Metamath Proof Explorer


Theorem kerunit

Description: If a unit element lies in the kernel of a ring homomorphism, then 0 = 1 , i.e. the target ring is the zero ring. (Contributed by Thierry Arnoux, 24-Oct-2017)

Ref Expression
Hypotheses kerunit.1 ⊢ U = Unit ⁡ R
kerunit.2 ⊢ 0 ˙ = 0 S
kerunit.3 ⊢ 1 ˙ = 1 S
Assertion kerunit ⊢ F ∈ R RingHom S ∧ U ∩ F -1 0 ˙ ≠ ∅ → 1 ˙ = 0 ˙

Proof

Step Hyp Ref Expression
1 kerunit.1 ⊢ U = Unit ⁡ R
2 kerunit.2 ⊢ 0 ˙ = 0 S
3 kerunit.3 ⊢ 1 ˙ = 1 S
4 elin ⊢ x ∈ U ∩ F -1 0 ˙ ↔ x ∈ U ∧ x ∈ F -1 0 ˙
5 4 bilani ⊢ F ∈ R RingHom S ∧ x ∈ U ∩ F -1 0 ˙ → x ∈ U ∧ x ∈ F -1 0 ˙
6 5 simpld ⊢ F ∈ R RingHom S ∧ x ∈ U ∩ F -1 0 ˙ → x ∈ U
7 rhmrcl1 ⊢ F ∈ R RingHom S → R ∈ Ring
8 eqid ⊢ inv r ⁡ R = inv r ⁡ R
9 eqid ⊢ ⋅ R = ⋅ R
10 eqid ⊢ 1 R = 1 R
11 1 8 9 10 unitlinv ⊢ R ∈ Ring ∧ x ∈ U → inv r ⁡ R ⁡ x ⋅ R x = 1 R
12 11 fveq2d ⊢ R ∈ Ring ∧ x ∈ U → F ⁡ inv r ⁡ R ⁡ x ⋅ R x = F ⁡ 1 R
13 7 12 sylan ⊢ F ∈ R RingHom S ∧ x ∈ U → F ⁡ inv r ⁡ R ⁡ x ⋅ R x = F ⁡ 1 R
14 6 13 syldan ⊢ F ∈ R RingHom S ∧ x ∈ U ∩ F -1 0 ˙ → F ⁡ inv r ⁡ R ⁡ x ⋅ R x = F ⁡ 1 R
15 simpl ⊢ F ∈ R RingHom S ∧ x ∈ U ∩ F -1 0 ˙ → F ∈ R RingHom S
16 7 adantr ⊢ F ∈ R RingHom S ∧ x ∈ U ∩ F -1 0 ˙ → R ∈ Ring
17 eqid ⊢ Base R = Base R
18 1 8 17 ringinvcl ⊢ R ∈ Ring ∧ x ∈ U → inv r ⁡ R ⁡ x ∈ Base R
19 16 6 18 syl2anc ⊢ F ∈ R RingHom S ∧ x ∈ U ∩ F -1 0 ˙ → inv r ⁡ R ⁡ x ∈ Base R
20 17 1 unitcl ⊢ x ∈ U → x ∈ Base R
21 6 20 syl ⊢ F ∈ R RingHom S ∧ x ∈ U ∩ F -1 0 ˙ → x ∈ Base R
22 eqid ⊢ ⋅ S = ⋅ S
23 17 9 22 rhmmul ⊢ F ∈ R RingHom S ∧ inv r ⁡ R ⁡ x ∈ Base R ∧ x ∈ Base R → F ⁡ inv r ⁡ R ⁡ x ⋅ R x = F ⁡ inv r ⁡ R ⁡ x ⋅ S F ⁡ x
24 15 19 21 23 syl3anc ⊢ F ∈ R RingHom S ∧ x ∈ U ∩ F -1 0 ˙ → F ⁡ inv r ⁡ R ⁡ x ⋅ R x = F ⁡ inv r ⁡ R ⁡ x ⋅ S F ⁡ x
25 5 simprd ⊢ F ∈ R RingHom S ∧ x ∈ U ∩ F -1 0 ˙ → x ∈ F -1 0 ˙
26 eqid ⊢ Base S = Base S
27 17 26 rhmf ⊢ F ∈ R RingHom S → F : Base R ⟶ Base S
28 ffn ⊢ F : Base R ⟶ Base S → F Fn Base R
29 elpreima ⊢ F Fn Base R → x ∈ F -1 0 ˙ ↔ x ∈ Base R ∧ F ⁡ x ∈ 0 ˙
30 27 28 29 3syl ⊢ F ∈ R RingHom S → x ∈ F -1 0 ˙ ↔ x ∈ Base R ∧ F ⁡ x ∈ 0 ˙
31 30 simplbda ⊢ F ∈ R RingHom S ∧ x ∈ F -1 0 ˙ → F ⁡ x ∈ 0 ˙
32 25 31 syldan ⊢ F ∈ R RingHom S ∧ x ∈ U ∩ F -1 0 ˙ → F ⁡ x ∈ 0 ˙
33 fvex ⊢ F ⁡ x ∈ V
34 33 elsn ⊢ F ⁡ x ∈ 0 ˙ ↔ F ⁡ x = 0 ˙
35 32 34 sylib ⊢ F ∈ R RingHom S ∧ x ∈ U ∩ F -1 0 ˙ → F ⁡ x = 0 ˙
36 35 oveq2d ⊢ F ∈ R RingHom S ∧ x ∈ U ∩ F -1 0 ˙ → F ⁡ inv r ⁡ R ⁡ x ⋅ S F ⁡ x = F ⁡ inv r ⁡ R ⁡ x ⋅ S 0 ˙
37 rhmrcl2 ⊢ F ∈ R RingHom S → S ∈ Ring
38 37 adantr ⊢ F ∈ R RingHom S ∧ x ∈ U ∩ F -1 0 ˙ → S ∈ Ring
39 27 adantr ⊢ F ∈ R RingHom S ∧ x ∈ U ∩ F -1 0 ˙ → F : Base R ⟶ Base S
40 39 19 ffvelcdmd ⊢ F ∈ R RingHom S ∧ x ∈ U ∩ F -1 0 ˙ → F ⁡ inv r ⁡ R ⁡ x ∈ Base S
41 26 22 2 ringrz ⊢ S ∈ Ring ∧ F ⁡ inv r ⁡ R ⁡ x ∈ Base S → F ⁡ inv r ⁡ R ⁡ x ⋅ S 0 ˙ = 0 ˙
42 38 40 41 syl2anc ⊢ F ∈ R RingHom S ∧ x ∈ U ∩ F -1 0 ˙ → F ⁡ inv r ⁡ R ⁡ x ⋅ S 0 ˙ = 0 ˙
43 24 36 42 3eqtrd ⊢ F ∈ R RingHom S ∧ x ∈ U ∩ F -1 0 ˙ → F ⁡ inv r ⁡ R ⁡ x ⋅ R x = 0 ˙
44 10 3 rhm1 ⊢ F ∈ R RingHom S → F ⁡ 1 R = 1 ˙
45 44 adantr ⊢ F ∈ R RingHom S ∧ x ∈ U ∩ F -1 0 ˙ → F ⁡ 1 R = 1 ˙
46 14 43 45 3eqtr3rd ⊢ F ∈ R RingHom S ∧ x ∈ U ∩ F -1 0 ˙ → 1 ˙ = 0 ˙
47 46 reximdva0 ⊢ F ∈ R RingHom S ∧ U ∩ F -1 0 ˙ ≠ ∅ → ∃ x ∈ U ∩ F -1 0 ˙ 1 ˙ = 0 ˙
48 id ⊢ 1 ˙ = 0 ˙ → 1 ˙ = 0 ˙
49 48 rexlimivw ⊢ ∃ x ∈ U ∩ F -1 0 ˙ 1 ˙ = 0 ˙ → 1 ˙ = 0 ˙
50 47 49 syl ⊢ F ∈ R RingHom S ∧ U ∩ F -1 0 ˙ ≠ ∅ → 1 ˙ = 0 ˙