Metamath Proof Explorer


Theorem kerlidl

Description: The kernel of a ring homomorphism is an ideal. (Contributed by Thierry Arnoux, 1-Jul-2024)

Ref Expression
Hypotheses kerlidl.i 𝐼 = ( LIdeal ‘ 𝑅 )
kerlidl.1 0 = ( 0g𝑆 )
Assertion kerlidl ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) → ( 𝐹 “ { 0 } ) ∈ 𝐼 )

Proof

Step Hyp Ref Expression
1 kerlidl.i 𝐼 = ( LIdeal ‘ 𝑅 )
2 kerlidl.1 0 = ( 0g𝑆 )
3 rhmrcl2 ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) → 𝑆 ∈ Ring )
4 eqid ( LIdeal ‘ 𝑆 ) = ( LIdeal ‘ 𝑆 )
5 4 2 lidl0 ( 𝑆 ∈ Ring → { 0 } ∈ ( LIdeal ‘ 𝑆 ) )
6 3 5 syl ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) → { 0 } ∈ ( LIdeal ‘ 𝑆 ) )
7 1 rhmpreimaidl ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ { 0 } ∈ ( LIdeal ‘ 𝑆 ) ) → ( 𝐹 “ { 0 } ) ∈ 𝐼 )
8 6 7 mpdan ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) → ( 𝐹 “ { 0 } ) ∈ 𝐼 )