Metamath Proof Explorer


Theorem rimrhm

Description: An isomorphism of rings is a homomorphism. (Contributed by AV, 22-Oct-2019)

Ref Expression
Hypotheses rhmf1o.b 𝐵 = ( Base ‘ 𝑅 )
rhmf1o.c 𝐶 = ( Base ‘ 𝑆 )
Assertion rimrhm ( 𝐹 ∈ ( 𝑅 RingIso 𝑆 ) → 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) )

Proof

Step Hyp Ref Expression
1 rhmf1o.b 𝐵 = ( Base ‘ 𝑅 )
2 rhmf1o.c 𝐶 = ( Base ‘ 𝑆 )
3 rimrcl ( 𝐹 ∈ ( 𝑅 RingIso 𝑆 ) → ( 𝑅 ∈ V ∧ 𝑆 ∈ V ) )
4 1 2 isrim ( ( 𝑅 ∈ V ∧ 𝑆 ∈ V ) → ( 𝐹 ∈ ( 𝑅 RingIso 𝑆 ) ↔ ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝐹 : 𝐵1-1-onto𝐶 ) ) )
5 simpl ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝐹 : 𝐵1-1-onto𝐶 ) → 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) )
6 4 5 syl6bi ( ( 𝑅 ∈ V ∧ 𝑆 ∈ V ) → ( 𝐹 ∈ ( 𝑅 RingIso 𝑆 ) → 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ) )
7 3 6 mpcom ( 𝐹 ∈ ( 𝑅 RingIso 𝑆 ) → 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) )