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 ˙ = 0 R
rhmkerinj.z ⊢ Z = 0 S
Assertion rhmkerinj ⊢ F ∈ R RingHom S → F : B ⟶ 1-1 C ↔ F -1 Z = 0 ˙

Proof

Step Hyp Ref Expression
1 rhmkerinj.b ⊢ B = Base R
2 rhmkerinj.c ⊢ C = Base S
3 rhmkerinj.0 ⊢ 0 ˙ = 0 R
4 rhmkerinj.z ⊢ Z = 0 S
5 rhmghm ⊢ F ∈ R RingHom S → F ∈ R GrpHom S
6 1 2 3 4 kerf1ghm ⊢ F ∈ R GrpHom S → F : B ⟶ 1-1 C ↔ F -1 Z = 0 ˙
7 5 6 syl ⊢ F ∈ R RingHom S → F : B ⟶ 1-1 C ↔ F -1 Z = 0 ˙