Do products exist in CRing

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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




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    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.
    – Rafay A.
    Aug 6 at 3:09










  • The usual direct product works.
    – 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.)
    – 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 $mathbfSet$.
    – Pece
    Aug 6 at 8:22






  • 1




    @DerekElkins Can you touch on what you mean by your comment about fields and inequations?
    – Randall
    Aug 6 at 14:19














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




share|cite|improve this question















  • 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.
    – Rafay A.
    Aug 6 at 3:09










  • The usual direct product works.
    – 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.)
    – 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 $mathbfSet$.
    – Pece
    Aug 6 at 8:22






  • 1




    @DerekElkins Can you touch on what you mean by your comment about fields and inequations?
    – Randall
    Aug 6 at 14:19












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




share|cite|improve this question











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





share|cite|improve this question










share|cite|improve this question




share|cite|improve this question









asked Aug 6 at 2:59









Skies burn

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  • 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.
    – Rafay A.
    Aug 6 at 3:09










  • The usual direct product works.
    – 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.)
    – 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 $mathbfSet$.
    – Pece
    Aug 6 at 8:22






  • 1




    @DerekElkins Can you touch on what you mean by your comment about fields and inequations?
    – Randall
    Aug 6 at 14:19












  • 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.
    – Rafay A.
    Aug 6 at 3:09










  • The usual direct product works.
    – 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.)
    – 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 $mathbfSet$.
    – Pece
    Aug 6 at 8:22






  • 1




    @DerekElkins Can you touch on what you mean by your comment about fields and inequations?
    – 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.
– 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.
– Rafay A.
Aug 6 at 3:09












The usual direct product works.
– Lord Shark the Unknown
Aug 6 at 3:15




The usual direct product works.
– 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.)
– 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.)
– 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 $mathbfSet$.
– 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 $mathbfSet$.
– Pece
Aug 6 at 8:22




1




1




@DerekElkins Can you touch on what you mean by your comment about fields and inequations?
– Randall
Aug 6 at 14:19




@DerekElkins Can you touch on what you mean by your comment about fields and inequations?
– Randall
Aug 6 at 14:19















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