Do products exist in CRing
Clash Royale CLAN TAG#URR8PPP
up vote
2
down vote
favorite
Do categorical products exist in CRing? I know they do in Ring, but that doesn't imply they exist here I believe.
ring-theory category-theory
 |Â
show 1 more comment
up vote
2
down vote
favorite
Do categorical products exist in CRing? I know they do in Ring, but that doesn't imply they exist here I believe.
ring-theory category-theory
2
Hint: The inclusion functor $i:textbfCRingtotextbfRing$ is full, so it suffices to show that the products defined in $textbfRing$ are commutative so long as the rings being multiplied are.
â Rafay A.
Aug 6 at 3:09
The usual direct product works.
â Lord Shark the Unknown
Aug 6 at 3:15
1
Products, and in fact all limits and colimits, exist in any category of algebraic objects, i.e. objects that can be described by an equational theory. (Fields, for example, don't count since they require inequations, but commutative rings in general certainly do count.)
â Derek Elkins
Aug 6 at 4:09
1
To add on Derek Elkins's comment, the limits are actually preserved and created by the forgetful functor to $mathbfSet$.
â Pece
Aug 6 at 8:22
1
@DerekElkins Can you touch on what you mean by your comment about fields and inequations?
â Randall
Aug 6 at 14:19
 |Â
show 1 more comment
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Do categorical products exist in CRing? I know they do in Ring, but that doesn't imply they exist here I believe.
ring-theory category-theory
Do categorical products exist in CRing? I know they do in Ring, but that doesn't imply they exist here I believe.
ring-theory category-theory
asked Aug 6 at 2:59
Skies burn
395113
395113
2
Hint: The inclusion functor $i:textbfCRingtotextbfRing$ is full, so it suffices to show that the products defined in $textbfRing$ are commutative so long as the rings being multiplied are.
â Rafay A.
Aug 6 at 3:09
The usual direct product works.
â Lord Shark the Unknown
Aug 6 at 3:15
1
Products, and in fact all limits and colimits, exist in any category of algebraic objects, i.e. objects that can be described by an equational theory. (Fields, for example, don't count since they require inequations, but commutative rings in general certainly do count.)
â Derek Elkins
Aug 6 at 4:09
1
To add on Derek Elkins's comment, the limits are actually preserved and created by the forgetful functor to $mathbfSet$.
â Pece
Aug 6 at 8:22
1
@DerekElkins Can you touch on what you mean by your comment about fields and inequations?
â Randall
Aug 6 at 14:19
 |Â
show 1 more comment
2
Hint: The inclusion functor $i:textbfCRingtotextbfRing$ is full, so it suffices to show that the products defined in $textbfRing$ are commutative so long as the rings being multiplied are.
â Rafay A.
Aug 6 at 3:09
The usual direct product works.
â Lord Shark the Unknown
Aug 6 at 3:15
1
Products, and in fact all limits and colimits, exist in any category of algebraic objects, i.e. objects that can be described by an equational theory. (Fields, for example, don't count since they require inequations, but commutative rings in general certainly do count.)
â Derek Elkins
Aug 6 at 4:09
1
To add on Derek Elkins's comment, the limits are actually preserved and created by the forgetful functor to $mathbfSet$.
â Pece
Aug 6 at 8:22
1
@DerekElkins Can you touch on what you mean by your comment about fields and inequations?
â Randall
Aug 6 at 14:19
2
2
Hint: The inclusion functor $i:textbfCRingtotextbfRing$ is full, so it suffices to show that the products defined in $textbfRing$ are commutative so long as the rings being multiplied are.
â Rafay A.
Aug 6 at 3:09
Hint: The inclusion functor $i:textbfCRingtotextbfRing$ is full, so it suffices to show that the products defined in $textbfRing$ are commutative so long as the rings being multiplied are.
â Rafay A.
Aug 6 at 3:09
The usual direct product works.
â Lord Shark the Unknown
Aug 6 at 3:15
The usual direct product works.
â Lord Shark the Unknown
Aug 6 at 3:15
1
1
Products, and in fact all limits and colimits, exist in any category of algebraic objects, i.e. objects that can be described by an equational theory. (Fields, for example, don't count since they require inequations, but commutative rings in general certainly do count.)
â Derek Elkins
Aug 6 at 4:09
Products, and in fact all limits and colimits, exist in any category of algebraic objects, i.e. objects that can be described by an equational theory. (Fields, for example, don't count since they require inequations, but commutative rings in general certainly do count.)
â Derek Elkins
Aug 6 at 4:09
1
1
To add on Derek Elkins's comment, the limits are actually preserved and created by the forgetful functor to $mathbfSet$.
â Pece
Aug 6 at 8:22
To add on Derek Elkins's comment, the limits are actually preserved and created by the forgetful functor to $mathbfSet$.
â Pece
Aug 6 at 8:22
1
1
@DerekElkins Can you touch on what you mean by your comment about fields and inequations?
â Randall
Aug 6 at 14:19
@DerekElkins Can you touch on what you mean by your comment about fields and inequations?
â Randall
Aug 6 at 14:19
 |Â
show 1 more comment
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Â
draft saved
draft discarded
Â
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2873552%2fdo-products-exist-in-cring%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
Hint: The inclusion functor $i:textbfCRingtotextbfRing$ is full, so it suffices to show that the products defined in $textbfRing$ are commutative so long as the rings being multiplied are.
â Rafay A.
Aug 6 at 3:09
The usual direct product works.
â Lord Shark the Unknown
Aug 6 at 3:15
1
Products, and in fact all limits and colimits, exist in any category of algebraic objects, i.e. objects that can be described by an equational theory. (Fields, for example, don't count since they require inequations, but commutative rings in general certainly do count.)
â Derek Elkins
Aug 6 at 4:09
1
To add on Derek Elkins's comment, the limits are actually preserved and created by the forgetful functor to $mathbfSet$.
â Pece
Aug 6 at 8:22
1
@DerekElkins Can you touch on what you mean by your comment about fields and inequations?
â Randall
Aug 6 at 14:19