Flat ring homomorphism but not injective.











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Let $Ato B$ be flat ring homomorphism(i.e. $B$ is flat $A$ module.)



If $B$ is faithfully flat, then $Ato B$ is injection.



$textbf{Q:}$ What is the example of flat but not injective ring homomorphism?(i.e. I want to fail faithfully flat but remain flat.) I think I need some ring $B$ as projective which realizes $B=F/N$ where $F$ is free $A-$module and this has to be compatible with ring structure as well. Clearly, I could not get this work over $A$ being a field.










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  • 1




    $A$ any ring, $B$ the zero ring?
    – Lord Shark the Unknown
    Oct 4 at 18:01










  • @LordSharktheUnknown $A,B$ are unital rings.
    – user45765
    Oct 4 at 18:02






  • 4




    The zero ring is unital....
    – Lord Shark the Unknown
    Oct 4 at 18:03










  • @LordSharktheUnknown Can I have a unital ring counter example rather than $0$ ring?
    – user45765
    Oct 4 at 18:05






  • 3




    Take $A=Boplus B$ and $Ato B$ the projection onto the first factor.
    – David Hill
    Oct 4 at 18:05















up vote
1
down vote

favorite












Let $Ato B$ be flat ring homomorphism(i.e. $B$ is flat $A$ module.)



If $B$ is faithfully flat, then $Ato B$ is injection.



$textbf{Q:}$ What is the example of flat but not injective ring homomorphism?(i.e. I want to fail faithfully flat but remain flat.) I think I need some ring $B$ as projective which realizes $B=F/N$ where $F$ is free $A-$module and this has to be compatible with ring structure as well. Clearly, I could not get this work over $A$ being a field.










share|cite|improve this question


















  • 1




    $A$ any ring, $B$ the zero ring?
    – Lord Shark the Unknown
    Oct 4 at 18:01










  • @LordSharktheUnknown $A,B$ are unital rings.
    – user45765
    Oct 4 at 18:02






  • 4




    The zero ring is unital....
    – Lord Shark the Unknown
    Oct 4 at 18:03










  • @LordSharktheUnknown Can I have a unital ring counter example rather than $0$ ring?
    – user45765
    Oct 4 at 18:05






  • 3




    Take $A=Boplus B$ and $Ato B$ the projection onto the first factor.
    – David Hill
    Oct 4 at 18:05













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $Ato B$ be flat ring homomorphism(i.e. $B$ is flat $A$ module.)



If $B$ is faithfully flat, then $Ato B$ is injection.



$textbf{Q:}$ What is the example of flat but not injective ring homomorphism?(i.e. I want to fail faithfully flat but remain flat.) I think I need some ring $B$ as projective which realizes $B=F/N$ where $F$ is free $A-$module and this has to be compatible with ring structure as well. Clearly, I could not get this work over $A$ being a field.










share|cite|improve this question













Let $Ato B$ be flat ring homomorphism(i.e. $B$ is flat $A$ module.)



If $B$ is faithfully flat, then $Ato B$ is injection.



$textbf{Q:}$ What is the example of flat but not injective ring homomorphism?(i.e. I want to fail faithfully flat but remain flat.) I think I need some ring $B$ as projective which realizes $B=F/N$ where $F$ is free $A-$module and this has to be compatible with ring structure as well. Clearly, I could not get this work over $A$ being a field.







abstract-algebra commutative-algebra






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asked Oct 4 at 17:54









user45765

2,4592721




2,4592721








  • 1




    $A$ any ring, $B$ the zero ring?
    – Lord Shark the Unknown
    Oct 4 at 18:01










  • @LordSharktheUnknown $A,B$ are unital rings.
    – user45765
    Oct 4 at 18:02






  • 4




    The zero ring is unital....
    – Lord Shark the Unknown
    Oct 4 at 18:03










  • @LordSharktheUnknown Can I have a unital ring counter example rather than $0$ ring?
    – user45765
    Oct 4 at 18:05






  • 3




    Take $A=Boplus B$ and $Ato B$ the projection onto the first factor.
    – David Hill
    Oct 4 at 18:05














  • 1




    $A$ any ring, $B$ the zero ring?
    – Lord Shark the Unknown
    Oct 4 at 18:01










  • @LordSharktheUnknown $A,B$ are unital rings.
    – user45765
    Oct 4 at 18:02






  • 4




    The zero ring is unital....
    – Lord Shark the Unknown
    Oct 4 at 18:03










  • @LordSharktheUnknown Can I have a unital ring counter example rather than $0$ ring?
    – user45765
    Oct 4 at 18:05






  • 3




    Take $A=Boplus B$ and $Ato B$ the projection onto the first factor.
    – David Hill
    Oct 4 at 18:05








1




1




$A$ any ring, $B$ the zero ring?
– Lord Shark the Unknown
Oct 4 at 18:01




$A$ any ring, $B$ the zero ring?
– Lord Shark the Unknown
Oct 4 at 18:01












@LordSharktheUnknown $A,B$ are unital rings.
– user45765
Oct 4 at 18:02




@LordSharktheUnknown $A,B$ are unital rings.
– user45765
Oct 4 at 18:02




4




4




The zero ring is unital....
– Lord Shark the Unknown
Oct 4 at 18:03




The zero ring is unital....
– Lord Shark the Unknown
Oct 4 at 18:03












@LordSharktheUnknown Can I have a unital ring counter example rather than $0$ ring?
– user45765
Oct 4 at 18:05




@LordSharktheUnknown Can I have a unital ring counter example rather than $0$ ring?
– user45765
Oct 4 at 18:05




3




3




Take $A=Boplus B$ and $Ato B$ the projection onto the first factor.
– David Hill
Oct 4 at 18:05




Take $A=Boplus B$ and $Ato B$ the projection onto the first factor.
– David Hill
Oct 4 at 18:05










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Let $A$ be a nonzero and absolutely flat ring, let $mathfrak{m}subseteq A$ be a maximal ideal, and let $B=A/mathfrak{m}$. Then, the canonical morphism $Arightarrow B$ is flat but not injective.






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    up vote
    0
    down vote













    Let $A$ be a nonzero and absolutely flat ring, let $mathfrak{m}subseteq A$ be a maximal ideal, and let $B=A/mathfrak{m}$. Then, the canonical morphism $Arightarrow B$ is flat but not injective.






    share|cite|improve this answer

























      up vote
      0
      down vote













      Let $A$ be a nonzero and absolutely flat ring, let $mathfrak{m}subseteq A$ be a maximal ideal, and let $B=A/mathfrak{m}$. Then, the canonical morphism $Arightarrow B$ is flat but not injective.






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        Let $A$ be a nonzero and absolutely flat ring, let $mathfrak{m}subseteq A$ be a maximal ideal, and let $B=A/mathfrak{m}$. Then, the canonical morphism $Arightarrow B$ is flat but not injective.






        share|cite|improve this answer












        Let $A$ be a nonzero and absolutely flat ring, let $mathfrak{m}subseteq A$ be a maximal ideal, and let $B=A/mathfrak{m}$. Then, the canonical morphism $Arightarrow B$ is flat but not injective.







        share|cite|improve this answer












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        share|cite|improve this answer










        answered Nov 22 at 18:13









        Fred Rohrer

        261110




        261110






























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