short exact sequence of abelian groups with $B cong A oplus C$
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– yamete kudasai
Dec 13 '18 at 14:42
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What do you mean by "drop the condition"?
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– positrón0802
Dec 13 '18 at 14:57
1
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I'm confused. Are you looking for another answer? How is this not answered by your linked question (2) (after all, abelian groups are $BbbZ$-modules)? Your answer doesn't prove that short exact sequences of finitely generated modules split when the middle module is isomorphic to the direct sum. It proves that they split when the modules have finite length. Finite generation is equivalent to finite length only when $R$ is Artinian. Also the answer to the linked question gives an example of a sequence of f.g. modules over a noncommutative ring that doesn't split.
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– jgon
2 days ago
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@jgon Thank you for your edit and kindly explaination, and sorry for my mistake. The linked questions completely answered my question. The answer below surely proves they split when the modules have finite length. I could't open that link when I edit hours ago so I made a big mistake. Thank you for your time and patience.
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– Andrews
2 days ago