Metamath Proof Explorer


Theorem cphsubrglem

Description: Lemma for cphsubrg . (Contributed by Mario Carneiro, 9-Oct-2015)

Ref Expression
Hypotheses cphsubrglem.k ⊢ K = Base F
cphsubrglem.1 ⊢ φ → F = ℂ fld ↾ 𝑠 A
cphsubrglem.2 ⊢ φ → F ∈ DivRing
Assertion cphsubrglem ⊢ φ → F = ℂ fld ↾ 𝑠 K ∧ K = A ∩ ℂ ∧ K ∈ SubRing ⁡ ℂ fld

Proof

Step Hyp Ref Expression
1 cphsubrglem.k ⊢ K = Base F
2 cphsubrglem.1 ⊢ φ → F = ℂ fld ↾ 𝑠 A
3 cphsubrglem.2 ⊢ φ → F ∈ DivRing
4 2 fveq2d ⊢ φ → Base F = Base ℂ fld ↾ 𝑠 A
5 drngring ⊢ F ∈ DivRing → F ∈ Ring
6 3 5 syl ⊢ φ → F ∈ Ring
7 2 6 eqeltrrd ⊢ φ → ℂ fld ↾ 𝑠 A ∈ Ring
8 eqid ⊢ Base ℂ fld ↾ 𝑠 A = Base ℂ fld ↾ 𝑠 A
9 eqid ⊢ 0 ℂ fld ↾ 𝑠 A = 0 ℂ fld ↾ 𝑠 A
10 8 9 ring0cl ⊢ ℂ fld ↾ 𝑠 A ∈ Ring → 0 ℂ fld ↾ 𝑠 A ∈ Base ℂ fld ↾ 𝑠 A
11 reldmress ⊢ Rel ⁡ dom ⁡ ↾ 𝑠
12 eqid ⊢ ℂ fld ↾ 𝑠 A = ℂ fld ↾ 𝑠 A
13 11 12 8 elbasov ⊢ 0 ℂ fld ↾ 𝑠 A ∈ Base ℂ fld ↾ 𝑠 A → ℂ fld ∈ V ∧ A ∈ V
14 7 10 13 3syl ⊢ φ → ℂ fld ∈ V ∧ A ∈ V
15 14 simprd ⊢ φ → A ∈ V
16 cnfldbas ⊢ ℂ = Base ℂ fld
17 12 16 ressbas ⊢ A ∈ V → A ∩ ℂ = Base ℂ fld ↾ 𝑠 A
18 15 17 syl ⊢ φ → A ∩ ℂ = Base ℂ fld ↾ 𝑠 A
19 4 18 eqtr4d ⊢ φ → Base F = A ∩ ℂ
20 1 19 eqtrid ⊢ φ → K = A ∩ ℂ
21 20 oveq2d ⊢ φ → ℂ fld ↾ 𝑠 K = ℂ fld ↾ 𝑠 A ∩ ℂ
22 16 ressinbas ⊢ A ∈ V → ℂ fld ↾ 𝑠 A = ℂ fld ↾ 𝑠 A ∩ ℂ
23 15 22 syl ⊢ φ → ℂ fld ↾ 𝑠 A = ℂ fld ↾ 𝑠 A ∩ ℂ
24 21 23 eqtr4d ⊢ φ → ℂ fld ↾ 𝑠 K = ℂ fld ↾ 𝑠 A
25 2 24 eqtr4d ⊢ φ → F = ℂ fld ↾ 𝑠 K
26 25 6 eqeltrrd ⊢ φ → ℂ fld ↾ 𝑠 K ∈ Ring
27 cnring ⊢ ℂ fld ∈ Ring
28 26 27 jctil ⊢ φ → ℂ fld ∈ Ring ∧ ℂ fld ↾ 𝑠 K ∈ Ring
29 12 16 ressbasss ⊢ Base ℂ fld ↾ 𝑠 A ⊆ ℂ
30 4 29 eqsstrdi ⊢ φ → Base F ⊆ ℂ
31 1 30 eqsstrid ⊢ φ → K ⊆ ℂ
32 eqid ⊢ 0 F = 0 F
33 eqid ⊢ 1 F = 1 F
34 32 33 drngunz ⊢ F ∈ DivRing → 1 F ≠ 0 F
35 3 34 syl ⊢ φ → 1 F ≠ 0 F
36 25 fveq2d ⊢ φ → 0 F = 0 ℂ fld ↾ 𝑠 K
37 ringgrp ⊢ ℂ fld ∈ Ring → ℂ fld ∈ Grp
38 27 37 mp1i ⊢ φ → ℂ fld ∈ Grp
39 ringgrp ⊢ ℂ fld ↾ 𝑠 K ∈ Ring → ℂ fld ↾ 𝑠 K ∈ Grp
40 26 39 syl ⊢ φ → ℂ fld ↾ 𝑠 K ∈ Grp
41 16 issubg ⊢ K ∈ SubGrp ⁡ ℂ fld ↔ ℂ fld ∈ Grp ∧ K ⊆ ℂ ∧ ℂ fld ↾ 𝑠 K ∈ Grp
42 38 31 40 41 syl3anbrc ⊢ φ → K ∈ SubGrp ⁡ ℂ fld
43 eqid ⊢ ℂ fld ↾ 𝑠 K = ℂ fld ↾ 𝑠 K
44 cnfld0 ⊢ 0 = 0 ℂ fld
45 43 44 subg0 ⊢ K ∈ SubGrp ⁡ ℂ fld → 0 = 0 ℂ fld ↾ 𝑠 K
46 42 45 syl ⊢ φ → 0 = 0 ℂ fld ↾ 𝑠 K
47 36 46 eqtr4d ⊢ φ → 0 F = 0
48 35 47 neeqtrd ⊢ φ → 1 F ≠ 0
49 48 neneqd ⊢ φ → ¬ 1 F = 0
50 1 33 ringidcl ⊢ F ∈ Ring → 1 F ∈ K
51 6 50 syl ⊢ φ → 1 F ∈ K
52 31 51 sseldd ⊢ φ → 1 F ∈ ℂ
53 52 sqvald ⊢ φ → 1 F 2 = 1 F ⁢ 1 F
54 25 fveq2d ⊢ φ → 1 F = 1 ℂ fld ↾ 𝑠 K
55 54 oveq1d ⊢ φ → 1 F ⁢ 1 F = 1 ℂ fld ↾ 𝑠 K ⁢ 1 F
56 25 fveq2d ⊢ φ → Base F = Base ℂ fld ↾ 𝑠 K
57 1 56 eqtrid ⊢ φ → K = Base ℂ fld ↾ 𝑠 K
58 51 57 eleqtrd ⊢ φ → 1 F ∈ Base ℂ fld ↾ 𝑠 K
59 eqid ⊢ Base ℂ fld ↾ 𝑠 K = Base ℂ fld ↾ 𝑠 K
60 1 fvexi ⊢ K ∈ V
61 cnfldmul ⊢ × = ⋅ ℂ fld
62 43 61 ressmulr ⊢ K ∈ V → × = ⋅ ℂ fld ↾ 𝑠 K
63 60 62 ax-mp ⊢ × = ⋅ ℂ fld ↾ 𝑠 K
64 eqid ⊢ 1 ℂ fld ↾ 𝑠 K = 1 ℂ fld ↾ 𝑠 K
65 59 63 64 ringlidm ⊢ ℂ fld ↾ 𝑠 K ∈ Ring ∧ 1 F ∈ Base ℂ fld ↾ 𝑠 K → 1 ℂ fld ↾ 𝑠 K ⁢ 1 F = 1 F
66 26 58 65 syl2anc ⊢ φ → 1 ℂ fld ↾ 𝑠 K ⁢ 1 F = 1 F
67 53 55 66 3eqtrd ⊢ φ → 1 F 2 = 1 F
68 sq01 ⊢ 1 F ∈ ℂ → 1 F 2 = 1 F ↔ 1 F = 0 ∨ 1 F = 1
69 52 68 syl ⊢ φ → 1 F 2 = 1 F ↔ 1 F = 0 ∨ 1 F = 1
70 67 69 mpbid ⊢ φ → 1 F = 0 ∨ 1 F = 1
71 70 ord ⊢ φ → ¬ 1 F = 0 → 1 F = 1
72 49 71 mpd ⊢ φ → 1 F = 1
73 72 51 eqeltrrd ⊢ φ → 1 ∈ K
74 31 73 jca ⊢ φ → K ⊆ ℂ ∧ 1 ∈ K
75 cnfld1 ⊢ 1 = 1 ℂ fld
76 16 75 issubrg ⊢ K ∈ SubRing ⁡ ℂ fld ↔ ℂ fld ∈ Ring ∧ ℂ fld ↾ 𝑠 K ∈ Ring ∧ K ⊆ ℂ ∧ 1 ∈ K
77 28 74 76 sylanbrc ⊢ φ → K ∈ SubRing ⁡ ℂ fld
78 25 20 77 3jca ⊢ φ → F = ℂ fld ↾ 𝑠 K ∧ K = A ∩ ℂ ∧ K ∈ SubRing ⁡ ℂ fld