Metamath Proof Explorer


Theorem ccfldextdgrr

Description: The degree of the field extension of the complex numbers over the real numbers is 2. (Suggested by GL, 4-Aug-2023.) (Contributed by Thierry Arnoux, 20-Aug-2023)

Ref Expression
Assertion ccfldextdgrr fld .:. fld = 2

Proof

Step Hyp Ref Expression
1 ccfldextrr fld /FldExt fld
2 extdgval fld /FldExt fld fld .:. fld = dim subringAlg fld Base fld
3 1 2 ax-mp fld .:. fld = dim subringAlg fld Base fld
4 rebase = Base fld
5 4 fveq2i subringAlg fld = subringAlg fld Base fld
6 5 fveq2i dim subringAlg fld = dim subringAlg fld Base fld
7 ccfldsrarelvec subringAlg fld LVec
8 df-pr 1 i = 1 i
9 eqid LSpan subringAlg fld = LSpan subringAlg fld
10 eqidd subringAlg fld = subringAlg fld
11 cnfld0 0 = 0 fld
12 11 a1i 0 = 0 fld
13 ax-resscn
14 cnfldbas = Base fld
15 13 14 sseqtri Base fld
16 15 a1i Base fld
17 10 12 16 sralmod0 0 = 0 subringAlg fld
18 17 mptru 0 = 0 subringAlg fld
19 7 a1i subringAlg fld LVec
20 ax-1cn 1
21 ax-1ne0 1 0
22 10 16 srabase Base fld = Base subringAlg fld
23 22 mptru Base fld = Base subringAlg fld
24 14 23 eqtri = Base subringAlg fld
25 24 18 lindssn subringAlg fld LVec 1 1 0 1 LIndS subringAlg fld
26 7 20 21 25 mp3an 1 LIndS subringAlg fld
27 26 a1i 1 LIndS subringAlg fld
28 ax-icn i
29 ine0 i 0
30 24 18 lindssn subringAlg fld LVec i i 0 i LIndS subringAlg fld
31 7 28 29 30 mp3an i LIndS subringAlg fld
32 31 a1i i LIndS subringAlg fld
33 lveclmod subringAlg fld LVec subringAlg fld LMod
34 7 33 ax-mp subringAlg fld LMod
35 df-refld fld = fld 𝑠
36 10 16 srasca fld 𝑠 = Scalar subringAlg fld
37 36 mptru fld 𝑠 = Scalar subringAlg fld
38 35 37 eqtri fld = Scalar subringAlg fld
39 cnfldmul × = fld
40 10 16 sravsca fld = subringAlg fld
41 40 mptru fld = subringAlg fld
42 39 41 eqtri × = subringAlg fld
43 38 4 24 42 9 lspsnel subringAlg fld LMod 1 z LSpan subringAlg fld 1 x z = x 1
44 34 20 43 mp2an z LSpan subringAlg fld 1 x z = x 1
45 38 4 24 42 9 lspsnel subringAlg fld LMod i z LSpan subringAlg fld i y z = y i
46 34 28 45 mp2an z LSpan subringAlg fld i y z = y i
47 44 46 anbi12i z LSpan subringAlg fld 1 z LSpan subringAlg fld i x z = x 1 y z = y i
48 reeanv x y z = x 1 z = y i x z = x 1 y z = y i
49 simprl x y z = x 1 z = y i z = x 1
50 simpll x y z = x 1 z = y i x
51 50 recnd x y z = x 1 z = y i x
52 51 mulid1d x y z = x 1 z = y i x 1 = x
53 49 52 eqtrd x y z = x 1 z = y i z = x
54 53 negeqd x y z = x 1 z = y i z = x
55 simprr x y z = x 1 z = y i z = y i
56 simplr x y z = x 1 z = y i y
57 56 recnd x y z = x 1 z = y i y
58 28 a1i x y z = x 1 z = y i i
59 57 58 mulcomd x y z = x 1 z = y i y i = i y
60 55 59 eqtrd x y z = x 1 z = y i z = i y
61 54 60 oveq12d x y z = x 1 z = y i - z + z = - x + i y
62 53 51 eqeltrd x y z = x 1 z = y i z
63 62 subidd x y z = x 1 z = y i z z = 0
64 63 negeqd x y z = x 1 z = y i z z = 0
65 62 62 negsubdid x y z = x 1 z = y i z z = - z + z
66 neg0 0 = 0
67 66 a1i x y z = x 1 z = y i 0 = 0
68 64 65 67 3eqtr3d x y z = x 1 z = y i - z + z = 0
69 61 68 eqtr3d x y z = x 1 z = y i - x + i y = 0
70 50 renegcld x y z = x 1 z = y i x
71 creq0 x y x = 0 y = 0 - x + i y = 0
72 70 56 71 syl2anc x y z = x 1 z = y i x = 0 y = 0 - x + i y = 0
73 69 72 mpbird x y z = x 1 z = y i x = 0 y = 0
74 73 simpld x y z = x 1 z = y i x = 0
75 51 74 negcon1ad x y z = x 1 z = y i 0 = x
76 53 75 67 3eqtr2d x y z = x 1 z = y i z = 0
77 76 ex x y z = x 1 z = y i z = 0
78 77 rexlimivv x y z = x 1 z = y i z = 0
79 0red z = 0 0
80 simpr z = 0 x = 0 x = 0
81 80 oveq1d z = 0 x = 0 x 1 = 0 1
82 81 eqeq2d z = 0 x = 0 z = x 1 z = 0 1
83 82 anbi1d z = 0 x = 0 z = x 1 z = y i z = 0 1 z = y i
84 83 rexbidv z = 0 x = 0 y z = x 1 z = y i y z = 0 1 z = y i
85 simpr z = 0 y = 0 y = 0
86 85 oveq1d z = 0 y = 0 y i = 0 i
87 86 eqeq2d z = 0 y = 0 z = y i z = 0 i
88 87 anbi2d z = 0 y = 0 z = 0 1 z = y i z = 0 1 z = 0 i
89 20 mul02i 0 1 = 0
90 89 eqeq2i z = 0 1 z = 0
91 90 biimpri z = 0 z = 0 1
92 28 mul02i 0 i = 0
93 92 eqeq2i z = 0 i z = 0
94 93 biimpri z = 0 z = 0 i
95 91 94 jca z = 0 z = 0 1 z = 0 i
96 79 88 95 rspcedvd z = 0 y z = 0 1 z = y i
97 79 84 96 rspcedvd z = 0 x y z = x 1 z = y i
98 78 97 impbii x y z = x 1 z = y i z = 0
99 47 48 98 3bitr2i z LSpan subringAlg fld 1 z LSpan subringAlg fld i z = 0
100 elin z LSpan subringAlg fld 1 LSpan subringAlg fld i z LSpan subringAlg fld 1 z LSpan subringAlg fld i
101 velsn z 0 z = 0
102 99 100 101 3bitr4i z LSpan subringAlg fld 1 LSpan subringAlg fld i z 0
103 102 eqriv LSpan subringAlg fld 1 LSpan subringAlg fld i = 0
104 103 a1i LSpan subringAlg fld 1 LSpan subringAlg fld i = 0
105 9 18 19 27 32 104 lindsun 1 i LIndS subringAlg fld
106 105 mptru 1 i LIndS subringAlg fld
107 8 106 eqeltri 1 i LIndS subringAlg fld
108 cnfldadd + = + fld
109 10 16 sraaddg + fld = + subringAlg fld
110 109 mptru + fld = + subringAlg fld
111 108 110 eqtri + = + subringAlg fld
112 34 a1i subringAlg fld LMod
113 1cnd 1
114 28 a1i i
115 24 111 38 4 42 9 112 113 114 lspprel z LSpan subringAlg fld 1 i x y z = x 1 + y i
116 115 mptru z LSpan subringAlg fld 1 i x y z = x 1 + y i
117 simpl x y x
118 117 recnd x y x
119 1cnd x y 1
120 118 119 mulcld x y x 1
121 simpr x y y
122 121 recnd x y y
123 28 a1i x y i
124 122 123 mulcld x y y i
125 120 124 addcld x y x 1 + y i
126 eleq1 z = x 1 + y i z x 1 + y i
127 125 126 syl5ibrcom x y z = x 1 + y i z
128 127 rexlimivv x y z = x 1 + y i z
129 recl z z
130 simpr z x = z x = z
131 130 oveq1d z x = z x 1 = z 1
132 131 oveq1d z x = z x 1 + y i = z 1 + y i
133 132 eqeq2d z x = z z = x 1 + y i z = z 1 + y i
134 133 rexbidv z x = z y z = x 1 + y i y z = z 1 + y i
135 imcl z z
136 simpr z y = z y = z
137 136 oveq1d z y = z y i = z i
138 137 oveq2d z y = z z 1 + y i = z 1 + z i
139 138 eqeq2d z y = z z = z 1 + y i z = z 1 + z i
140 replim z z = z + i z
141 129 recnd z z
142 141 mulid1d z z 1 = z
143 135 recnd z z
144 28 a1i z i
145 143 144 mulcomd z z i = i z
146 142 145 oveq12d z z 1 + z i = z + i z
147 140 146 eqtr4d z z = z 1 + z i
148 135 139 147 rspcedvd z y z = z 1 + y i
149 129 134 148 rspcedvd z x y z = x 1 + y i
150 128 149 impbii x y z = x 1 + y i z
151 116 150 bitri z LSpan subringAlg fld 1 i z
152 151 eqriv LSpan subringAlg fld 1 i =
153 eqid LBasis subringAlg fld = LBasis subringAlg fld
154 24 153 9 islbs4 1 i LBasis subringAlg fld 1 i LIndS subringAlg fld LSpan subringAlg fld 1 i =
155 107 152 154 mpbir2an 1 i LBasis subringAlg fld
156 153 dimval subringAlg fld LVec 1 i LBasis subringAlg fld dim subringAlg fld = 1 i
157 7 155 156 mp2an dim subringAlg fld = 1 i
158 1nei 1 i
159 hashprg 1 i 1 i 1 i = 2
160 20 28 159 mp2an 1 i 1 i = 2
161 158 160 mpbi 1 i = 2
162 157 161 eqtri dim subringAlg fld = 2
163 3 6 162 3eqtr2i fld .:. fld = 2