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 = ( 0g𝑆 )
kerunit.3 1 = ( 1r𝑆 )
Assertion kerunit ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) ≠ ∅ ) → 1 = 0 )

Proof

Step Hyp Ref Expression
1 kerunit.1 𝑈 = ( Unit ‘ 𝑅 )
2 kerunit.2 0 = ( 0g𝑆 )
3 kerunit.3 1 = ( 1r𝑆 )
4 elin ( 𝑥 ∈ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) ↔ ( 𝑥𝑈𝑥 ∈ ( 𝐹 “ { 0 } ) ) )
5 4 bilani ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝑥 ∈ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) ) → ( 𝑥𝑈𝑥 ∈ ( 𝐹 “ { 0 } ) ) )
6 5 simpld ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝑥 ∈ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) ) → 𝑥𝑈 )
7 rhmrcl1 ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) → 𝑅 ∈ Ring )
8 eqid ( invr𝑅 ) = ( invr𝑅 )
9 eqid ( .r𝑅 ) = ( .r𝑅 )
10 eqid ( 1r𝑅 ) = ( 1r𝑅 )
11 1 8 9 10 unitlinv ( ( 𝑅 ∈ Ring ∧ 𝑥𝑈 ) → ( ( ( invr𝑅 ) ‘ 𝑥 ) ( .r𝑅 ) 𝑥 ) = ( 1r𝑅 ) )
12 11 fveq2d ( ( 𝑅 ∈ Ring ∧ 𝑥𝑈 ) → ( 𝐹 ‘ ( ( ( invr𝑅 ) ‘ 𝑥 ) ( .r𝑅 ) 𝑥 ) ) = ( 𝐹 ‘ ( 1r𝑅 ) ) )
13 7 12 sylan ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝑥𝑈 ) → ( 𝐹 ‘ ( ( ( invr𝑅 ) ‘ 𝑥 ) ( .r𝑅 ) 𝑥 ) ) = ( 𝐹 ‘ ( 1r𝑅 ) ) )
14 6 13 syldan ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝑥 ∈ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) ) → ( 𝐹 ‘ ( ( ( invr𝑅 ) ‘ 𝑥 ) ( .r𝑅 ) 𝑥 ) ) = ( 𝐹 ‘ ( 1r𝑅 ) ) )
15 simpl ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝑥 ∈ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) ) → 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) )
16 7 adantr ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝑥 ∈ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) ) → 𝑅 ∈ Ring )
17 eqid ( Base ‘ 𝑅 ) = ( Base ‘ 𝑅 )
18 1 8 17 ringinvcl ( ( 𝑅 ∈ Ring ∧ 𝑥𝑈 ) → ( ( invr𝑅 ) ‘ 𝑥 ) ∈ ( Base ‘ 𝑅 ) )
19 16 6 18 syl2anc ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝑥 ∈ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) ) → ( ( invr𝑅 ) ‘ 𝑥 ) ∈ ( Base ‘ 𝑅 ) )
20 17 1 unitcl ( 𝑥𝑈𝑥 ∈ ( Base ‘ 𝑅 ) )
21 6 20 syl ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝑥 ∈ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) ) → 𝑥 ∈ ( Base ‘ 𝑅 ) )
22 eqid ( .r𝑆 ) = ( .r𝑆 )
23 17 9 22 rhmmul ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ ( ( invr𝑅 ) ‘ 𝑥 ) ∈ ( Base ‘ 𝑅 ) ∧ 𝑥 ∈ ( Base ‘ 𝑅 ) ) → ( 𝐹 ‘ ( ( ( invr𝑅 ) ‘ 𝑥 ) ( .r𝑅 ) 𝑥 ) ) = ( ( 𝐹 ‘ ( ( invr𝑅 ) ‘ 𝑥 ) ) ( .r𝑆 ) ( 𝐹𝑥 ) ) )
24 15 19 21 23 syl3anc ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝑥 ∈ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) ) → ( 𝐹 ‘ ( ( ( invr𝑅 ) ‘ 𝑥 ) ( .r𝑅 ) 𝑥 ) ) = ( ( 𝐹 ‘ ( ( invr𝑅 ) ‘ 𝑥 ) ) ( .r𝑆 ) ( 𝐹𝑥 ) ) )
25 5 simprd ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝑥 ∈ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) ) → 𝑥 ∈ ( 𝐹 “ { 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 “ { 0 } ) ↔ ( 𝑥 ∈ ( Base ‘ 𝑅 ) ∧ ( 𝐹𝑥 ) ∈ { 0 } ) ) )
30 27 28 29 3syl ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) → ( 𝑥 ∈ ( 𝐹 “ { 0 } ) ↔ ( 𝑥 ∈ ( Base ‘ 𝑅 ) ∧ ( 𝐹𝑥 ) ∈ { 0 } ) ) )
31 30 simplbda ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝑥 ∈ ( 𝐹 “ { 0 } ) ) → ( 𝐹𝑥 ) ∈ { 0 } )
32 25 31 syldan ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝑥 ∈ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) ) → ( 𝐹𝑥 ) ∈ { 0 } )
33 fvex ( 𝐹𝑥 ) ∈ V
34 33 elsn ( ( 𝐹𝑥 ) ∈ { 0 } ↔ ( 𝐹𝑥 ) = 0 )
35 32 34 sylib ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝑥 ∈ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) ) → ( 𝐹𝑥 ) = 0 )
36 35 oveq2d ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝑥 ∈ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) ) → ( ( 𝐹 ‘ ( ( invr𝑅 ) ‘ 𝑥 ) ) ( .r𝑆 ) ( 𝐹𝑥 ) ) = ( ( 𝐹 ‘ ( ( invr𝑅 ) ‘ 𝑥 ) ) ( .r𝑆 ) 0 ) )
37 rhmrcl2 ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) → 𝑆 ∈ Ring )
38 37 adantr ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝑥 ∈ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) ) → 𝑆 ∈ Ring )
39 27 adantr ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝑥 ∈ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) ) → 𝐹 : ( Base ‘ 𝑅 ) ⟶ ( Base ‘ 𝑆 ) )
40 39 19 ffvelcdmd ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝑥 ∈ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) ) → ( 𝐹 ‘ ( ( invr𝑅 ) ‘ 𝑥 ) ) ∈ ( Base ‘ 𝑆 ) )
41 26 22 2 ringrz ( ( 𝑆 ∈ Ring ∧ ( 𝐹 ‘ ( ( invr𝑅 ) ‘ 𝑥 ) ) ∈ ( Base ‘ 𝑆 ) ) → ( ( 𝐹 ‘ ( ( invr𝑅 ) ‘ 𝑥 ) ) ( .r𝑆 ) 0 ) = 0 )
42 38 40 41 syl2anc ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝑥 ∈ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) ) → ( ( 𝐹 ‘ ( ( invr𝑅 ) ‘ 𝑥 ) ) ( .r𝑆 ) 0 ) = 0 )
43 24 36 42 3eqtrd ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝑥 ∈ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) ) → ( 𝐹 ‘ ( ( ( invr𝑅 ) ‘ 𝑥 ) ( .r𝑅 ) 𝑥 ) ) = 0 )
44 10 3 rhm1 ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) → ( 𝐹 ‘ ( 1r𝑅 ) ) = 1 )
45 44 adantr ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝑥 ∈ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) ) → ( 𝐹 ‘ ( 1r𝑅 ) ) = 1 )
46 14 43 45 3eqtr3rd ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ 𝑥 ∈ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) ) → 1 = 0 )
47 46 reximdva0 ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) ≠ ∅ ) → ∃ 𝑥 ∈ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) 1 = 0 )
48 id ( 1 = 01 = 0 )
49 48 rexlimivw ( ∃ 𝑥 ∈ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) 1 = 01 = 0 )
50 47 49 syl ( ( 𝐹 ∈ ( 𝑅 RingHom 𝑆 ) ∧ ( 𝑈 ∩ ( 𝐹 “ { 0 } ) ) ≠ ∅ ) → 1 = 0 )