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 = fld 𝑠 A
cphsubrglem.2 φ F DivRing
Assertion cphsubrglem φ F = fld 𝑠 K K = A K SubRing fld

Proof

Step Hyp Ref Expression
1 cphsubrglem.k K = Base F
2 cphsubrglem.1 φ F = fld 𝑠 A
3 cphsubrglem.2 φ F DivRing
4 2 fveq2d φ Base F = Base fld 𝑠 A
5 drngring F DivRing F Ring
6 3 5 syl φ F Ring
7 2 6 eqeltrrd φ fld 𝑠 A Ring
8 eqid Base fld 𝑠 A = Base fld 𝑠 A
9 eqid 0 fld 𝑠 A = 0 fld 𝑠 A
10 8 9 ring0cl fld 𝑠 A Ring 0 fld 𝑠 A Base fld 𝑠 A
11 reldmress Rel dom 𝑠
12 eqid fld 𝑠 A = fld 𝑠 A
13 11 12 8 elbasov 0 fld 𝑠 A Base fld 𝑠 A fld V A V
14 7 10 13 3syl φ fld V A V
15 14 simprd φ A V
16 cnfldbas = Base fld
17 12 16 ressbas A V A = Base fld 𝑠 A
18 15 17 syl φ A = Base fld 𝑠 A
19 4 18 eqtr4d φ Base F = A
20 1 19 syl5eq φ K = A
21 20 oveq2d φ fld 𝑠 K = fld 𝑠 A
22 16 ressinbas A V fld 𝑠 A = fld 𝑠 A
23 15 22 syl φ fld 𝑠 A = fld 𝑠 A
24 21 23 eqtr4d φ fld 𝑠 K = fld 𝑠 A
25 2 24 eqtr4d φ F = fld 𝑠 K
26 25 6 eqeltrrd φ fld 𝑠 K Ring
27 cnring fld Ring
28 26 27 jctil φ fld Ring fld 𝑠 K Ring
29 12 16 ressbasss Base fld 𝑠 A
30 4 29 eqsstrdi φ Base F
31 1 30 eqsstrid φ K
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 fld 𝑠 K
37 ringgrp fld Ring fld Grp
38 27 37 mp1i φ fld Grp
39 ringgrp fld 𝑠 K Ring fld 𝑠 K Grp
40 26 39 syl φ fld 𝑠 K Grp
41 16 issubg K SubGrp fld fld Grp K fld 𝑠 K Grp
42 38 31 40 41 syl3anbrc φ K SubGrp fld
43 eqid fld 𝑠 K = fld 𝑠 K
44 cnfld0 0 = 0 fld
45 43 44 subg0 K SubGrp fld 0 = 0 fld 𝑠 K
46 42 45 syl φ 0 = 0 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
53 52 sqvald φ 1 F 2 = 1 F 1 F
54 25 fveq2d φ 1 F = 1 fld 𝑠 K
55 54 oveq1d φ 1 F 1 F = 1 fld 𝑠 K 1 F
56 25 fveq2d φ Base F = Base fld 𝑠 K
57 1 56 syl5eq φ K = Base fld 𝑠 K
58 51 57 eleqtrd φ 1 F Base fld 𝑠 K
59 eqid Base fld 𝑠 K = Base fld 𝑠 K
60 1 fvexi K V
61 cnfldmul × = fld
62 43 61 ressmulr K V × = fld 𝑠 K
63 60 62 ax-mp × = fld 𝑠 K
64 eqid 1 fld 𝑠 K = 1 fld 𝑠 K
65 59 63 64 ringlidm fld 𝑠 K Ring 1 F Base fld 𝑠 K 1 fld 𝑠 K 1 F = 1 F
66 26 58 65 syl2anc φ 1 fld 𝑠 K 1 F = 1 F
67 53 55 66 3eqtrd φ 1 F 2 = 1 F
68 sq01 1 F 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 1 K
75 cnfld1 1 = 1 fld
76 16 75 issubrg K SubRing fld fld Ring fld 𝑠 K Ring K 1 K
77 28 74 76 sylanbrc φ K SubRing fld
78 25 20 77 3jca φ F = fld 𝑠 K K = A K SubRing fld