Mixing
Full subcategories of $mathsf{Mod}_R$ which are “almost” abelian
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Is there an example of a full subcategory of $mathsf{Mod}_R$ (namely left $R$-modules) which are not abelian categories, but are "almost" abelian -- that is, they satisfy all of the prerequisites of an abelian category (additive + kernels/cokernels) except that the canonical map $coim(f) to im(f)$ is not an isomorphism.
There are of course many examples of those that are, e.g. torsion $R$-modules form a Serre subcategory of $mathsf{Mod}_R$ and coherent $R$-modules form a weak Serre subcategory of $mathsf{Mod}_R$.
commutative-algebra abelian-categories
asked Oct 11 '17 at 13:11
MCTMCT
14.5k42668
$endgroup$
add a comment |
$begingroup$
Is there an example of a full subcategory of $mathsf{Mod}_R$ (namely left $R$-modules) which are not abelian categories, but are "almost" abelian -- that is, they satisfy all of the prerequisites of an abelian category (additive + kernels/cokernels) except that the canonical map $coim(f) to im(f)$ is not an isomorphism.
There are of course many examples of those that are, e.g. torsion $R$-modules form a Serre subcategory of $mathsf{Mod}_R$ and coherent $R$-modules form a weak Serre subcategory of $mathsf{Mod}_R$.
commutative-algebra abelian-categories
asked Oct 11 '17 at 13:11
MCTMCT
14.5k42668
$endgroup$
1
$begingroup$
So you would need to choose the subcategory in such a way that at least one of those objects is different from what it was in the original category. That seems tricky.
$endgroup$
– Tobias Kildetoft
Oct 11 '17 at 13:16
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Yep that's exactly right.
$endgroup$
– MCT
Oct 11 '17 at 13:32
add a comment |
$begingroup$
Is there an example of a full subcategory of $mathsf{Mod}_R$ (namely left $R$-modules) which are not abelian categories, but are "almost" abelian -- that is, they satisfy all of the prerequisites of an abelian category (additive + kernels/cokernels) except that the canonical map $coim(f) to im(f)$ is not an isomorphism.
There are of course many examples of those that are, e.g. torsion $R$-modules form a Serre subcategory of $mathsf{Mod}_R$ and coherent $R$-modules form a weak Serre subcategory of $mathsf{Mod}_R$.
commutative-algebra abelian-categories
asked Oct 11 '17 at 13:11
MCTMCT
14.5k42668
$endgroup$
Is there an example of a full subcategory of $mathsf{Mod}_R$ (namely left $R$-modules) which are not abelian categories, but are "almost" abelian -- that is, they satisfy all of the prerequisites of an abelian category (additive + kernels/cokernels) except that the canonical map $coim(f) to im(f)$ is not an isomorphism.
There are of course many examples of those that are, e.g. torsion $R$-modules form a Serre subcategory of $mathsf{Mod}_R$ and coherent $R$-modules form a weak Serre subcategory of $mathsf{Mod}_R$.
commutative-algebra abelian-categories
commutative-algebra abelian-categories
asked Oct 11 '17 at 13:11
MCTMCT
14.5k42668
asked Oct 11 '17 at 13:11
MCTMCT
14.5k42668
edited Oct 11 '17 at 13:22
MCT
asked Oct 11 '17 at 13:11
MCTMCT
14.5k42668
asked Oct 11 '17 at 13:11
MCTMCT
14.5k42668
asked Oct 11 '17 at 13:11
MCTMCT
14.5k42668
14.5k42668
1
$begingroup$
So you would need to choose the subcategory in such a way that at least one of those objects is different from what it was in the original category. That seems tricky.
$endgroup$
– Tobias Kildetoft
Oct 11 '17 at 13:16
$begingroup$
Yep that's exactly right.
$endgroup$
– MCT
Oct 11 '17 at 13:32
add a comment |
1
$begingroup$
So you would need to choose the subcategory in such a way that at least one of those objects is different from what it was in the original category. That seems tricky.
$endgroup$
– Tobias Kildetoft
Oct 11 '17 at 13:16
$begingroup$
Yep that's exactly right.
$endgroup$
– MCT
Oct 11 '17 at 13:32
1
1
$begingroup$
So you would need to choose the subcategory in such a way that at least one of those objects is different from what it was in the original category. That seems tricky.
$endgroup$
– Tobias Kildetoft
Oct 11 '17 at 13:16
$begingroup$
So you would need to choose the subcategory in such a way that at least one of those objects is different from what it was in the original category. That seems tricky.
$endgroup$
– Tobias Kildetoft
Oct 11 '17 at 13:16
$begingroup$
Yep that's exactly right.
$endgroup$
– MCT
Oct 11 '17 at 13:32
$begingroup$
Yep that's exactly right.
$endgroup$
– MCT
Oct 11 '17 at 13:32
add a comment |
1 Answer
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For the case where $R=Bbb Z$ : the full subcategory $mathbf{Ab}_{t-f}$ of torsion-free abelian groups is additive, has kernels (because a subgroup of a torsion-free abelian group is torsion free), and it also has cokernels : given a morphism $f:Ato B$, its cokernel in $mathbf{Ab}_{t-f}$ is the quotient of $frac{B}{f(A)}$ by its torsion subgroup $Tleft(frac{B}{A}right)$, or equivalently the quotient $frac{B}{widehat{f(A)}}$, where
$$widehat{f(A)}={bin Bmid exists min Bbb Zsetminus0 text{ such that } mcdot bin f(A) }.$$
Indeed, any morphism $g:Bto C$ such that $gf=0$ factorizes in $mathbf{Ab}$ through $frac{B}{f(A)}$, and then if $C$ is torsion-free the corresponding morphism $frac{B}{f(A)}to C$ should be $0$ on the torsion of $frac{B}{f(A)}$.
Now this is not an abelian subcategory because for example the inclusion $2Bbb Zto Bbb Z$ has both its kernel and cokernel equal to $0$, but is not an isomorphism.
answered Oct 11 '17 at 13:40
Arnaud D.Arnaud D.
16.2k52445
$endgroup$
$begingroup$
Great example. You could even take the subcategory to be finitely generated, since the ring is Noetherian the kernel will be the same.
$endgroup$
– MCT
Oct 11 '17 at 13:47
add a comment |
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$begingroup$
For the case where $R=Bbb Z$ : the full subcategory $mathbf{Ab}_{t-f}$ of torsion-free abelian groups is additive, has kernels (because a subgroup of a torsion-free abelian group is torsion free), and it also has cokernels : given a morphism $f:Ato B$, its cokernel in $mathbf{Ab}_{t-f}$ is the quotient of $frac{B}{f(A)}$ by its torsion subgroup $Tleft(frac{B}{A}right)$, or equivalently the quotient $frac{B}{widehat{f(A)}}$, where
$$widehat{f(A)}={bin Bmid exists min Bbb Zsetminus0 text{ such that } mcdot bin f(A) }.$$
Indeed, any morphism $g:Bto C$ such that $gf=0$ factorizes in $mathbf{Ab}$ through $frac{B}{f(A)}$, and then if $C$ is torsion-free the corresponding morphism $frac{B}{f(A)}to C$ should be $0$ on the torsion of $frac{B}{f(A)}$.
Now this is not an abelian subcategory because for example the inclusion $2Bbb Zto Bbb Z$ has both its kernel and cokernel equal to $0$, but is not an isomorphism.
answered Oct 11 '17 at 13:40
Arnaud D.Arnaud D.
16.2k52445
$endgroup$
$begingroup$
Great example. You could even take the subcategory to be finitely generated, since the ring is Noetherian the kernel will be the same.
$endgroup$
– MCT
Oct 11 '17 at 13:47
add a comment |
$begingroup$
For the case where $R=Bbb Z$ : the full subcategory $mathbf{Ab}_{t-f}$ of torsion-free abelian groups is additive, has kernels (because a subgroup of a torsion-free abelian group is torsion free), and it also has cokernels : given a morphism $f:Ato B$, its cokernel in $mathbf{Ab}_{t-f}$ is the quotient of $frac{B}{f(A)}$ by its torsion subgroup $Tleft(frac{B}{A}right)$, or equivalently the quotient $frac{B}{widehat{f(A)}}$, where
$$widehat{f(A)}={bin Bmid exists min Bbb Zsetminus0 text{ such that } mcdot bin f(A) }.$$
Indeed, any morphism $g:Bto C$ such that $gf=0$ factorizes in $mathbf{Ab}$ through $frac{B}{f(A)}$, and then if $C$ is torsion-free the corresponding morphism $frac{B}{f(A)}to C$ should be $0$ on the torsion of $frac{B}{f(A)}$.
Now this is not an abelian subcategory because for example the inclusion $2Bbb Zto Bbb Z$ has both its kernel and cokernel equal to $0$, but is not an isomorphism.
answered Oct 11 '17 at 13:40
Arnaud D.Arnaud D.
16.2k52445
$endgroup$
$begingroup$
Great example. You could even take the subcategory to be finitely generated, since the ring is Noetherian the kernel will be the same.
$endgroup$
– MCT
Oct 11 '17 at 13:47
add a comment |
$begingroup$
For the case where $R=Bbb Z$ : the full subcategory $mathbf{Ab}_{t-f}$ of torsion-free abelian groups is additive, has kernels (because a subgroup of a torsion-free abelian group is torsion free), and it also has cokernels : given a morphism $f:Ato B$, its cokernel in $mathbf{Ab}_{t-f}$ is the quotient of $frac{B}{f(A)}$ by its torsion subgroup $Tleft(frac{B}{A}right)$, or equivalently the quotient $frac{B}{widehat{f(A)}}$, where
$$widehat{f(A)}={bin Bmid exists min Bbb Zsetminus0 text{ such that } mcdot bin f(A) }.$$
Indeed, any morphism $g:Bto C$ such that $gf=0$ factorizes in $mathbf{Ab}$ through $frac{B}{f(A)}$, and then if $C$ is torsion-free the corresponding morphism $frac{B}{f(A)}to C$ should be $0$ on the torsion of $frac{B}{f(A)}$.
Now this is not an abelian subcategory because for example the inclusion $2Bbb Zto Bbb Z$ has both its kernel and cokernel equal to $0$, but is not an isomorphism.
answered Oct 11 '17 at 13:40
Arnaud D.Arnaud D.
16.2k52445
$endgroup$
For the case where $R=Bbb Z$ : the full subcategory $mathbf{Ab}_{t-f}$ of torsion-free abelian groups is additive, has kernels (because a subgroup of a torsion-free abelian group is torsion free), and it also has cokernels : given a morphism $f:Ato B$, its cokernel in $mathbf{Ab}_{t-f}$ is the quotient of $frac{B}{f(A)}$ by its torsion subgroup $Tleft(frac{B}{A}right)$, or equivalently the quotient $frac{B}{widehat{f(A)}}$, where
$$widehat{f(A)}={bin Bmid exists min Bbb Zsetminus0 text{ such that } mcdot bin f(A) }.$$
Indeed, any morphism $g:Bto C$ such that $gf=0$ factorizes in $mathbf{Ab}$ through $frac{B}{f(A)}$, and then if $C$ is torsion-free the corresponding morphism $frac{B}{f(A)}to C$ should be $0$ on the torsion of $frac{B}{f(A)}$.
Now this is not an abelian subcategory because for example the inclusion $2Bbb Zto Bbb Z$ has both its kernel and cokernel equal to $0$, but is not an isomorphism.
answered Oct 11 '17 at 13:40
Arnaud D.Arnaud D.
16.2k52445
edited Feb 1 at 19:08
answered Oct 11 '17 at 13:40
Arnaud D.Arnaud D.
16.2k52445
answered Oct 11 '17 at 13:40
Arnaud D.Arnaud D.
16.2k52445
answered Oct 11 '17 at 13:40
Arnaud D.Arnaud D.
16.2k52445
16.2k52445
$begingroup$
Great example. You could even take the subcategory to be finitely generated, since the ring is Noetherian the kernel will be the same.
$endgroup$
– MCT
Oct 11 '17 at 13:47
add a comment |
$begingroup$
Great example. You could even take the subcategory to be finitely generated, since the ring is Noetherian the kernel will be the same.
$endgroup$
– MCT
Oct 11 '17 at 13:47
$begingroup$
Great example. You could even take the subcategory to be finitely generated, since the ring is Noetherian the kernel will be the same.
$endgroup$
– MCT
Oct 11 '17 at 13:47
$begingroup$
Great example. You could even take the subcategory to be finitely generated, since the ring is Noetherian the kernel will be the same.
$endgroup$
– MCT
Oct 11 '17 at 13:47
add a comment |
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$begingroup$
So you would need to choose the subcategory in such a way that at least one of those objects is different from what it was in the original category. That seems tricky.
$endgroup$
– Tobias Kildetoft
Oct 11 '17 at 13:16
$begingroup$
Yep that's exactly right.
$endgroup$
– MCT
Oct 11 '17 at 13:32