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Categorical definitions of ring properties
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Looking at the interesting list of ring properties that are inherited from a ring $mathcalR$ by its polynomial ring $mathcalR$[X] and remembering a question I once asked I want to repeat the latter in a more general way:
Can you give ring properties with catchy categorical
definitions like these:
A ring $mathcalR$ has the structure of $mathbbZ$ iff it is an initial object in the category of rings.
A ring $mathcalR$ has characteristic $0$ iff the morphism from $mathbbZ$ is a monomorphism.
What about being commutative, factorial, Noetherian, Abelian, or an integral domain?
[Note that the property of having a multiplicative identity (i.e. of being unital) doesn't have to be defined, because it's presupposed in the category of rings.]
List of definitions from the answers below:
- A ring $mathcalR$ is Noetherian iff every ascending chain of strong epimorphisms is stationary.
ring-theory category-theory definition big-list
asked 2 hours ago
Hans Stricker
4,90613881
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up vote
5
down vote
favorite
Looking at the interesting list of ring properties that are inherited from a ring $mathcalR$ by its polynomial ring $mathcalR$[X] and remembering a question I once asked I want to repeat the latter in a more general way:
Can you give ring properties with catchy categorical
definitions like these:
A ring $mathcalR$ has the structure of $mathbbZ$ iff it is an initial object in the category of rings.
A ring $mathcalR$ has characteristic $0$ iff the morphism from $mathbbZ$ is a monomorphism.
What about being commutative, factorial, Noetherian, Abelian, or an integral domain?
[Note that the property of having a multiplicative identity (i.e. of being unital) doesn't have to be defined, because it's presupposed in the category of rings.]
List of definitions from the answers below:
- A ring $mathcalR$ is Noetherian iff every ascending chain of strong epimorphisms is stationary.
ring-theory category-theory definition big-list
asked 2 hours ago
Hans Stricker
4,90613881
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
Looking at the interesting list of ring properties that are inherited from a ring $mathcalR$ by its polynomial ring $mathcalR$[X] and remembering a question I once asked I want to repeat the latter in a more general way:
Can you give ring properties with catchy categorical
definitions like these:
A ring $mathcalR$ has the structure of $mathbbZ$ iff it is an initial object in the category of rings.
A ring $mathcalR$ has characteristic $0$ iff the morphism from $mathbbZ$ is a monomorphism.
What about being commutative, factorial, Noetherian, Abelian, or an integral domain?
[Note that the property of having a multiplicative identity (i.e. of being unital) doesn't have to be defined, because it's presupposed in the category of rings.]
List of definitions from the answers below:
- A ring $mathcalR$ is Noetherian iff every ascending chain of strong epimorphisms is stationary.
ring-theory category-theory definition big-list
asked 2 hours ago
Hans Stricker
4,90613881
Looking at the interesting list of ring properties that are inherited from a ring $mathcalR$ by its polynomial ring $mathcalR$[X] and remembering a question I once asked I want to repeat the latter in a more general way:
Can you give ring properties with catchy categorical
definitions like these:
A ring $mathcalR$ has the structure of $mathbbZ$ iff it is an initial object in the category of rings.
A ring $mathcalR$ has characteristic $0$ iff the morphism from $mathbbZ$ is a monomorphism.
What about being commutative, factorial, Noetherian, Abelian, or an integral domain?
[Note that the property of having a multiplicative identity (i.e. of being unital) doesn't have to be defined, because it's presupposed in the category of rings.]
List of definitions from the answers below:
- A ring $mathcalR$ is Noetherian iff every ascending chain of strong epimorphisms is stationary.
ring-theory category-theory definition big-list
ring-theory category-theory definition big-list
asked 2 hours ago
Hans Stricker
4,90613881
asked 2 hours ago
Hans Stricker
4,90613881
edited 6 mins ago
asked 2 hours ago
Hans Stricker
4,90613881
asked 2 hours ago
Hans Stricker
4,90613881
asked 2 hours ago
Hans Stricker
4,90613881
4,90613881
add a comment |Â
add a comment |Â
1 Answer
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There is a notion of noetherian object in any category, which is simply that every ascending chain of subobjects is stationnary. With this definition, noetherian objects in the category of modules over a ring are simply noetherian modules; however, ideals are not subrings (since in general they do not contain the identity element), so noetherian objects in the category of unital rings are not noetherian rings.
This is easy to fix, however : every ascending chain of ideals
$$I_0subset I_1subsetdots subset I_nsubset I_n+1subset dots,$$
in a ring $R$ induces an ascending chain of quotient rings
$$R/I_0to R/I_1todotsto R/I_nto R/I_n+1to dots$$
and the original chain of ideals is stationnary if and only if the chain of quotients is. Now quotient maps are just the surjective maps, which are the same thing as strong or regular epimorphisms in the category of rings (or in any "algebraic" category). Thus noetherian rings are precisely those for which every ascending chain of strong epimorphisms is stationnary, which one might call "strongly co-noetherian objects".
answered 1 hour ago
Arnaud D.
15k52242
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
There is a notion of noetherian object in any category, which is simply that every ascending chain of subobjects is stationnary. With this definition, noetherian objects in the category of modules over a ring are simply noetherian modules; however, ideals are not subrings (since in general they do not contain the identity element), so noetherian objects in the category of unital rings are not noetherian rings.
This is easy to fix, however : every ascending chain of ideals
$$I_0subset I_1subsetdots subset I_nsubset I_n+1subset dots,$$
in a ring $R$ induces an ascending chain of quotient rings
$$R/I_0to R/I_1todotsto R/I_nto R/I_n+1to dots$$
and the original chain of ideals is stationnary if and only if the chain of quotients is. Now quotient maps are just the surjective maps, which are the same thing as strong or regular epimorphisms in the category of rings (or in any "algebraic" category). Thus noetherian rings are precisely those for which every ascending chain of strong epimorphisms is stationnary, which one might call "strongly co-noetherian objects".
answered 1 hour ago
Arnaud D.
15k52242
add a comment |Â
up vote
4
down vote
There is a notion of noetherian object in any category, which is simply that every ascending chain of subobjects is stationnary. With this definition, noetherian objects in the category of modules over a ring are simply noetherian modules; however, ideals are not subrings (since in general they do not contain the identity element), so noetherian objects in the category of unital rings are not noetherian rings.
This is easy to fix, however : every ascending chain of ideals
$$I_0subset I_1subsetdots subset I_nsubset I_n+1subset dots,$$
in a ring $R$ induces an ascending chain of quotient rings
$$R/I_0to R/I_1todotsto R/I_nto R/I_n+1to dots$$
and the original chain of ideals is stationnary if and only if the chain of quotients is. Now quotient maps are just the surjective maps, which are the same thing as strong or regular epimorphisms in the category of rings (or in any "algebraic" category). Thus noetherian rings are precisely those for which every ascending chain of strong epimorphisms is stationnary, which one might call "strongly co-noetherian objects".
answered 1 hour ago
Arnaud D.
15k52242
add a comment |Â
up vote
4
down vote
up vote
4
down vote
There is a notion of noetherian object in any category, which is simply that every ascending chain of subobjects is stationnary. With this definition, noetherian objects in the category of modules over a ring are simply noetherian modules; however, ideals are not subrings (since in general they do not contain the identity element), so noetherian objects in the category of unital rings are not noetherian rings.
This is easy to fix, however : every ascending chain of ideals
$$I_0subset I_1subsetdots subset I_nsubset I_n+1subset dots,$$
in a ring $R$ induces an ascending chain of quotient rings
$$R/I_0to R/I_1todotsto R/I_nto R/I_n+1to dots$$
and the original chain of ideals is stationnary if and only if the chain of quotients is. Now quotient maps are just the surjective maps, which are the same thing as strong or regular epimorphisms in the category of rings (or in any "algebraic" category). Thus noetherian rings are precisely those for which every ascending chain of strong epimorphisms is stationnary, which one might call "strongly co-noetherian objects".
answered 1 hour ago
Arnaud D.
15k52242
There is a notion of noetherian object in any category, which is simply that every ascending chain of subobjects is stationnary. With this definition, noetherian objects in the category of modules over a ring are simply noetherian modules; however, ideals are not subrings (since in general they do not contain the identity element), so noetherian objects in the category of unital rings are not noetherian rings.
This is easy to fix, however : every ascending chain of ideals
$$I_0subset I_1subsetdots subset I_nsubset I_n+1subset dots,$$
in a ring $R$ induces an ascending chain of quotient rings
$$R/I_0to R/I_1todotsto R/I_nto R/I_n+1to dots$$
and the original chain of ideals is stationnary if and only if the chain of quotients is. Now quotient maps are just the surjective maps, which are the same thing as strong or regular epimorphisms in the category of rings (or in any "algebraic" category). Thus noetherian rings are precisely those for which every ascending chain of strong epimorphisms is stationnary, which one might call "strongly co-noetherian objects".
answered 1 hour ago
Arnaud D.
15k52242
answered 1 hour ago
Arnaud D.
15k52242
answered 1 hour ago
Arnaud D.
15k52242
answered 1 hour ago
Arnaud D.
15k52242
15k52242
add a comment |Â
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