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.0 ⊢ 0 ˙ = 0 S
Assertion kerlidl ⊢ F ∈ R RingHom S → F -1 0 ˙ ∈ I

Proof

Step Hyp Ref Expression
1 kerlidl.i ⊢ I = LIdeal ⁡ R
2 kerlidl.0 ⊢ 0 ˙ = 0 S
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 -1 0 ˙ ∈ I
8 6 7 mpdan ⊢ F ∈ R RingHom S → F -1 0 ˙ ∈ I