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“categorical†proof of a seemingly symmetric statement about Noetherian/Artinian modules
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There are two statements which to me seem rather symmetric: Let $A$ be a ring, $M$ an $A$-module, and $f : M to M$.
If $M$ is Noetherian and $f$ is surjective, then $f$ is injective.
If $M$ is Artinian and $f$ is injective, then $f$ is surjective.
The proofs also seem symmetric in a sense: in the first case one constructs the increasing chain of ideals $0 subset ker f subset ker f^2 subset dots$ which is strict when $f$ is surjective but not injective. In the second case one uses the injectivity of $f$ to construct the decreasing chain of ideals $M supset im , f supset im , f^2 supset dots$ which is strict when $f$ is injective but not surjective. However, some symmetry is lost in the assertion of the last part ("which is strict when $f$ is __ but not __"). In the first case I use the fact $ker f^n = ker f^n+1$ implies that $f$ is injective on $im , f^n = M$. In the second case I use the fact that $M supsetneq im , f$ would imply that $im , f^n supsetneq im , f^n+1$ because injective maps preserve strict inclusions.
My question is, is there a way to prove one of the statements in the appropriate category/framework such that the other follows from some kind of formulaic reversal of arrows? This is definitely more of a soft question because I'm not sure what this might mean, but the two situations seem symmetric enough that this might be plausible.
commutative-algebra category-theory
asked 2 hours ago
MCT
14.1k42564
add a comment |Â
up vote
5
down vote
favorite
There are two statements which to me seem rather symmetric: Let $A$ be a ring, $M$ an $A$-module, and $f : M to M$.
If $M$ is Noetherian and $f$ is surjective, then $f$ is injective.
If $M$ is Artinian and $f$ is injective, then $f$ is surjective.
The proofs also seem symmetric in a sense: in the first case one constructs the increasing chain of ideals $0 subset ker f subset ker f^2 subset dots$ which is strict when $f$ is surjective but not injective. In the second case one uses the injectivity of $f$ to construct the decreasing chain of ideals $M supset im , f supset im , f^2 supset dots$ which is strict when $f$ is injective but not surjective. However, some symmetry is lost in the assertion of the last part ("which is strict when $f$ is __ but not __"). In the first case I use the fact $ker f^n = ker f^n+1$ implies that $f$ is injective on $im , f^n = M$. In the second case I use the fact that $M supsetneq im , f$ would imply that $im , f^n supsetneq im , f^n+1$ because injective maps preserve strict inclusions.
My question is, is there a way to prove one of the statements in the appropriate category/framework such that the other follows from some kind of formulaic reversal of arrows? This is definitely more of a soft question because I'm not sure what this might mean, but the two situations seem symmetric enough that this might be plausible.
commutative-algebra category-theory
asked 2 hours ago
MCT
14.1k42564
Just prove one in R-mod and turn to the opposite category R-mod^op to get the dual?
– user10354138
1 hour ago
@user10354138 the opp isn't usually a category of modules so not sure why that works
– MCT
58 mins ago
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
There are two statements which to me seem rather symmetric: Let $A$ be a ring, $M$ an $A$-module, and $f : M to M$.
If $M$ is Noetherian and $f$ is surjective, then $f$ is injective.
If $M$ is Artinian and $f$ is injective, then $f$ is surjective.
The proofs also seem symmetric in a sense: in the first case one constructs the increasing chain of ideals $0 subset ker f subset ker f^2 subset dots$ which is strict when $f$ is surjective but not injective. In the second case one uses the injectivity of $f$ to construct the decreasing chain of ideals $M supset im , f supset im , f^2 supset dots$ which is strict when $f$ is injective but not surjective. However, some symmetry is lost in the assertion of the last part ("which is strict when $f$ is __ but not __"). In the first case I use the fact $ker f^n = ker f^n+1$ implies that $f$ is injective on $im , f^n = M$. In the second case I use the fact that $M supsetneq im , f$ would imply that $im , f^n supsetneq im , f^n+1$ because injective maps preserve strict inclusions.
My question is, is there a way to prove one of the statements in the appropriate category/framework such that the other follows from some kind of formulaic reversal of arrows? This is definitely more of a soft question because I'm not sure what this might mean, but the two situations seem symmetric enough that this might be plausible.
commutative-algebra category-theory
asked 2 hours ago
MCT
14.1k42564
There are two statements which to me seem rather symmetric: Let $A$ be a ring, $M$ an $A$-module, and $f : M to M$.
If $M$ is Noetherian and $f$ is surjective, then $f$ is injective.
If $M$ is Artinian and $f$ is injective, then $f$ is surjective.
The proofs also seem symmetric in a sense: in the first case one constructs the increasing chain of ideals $0 subset ker f subset ker f^2 subset dots$ which is strict when $f$ is surjective but not injective. In the second case one uses the injectivity of $f$ to construct the decreasing chain of ideals $M supset im , f supset im , f^2 supset dots$ which is strict when $f$ is injective but not surjective. However, some symmetry is lost in the assertion of the last part ("which is strict when $f$ is __ but not __"). In the first case I use the fact $ker f^n = ker f^n+1$ implies that $f$ is injective on $im , f^n = M$. In the second case I use the fact that $M supsetneq im , f$ would imply that $im , f^n supsetneq im , f^n+1$ because injective maps preserve strict inclusions.
My question is, is there a way to prove one of the statements in the appropriate category/framework such that the other follows from some kind of formulaic reversal of arrows? This is definitely more of a soft question because I'm not sure what this might mean, but the two situations seem symmetric enough that this might be plausible.
commutative-algebra category-theory
commutative-algebra category-theory
asked 2 hours ago
MCT
14.1k42564
asked 2 hours ago
MCT
14.1k42564
asked 2 hours ago
MCT
14.1k42564
asked 2 hours ago
MCT
14.1k42564
asked 2 hours ago
MCT
14.1k42564
14.1k42564
Just prove one in R-mod and turn to the opposite category R-mod^op to get the dual?
– user10354138
1 hour ago
@user10354138 the opp isn't usually a category of modules so not sure why that works
– MCT
58 mins ago
add a comment |Â
Just prove one in R-mod and turn to the opposite category R-mod^op to get the dual?
– user10354138
1 hour ago
@user10354138 the opp isn't usually a category of modules so not sure why that works
– MCT
58 mins ago
Just prove one in R-mod and turn to the opposite category R-mod^op to get the dual?
– user10354138
1 hour ago
Just prove one in R-mod and turn to the opposite category R-mod^op to get the dual?
– user10354138
1 hour ago
@user10354138 the opp isn't usually a category of modules so not sure why that works
– MCT
58 mins ago
@user10354138 the opp isn't usually a category of modules so not sure why that works
– MCT
58 mins ago
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
Yes! These are both special cases of a general statement:
If $M$ is a Noetherian object in an abelian category and $f:Mto M$ is an epimorphism, then $f$ is a monomorphism.
Here an object is "Noetherian" if every ascending chain of subobjects stabilizes. The proof is exactly the same as in the case of modules: look at the ascending chain $0 subset ker f subset ker f^2 subset dots$ (though it takes a little more work to prove this chain is strictly ascending in an abstract abelian category than in the case of modules).
Now, how does this imply the Artinian version? Well, the opposite category of $A$-modules is also an abelian category, so we can apply the result in that category. What does it mean for $M$ to be a Noetherian object in the opposite category of $A$-modules? Well, a subobject is a monomorphism $Nto M$ (up to isomorphism), which would be an epimorphism $Mto N$ in the original category. But such an epimorphism is determined (up to isomorphism) by its kernel, which is a subobject of $M$. So subobjects of $M$ in the opposite category are naturally in bijection with subobjects in the original category.
However, this bijection reverses the inclusion order on subobjects. Indeed, suppose $Nto M$ and $Pto M$ are two subobjects of $M$ in the opposite category, with $N$ contained in $P$. That means we can factor the map $Nto M$ as $Nto Pto M$. In the original category, then, this means we can factor the quotient map $Mto N$ as $Mto Pto N$. This is possible if and only if the kernel of $Mto N$ contains the kernel of $Mto P$. In other words, $N$ is contained in $P$ as subobjects in the opposite category iff the subobject in the original category corresponding to $P$ is contained in the subobject in the original category corresponding to $N$.
This means that $M$ is Noetherian in the opposite category iff $M$ is Artinian in the original category, since the order on subobjects has been reversed. Applying the result in the opposite category, we conclude that if $M$ is Artinian and if $f:Mto M$ is a monomorphism, then $f$ is an epimorphism.
answered 1 hour ago
Eric Wofsey
167k12196310
1
Great. As an aside, a reason I was not sure this construction existed is that the proof of both cases relied on looking at subobjects, when one case should be subs and the other case quotients. This is fixed by the viewpoint of looking at the cokernel instead of the image in the Artinian case, with chains being compositions of surjective maps in the other direction.
– MCT
1 hour ago
add a comment |Â
up vote
2
down vote
If you look at the proof in the Noetherian case, you see that it
is valid in a general Abelian category. (A Noetherian object in an Abelian
category is one with ACC on subobjects). Now use the categorical
principle of duality. The opposite of an Abelian category is an Abelian
category, Noetherian becomes Artinian, injective becomes surjective etc.
So the Artinian case follows from the Noetherian case and vice versa.
Of course, algebra textbooks don't do this, largely because the opposite
of a module category is rarely also a module category.
answered 1 hour ago
Lord Shark the Unknown
90.2k955117
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Yes! These are both special cases of a general statement:
If $M$ is a Noetherian object in an abelian category and $f:Mto M$ is an epimorphism, then $f$ is a monomorphism.
Here an object is "Noetherian" if every ascending chain of subobjects stabilizes. The proof is exactly the same as in the case of modules: look at the ascending chain $0 subset ker f subset ker f^2 subset dots$ (though it takes a little more work to prove this chain is strictly ascending in an abstract abelian category than in the case of modules).
Now, how does this imply the Artinian version? Well, the opposite category of $A$-modules is also an abelian category, so we can apply the result in that category. What does it mean for $M$ to be a Noetherian object in the opposite category of $A$-modules? Well, a subobject is a monomorphism $Nto M$ (up to isomorphism), which would be an epimorphism $Mto N$ in the original category. But such an epimorphism is determined (up to isomorphism) by its kernel, which is a subobject of $M$. So subobjects of $M$ in the opposite category are naturally in bijection with subobjects in the original category.
However, this bijection reverses the inclusion order on subobjects. Indeed, suppose $Nto M$ and $Pto M$ are two subobjects of $M$ in the opposite category, with $N$ contained in $P$. That means we can factor the map $Nto M$ as $Nto Pto M$. In the original category, then, this means we can factor the quotient map $Mto N$ as $Mto Pto N$. This is possible if and only if the kernel of $Mto N$ contains the kernel of $Mto P$. In other words, $N$ is contained in $P$ as subobjects in the opposite category iff the subobject in the original category corresponding to $P$ is contained in the subobject in the original category corresponding to $N$.
This means that $M$ is Noetherian in the opposite category iff $M$ is Artinian in the original category, since the order on subobjects has been reversed. Applying the result in the opposite category, we conclude that if $M$ is Artinian and if $f:Mto M$ is a monomorphism, then $f$ is an epimorphism.
answered 1 hour ago
Eric Wofsey
167k12196310
1
Great. As an aside, a reason I was not sure this construction existed is that the proof of both cases relied on looking at subobjects, when one case should be subs and the other case quotients. This is fixed by the viewpoint of looking at the cokernel instead of the image in the Artinian case, with chains being compositions of surjective maps in the other direction.
– MCT
1 hour ago
add a comment |Â
up vote
3
down vote
Yes! These are both special cases of a general statement:
If $M$ is a Noetherian object in an abelian category and $f:Mto M$ is an epimorphism, then $f$ is a monomorphism.
Here an object is "Noetherian" if every ascending chain of subobjects stabilizes. The proof is exactly the same as in the case of modules: look at the ascending chain $0 subset ker f subset ker f^2 subset dots$ (though it takes a little more work to prove this chain is strictly ascending in an abstract abelian category than in the case of modules).
Now, how does this imply the Artinian version? Well, the opposite category of $A$-modules is also an abelian category, so we can apply the result in that category. What does it mean for $M$ to be a Noetherian object in the opposite category of $A$-modules? Well, a subobject is a monomorphism $Nto M$ (up to isomorphism), which would be an epimorphism $Mto N$ in the original category. But such an epimorphism is determined (up to isomorphism) by its kernel, which is a subobject of $M$. So subobjects of $M$ in the opposite category are naturally in bijection with subobjects in the original category.
However, this bijection reverses the inclusion order on subobjects. Indeed, suppose $Nto M$ and $Pto M$ are two subobjects of $M$ in the opposite category, with $N$ contained in $P$. That means we can factor the map $Nto M$ as $Nto Pto M$. In the original category, then, this means we can factor the quotient map $Mto N$ as $Mto Pto N$. This is possible if and only if the kernel of $Mto N$ contains the kernel of $Mto P$. In other words, $N$ is contained in $P$ as subobjects in the opposite category iff the subobject in the original category corresponding to $P$ is contained in the subobject in the original category corresponding to $N$.
This means that $M$ is Noetherian in the opposite category iff $M$ is Artinian in the original category, since the order on subobjects has been reversed. Applying the result in the opposite category, we conclude that if $M$ is Artinian and if $f:Mto M$ is a monomorphism, then $f$ is an epimorphism.
answered 1 hour ago
Eric Wofsey
167k12196310
1
Great. As an aside, a reason I was not sure this construction existed is that the proof of both cases relied on looking at subobjects, when one case should be subs and the other case quotients. This is fixed by the viewpoint of looking at the cokernel instead of the image in the Artinian case, with chains being compositions of surjective maps in the other direction.
– MCT
1 hour ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Yes! These are both special cases of a general statement:
If $M$ is a Noetherian object in an abelian category and $f:Mto M$ is an epimorphism, then $f$ is a monomorphism.
Here an object is "Noetherian" if every ascending chain of subobjects stabilizes. The proof is exactly the same as in the case of modules: look at the ascending chain $0 subset ker f subset ker f^2 subset dots$ (though it takes a little more work to prove this chain is strictly ascending in an abstract abelian category than in the case of modules).
Now, how does this imply the Artinian version? Well, the opposite category of $A$-modules is also an abelian category, so we can apply the result in that category. What does it mean for $M$ to be a Noetherian object in the opposite category of $A$-modules? Well, a subobject is a monomorphism $Nto M$ (up to isomorphism), which would be an epimorphism $Mto N$ in the original category. But such an epimorphism is determined (up to isomorphism) by its kernel, which is a subobject of $M$. So subobjects of $M$ in the opposite category are naturally in bijection with subobjects in the original category.
However, this bijection reverses the inclusion order on subobjects. Indeed, suppose $Nto M$ and $Pto M$ are two subobjects of $M$ in the opposite category, with $N$ contained in $P$. That means we can factor the map $Nto M$ as $Nto Pto M$. In the original category, then, this means we can factor the quotient map $Mto N$ as $Mto Pto N$. This is possible if and only if the kernel of $Mto N$ contains the kernel of $Mto P$. In other words, $N$ is contained in $P$ as subobjects in the opposite category iff the subobject in the original category corresponding to $P$ is contained in the subobject in the original category corresponding to $N$.
This means that $M$ is Noetherian in the opposite category iff $M$ is Artinian in the original category, since the order on subobjects has been reversed. Applying the result in the opposite category, we conclude that if $M$ is Artinian and if $f:Mto M$ is a monomorphism, then $f$ is an epimorphism.
answered 1 hour ago
Eric Wofsey
167k12196310
Yes! These are both special cases of a general statement:
If $M$ is a Noetherian object in an abelian category and $f:Mto M$ is an epimorphism, then $f$ is a monomorphism.
Here an object is "Noetherian" if every ascending chain of subobjects stabilizes. The proof is exactly the same as in the case of modules: look at the ascending chain $0 subset ker f subset ker f^2 subset dots$ (though it takes a little more work to prove this chain is strictly ascending in an abstract abelian category than in the case of modules).
Now, how does this imply the Artinian version? Well, the opposite category of $A$-modules is also an abelian category, so we can apply the result in that category. What does it mean for $M$ to be a Noetherian object in the opposite category of $A$-modules? Well, a subobject is a monomorphism $Nto M$ (up to isomorphism), which would be an epimorphism $Mto N$ in the original category. But such an epimorphism is determined (up to isomorphism) by its kernel, which is a subobject of $M$. So subobjects of $M$ in the opposite category are naturally in bijection with subobjects in the original category.
However, this bijection reverses the inclusion order on subobjects. Indeed, suppose $Nto M$ and $Pto M$ are two subobjects of $M$ in the opposite category, with $N$ contained in $P$. That means we can factor the map $Nto M$ as $Nto Pto M$. In the original category, then, this means we can factor the quotient map $Mto N$ as $Mto Pto N$. This is possible if and only if the kernel of $Mto N$ contains the kernel of $Mto P$. In other words, $N$ is contained in $P$ as subobjects in the opposite category iff the subobject in the original category corresponding to $P$ is contained in the subobject in the original category corresponding to $N$.
This means that $M$ is Noetherian in the opposite category iff $M$ is Artinian in the original category, since the order on subobjects has been reversed. Applying the result in the opposite category, we conclude that if $M$ is Artinian and if $f:Mto M$ is a monomorphism, then $f$ is an epimorphism.
answered 1 hour ago
Eric Wofsey
167k12196310
answered 1 hour ago
Eric Wofsey
167k12196310
answered 1 hour ago
Eric Wofsey
167k12196310
answered 1 hour ago
Eric Wofsey
167k12196310
167k12196310
1
Great. As an aside, a reason I was not sure this construction existed is that the proof of both cases relied on looking at subobjects, when one case should be subs and the other case quotients. This is fixed by the viewpoint of looking at the cokernel instead of the image in the Artinian case, with chains being compositions of surjective maps in the other direction.
– MCT
1 hour ago
add a comment |Â
1
Great. As an aside, a reason I was not sure this construction existed is that the proof of both cases relied on looking at subobjects, when one case should be subs and the other case quotients. This is fixed by the viewpoint of looking at the cokernel instead of the image in the Artinian case, with chains being compositions of surjective maps in the other direction.
– MCT
1 hour ago
1
1
Great. As an aside, a reason I was not sure this construction existed is that the proof of both cases relied on looking at subobjects, when one case should be subs and the other case quotients. This is fixed by the viewpoint of looking at the cokernel instead of the image in the Artinian case, with chains being compositions of surjective maps in the other direction.
– MCT
1 hour ago
Great. As an aside, a reason I was not sure this construction existed is that the proof of both cases relied on looking at subobjects, when one case should be subs and the other case quotients. This is fixed by the viewpoint of looking at the cokernel instead of the image in the Artinian case, with chains being compositions of surjective maps in the other direction.
– MCT
1 hour ago
add a comment |Â
up vote
2
down vote
If you look at the proof in the Noetherian case, you see that it
is valid in a general Abelian category. (A Noetherian object in an Abelian
category is one with ACC on subobjects). Now use the categorical
principle of duality. The opposite of an Abelian category is an Abelian
category, Noetherian becomes Artinian, injective becomes surjective etc.
So the Artinian case follows from the Noetherian case and vice versa.
Of course, algebra textbooks don't do this, largely because the opposite
of a module category is rarely also a module category.
answered 1 hour ago
Lord Shark the Unknown
90.2k955117
add a comment |Â
up vote
2
down vote
If you look at the proof in the Noetherian case, you see that it
is valid in a general Abelian category. (A Noetherian object in an Abelian
category is one with ACC on subobjects). Now use the categorical
principle of duality. The opposite of an Abelian category is an Abelian
category, Noetherian becomes Artinian, injective becomes surjective etc.
So the Artinian case follows from the Noetherian case and vice versa.
Of course, algebra textbooks don't do this, largely because the opposite
of a module category is rarely also a module category.
answered 1 hour ago
Lord Shark the Unknown
90.2k955117
add a comment |Â
up vote
2
down vote
up vote
2
down vote
If you look at the proof in the Noetherian case, you see that it
is valid in a general Abelian category. (A Noetherian object in an Abelian
category is one with ACC on subobjects). Now use the categorical
principle of duality. The opposite of an Abelian category is an Abelian
category, Noetherian becomes Artinian, injective becomes surjective etc.
So the Artinian case follows from the Noetherian case and vice versa.
Of course, algebra textbooks don't do this, largely because the opposite
of a module category is rarely also a module category.
answered 1 hour ago
Lord Shark the Unknown
90.2k955117
If you look at the proof in the Noetherian case, you see that it
is valid in a general Abelian category. (A Noetherian object in an Abelian
category is one with ACC on subobjects). Now use the categorical
principle of duality. The opposite of an Abelian category is an Abelian
category, Noetherian becomes Artinian, injective becomes surjective etc.
So the Artinian case follows from the Noetherian case and vice versa.
Of course, algebra textbooks don't do this, largely because the opposite
of a module category is rarely also a module category.
answered 1 hour ago
Lord Shark the Unknown
90.2k955117
answered 1 hour ago
Lord Shark the Unknown
90.2k955117
answered 1 hour ago
Lord Shark the Unknown
90.2k955117
answered 1 hour ago
Lord Shark the Unknown
90.2k955117
90.2k955117
add a comment |Â
add a comment |Â
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Just prove one in R-mod and turn to the opposite category R-mod^op to get the dual?
– user10354138
1 hour ago
@user10354138 the opp isn't usually a category of modules so not sure why that works
– MCT
58 mins ago