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 = ( 0g𝑅 )
rhmkerinj.z 𝑍 = ( 0g𝑆 )
Assertion rhmkerinj ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) → ( 𝐹 : 𝐵1-1𝐶 ↔ ( 𝐹 “ { 𝑍 } ) = { 0 } ) )

Proof

Step Hyp Ref Expression
1 rhmkerinj.b 𝐵 = ( Base ‘ 𝑅 )
2 rhmkerinj.c 𝐶 = ( Base ‘ 𝑆 )
3 rhmkerinj.0 0 = ( 0g𝑅 )
4 rhmkerinj.z 𝑍 = ( 0g𝑆 )
5 rhmghm ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) → 𝐹 ∈ ( 𝑅 GrpHom 𝑆 ) )
6 1 2 3 4 kerf1ghm ( 𝐹 ∈ ( 𝑅 GrpHom 𝑆 ) → ( 𝐹 : 𝐵1-1𝐶 ↔ ( 𝐹 “ { 𝑍 } ) = { 0 } ) )
7 5 6 syl ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) → ( 𝐹 : 𝐵1-1𝐶 ↔ ( 𝐹 “ { 𝑍 } ) = { 0 } ) )