Metamath Proof Explorer


Theorem rhmkerinj

Description: A ring homomorphism is injective if and only if its kernel is zero. (Contributed by Jeff Madsen, 16-Jun-2011) (Revised by AV, 24-Jul-2026)

Ref Expression
Hypotheses rhmkerinj.b
|- B = ( Base ` R )
rhmkerinj.c
|- C = ( Base ` S )
rhmkerinj.0
|- .0. = ( 0g ` R )
rhmkerinj.z
|- Z = ( 0g ` S )
Assertion rhmkerinj
|- ( F e. ( R RingHom S ) -> ( F : B -1-1-> C <-> ( `' F " { Z } ) = { .0. } ) )

Proof

Step Hyp Ref Expression
1 rhmkerinj.b
 |-  B = ( Base ` R )
2 rhmkerinj.c
 |-  C = ( Base ` S )
3 rhmkerinj.0
 |-  .0. = ( 0g ` R )
4 rhmkerinj.z
 |-  Z = ( 0g ` S )
5 rhmghm
 |-  ( F e. ( R RingHom S ) -> F e. ( R GrpHom S ) )
6 1 2 3 4 kerf1ghm
 |-  ( F e. ( R GrpHom S ) -> ( F : B -1-1-> C <-> ( `' F " { Z } ) = { .0. } ) )
7 5 6 syl
 |-  ( F e. ( R RingHom S ) -> ( F : B -1-1-> C <-> ( `' F " { Z } ) = { .0. } ) )