Metamath Proof Explorer
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 ∈ ( 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 ∈ ( 𝑅 RingHom 𝑆 ) → 𝐹 ∈ ( 𝑅 GrpHom 𝑆 ) ) |
| 6 |
1 2 3 4
|
kerf1ghm |
⊢ ( 𝐹 ∈ ( 𝑅 GrpHom 𝑆 ) → ( 𝐹 : 𝐵 –1-1→ 𝐶 ↔ ( ◡ 𝐹 “ { 𝑍 } ) = { 0 } ) ) |
| 7 |
5 6
|
syl |
⊢ ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) → ( 𝐹 : 𝐵 –1-1→ 𝐶 ↔ ( ◡ 𝐹 “ { 𝑍 } ) = { 0 } ) ) |