Metamath Proof Explorer


Theorem aks6d1c5lem1

Description: Lemma for claim 5, evaluate the linear factor at -c to get a root. (Contributed by metakunt, 5-May-2025)

Ref Expression
Hypotheses aks6d1p5.1 φ K Field
aks6d1p5.2 φ P
aks6d1c5.3 P = chr K
aks6d1c5.4 φ A 0
aks6d1c5.5 φ A < P
aks6d1c5.6 X = var 1 K
aks6d1c5.7 × ˙ = mulGrp Poly 1 K
aks6d1c5.8 G = g 0 0 A mulGrp Poly 1 K i = 0 A g i × ˙ X + Poly 1 K algSc Poly 1 K ℤRHom K i
aks6d1c5p1.1 φ B 0 A
aks6d1c5p1.2 φ C 0 A
Assertion aks6d1c5lem1 φ B = C eval 1 K X + Poly 1 K algSc Poly 1 K ℤRHom K B ℤRHom K 0 C = 0 K

Proof

Step Hyp Ref Expression
1 aks6d1p5.1 φ K Field
2 aks6d1p5.2 φ P
3 aks6d1c5.3 P = chr K
4 aks6d1c5.4 φ A 0
5 aks6d1c5.5 φ A < P
6 aks6d1c5.6 X = var 1 K
7 aks6d1c5.7 × ˙ = mulGrp Poly 1 K
8 aks6d1c5.8 G = g 0 0 A mulGrp Poly 1 K i = 0 A g i × ˙ X + Poly 1 K algSc Poly 1 K ℤRHom K i
9 aks6d1c5p1.1 φ B 0 A
10 aks6d1c5p1.2 φ C 0 A
11 zringplusg + = + ring
12 11 eqcomi + ring = +
13 12 a1i φ + ring = +
14 13 oveqd φ 0 C + ring B = 0 - C + B
15 0cnd φ 0
16 10 elfzelzd φ C
17 16 zcnd φ C
18 9 elfzelzd φ B
19 18 zcnd φ B
20 15 17 19 subadd23d φ 0 - C + B = 0 + B - C
21 19 17 subcld φ B C
22 21 addlidd φ 0 + B - C = B C
23 20 22 eqtrd φ 0 - C + B = B C
24 14 23 eqtrd φ 0 C + ring B = B C
25 24 fveq2d φ ℤRHom K 0 C + ring B = ℤRHom K B C
26 25 eqeq1d φ ℤRHom K 0 C + ring B = 0 K ℤRHom K B C = 0 K
27 2 adantr φ B = C P
28 prmnn P P
29 27 28 syl φ B = C P
30 29 nnzd φ B = C P
31 dvds0 P P 0
32 30 31 syl φ B = C P 0
33 19 adantr φ B = C B
34 33 subidd φ B = C B B = 0
35 34 eqcomd φ B = C 0 = B B
36 simpr φ B = C B = C
37 36 oveq2d φ B = C B B = B C
38 35 37 eqtrd φ B = C 0 = B C
39 32 38 breqtrd φ B = C P B C
40 39 ex φ B = C P B C
41 2 28 syl φ P
42 41 adantr φ ¬ B = C P
43 42 adantr φ ¬ B = C C < B P
44 1zzd φ ¬ B = C C < B 1
45 43 nnzd φ ¬ B = C C < B P
46 45 44 zsubcld φ ¬ B = C C < B P 1
47 18 16 zsubcld φ B C
48 47 ad2antrr φ ¬ B = C C < B B C
49 1e0p1 1 = 0 + 1
50 49 a1i φ ¬ B = C C < B 1 = 0 + 1
51 simpr φ ¬ B = C C < B C < B
52 16 zred φ C
53 52 adantr φ ¬ B = C C
54 53 adantr φ ¬ B = C C < B C
55 18 zred φ B
56 55 adantr φ ¬ B = C B
57 56 adantr φ ¬ B = C C < B B
58 54 57 posdifd φ ¬ B = C C < B C < B 0 < B C
59 51 58 mpbid φ ¬ B = C C < B 0 < B C
60 0zd φ ¬ B = C C < B 0
61 60 48 zltp1led φ ¬ B = C C < B 0 < B C 0 + 1 B C
62 59 61 mpbid φ ¬ B = C C < B 0 + 1 B C
63 50 62 eqbrtrd φ ¬ B = C C < B 1 B C
64 48 zred φ ¬ B = C C < B B C
65 43 nnred φ ¬ B = C C < B P
66 elfzle1 C 0 A 0 C
67 10 66 syl φ 0 C
68 67 adantr φ ¬ B = C 0 C
69 68 adantr φ ¬ B = C C < B 0 C
70 57 54 subge02d φ ¬ B = C C < B 0 C B C B
71 69 70 mpbid φ ¬ B = C C < B B C B
72 4 nn0red φ A
73 41 nnred φ P
74 elfzle2 B 0 A B A
75 9 74 syl φ B A
76 55 72 73 75 5 lelttrd φ B < P
77 76 adantr φ ¬ B = C B < P
78 77 adantr φ ¬ B = C C < B B < P
79 64 57 65 71 78 lelttrd φ ¬ B = C C < B B C < P
80 48 45 zltlem1d φ ¬ B = C C < B B C < P B C P 1
81 79 80 mpbid φ ¬ B = C C < B B C P 1
82 44 46 48 63 81 elfzd φ ¬ B = C C < B B C 1 P 1
83 fzm1ndvds P B C 1 P 1 ¬ P B C
84 43 82 83 syl2anc φ ¬ B = C C < B ¬ P B C
85 simpll φ ¬ B = C ¬ C < B φ
86 axlttri B C B < C ¬ B = C C < B
87 55 52 86 syl2anc φ B < C ¬ B = C C < B
88 ioran ¬ B = C C < B ¬ B = C ¬ C < B
89 88 a1i φ ¬ B = C C < B ¬ B = C ¬ C < B
90 87 89 bitr2d φ ¬ B = C ¬ C < B B < C
91 90 biimpd φ ¬ B = C ¬ C < B B < C
92 91 imp φ ¬ B = C ¬ C < B B < C
93 92 anassrs φ ¬ B = C ¬ C < B B < C
94 85 93 jca φ ¬ B = C ¬ C < B φ B < C
95 41 adantr φ B < C P
96 1zzd φ B < C 1
97 41 nnzd φ P
98 97 adantr φ B < C P
99 98 96 zsubcld φ B < C P 1
100 16 adantr φ B < C C
101 18 adantr φ B < C B
102 100 101 zsubcld φ B < C C B
103 49 a1i φ B < C 1 = 0 + 1
104 55 52 posdifd φ B < C 0 < C B
105 104 biimpd φ B < C 0 < C B
106 105 imp φ B < C 0 < C B
107 0zd φ B < C 0
108 107 102 zltp1led φ B < C 0 < C B 0 + 1 C B
109 106 108 mpbid φ B < C 0 + 1 C B
110 103 109 eqbrtrd φ B < C 1 C B
111 102 zred φ B < C C B
112 52 adantr φ B < C C
113 73 adantr φ B < C P
114 9 adantr φ B < C B 0 A
115 elfzle1 B 0 A 0 B
116 114 115 syl φ B < C 0 B
117 55 adantr φ B < C B
118 112 117 subge02d φ B < C 0 B C B C
119 116 118 mpbid φ B < C C B C
120 72 adantr φ B < C A
121 elfzle2 C 0 A C A
122 10 121 syl φ C A
123 122 adantr φ B < C C A
124 5 adantr φ B < C A < P
125 112 120 113 123 124 lelttrd φ B < C C < P
126 111 112 113 119 125 lelttrd φ B < C C B < P
127 102 98 zltlem1d φ B < C C B < P C B P 1
128 126 127 mpbid φ B < C C B P 1
129 96 99 102 110 128 elfzd φ B < C C B 1 P 1
130 fzm1ndvds P C B 1 P 1 ¬ P C B
131 95 129 130 syl2anc φ B < C ¬ P C B
132 dvdsnegb P B C P B C P B C
133 97 47 132 syl2anc φ P B C P B C
134 19 17 negsubdi2d φ B C = C B
135 134 breq2d φ P B C P C B
136 133 135 bitrd φ P B C P C B
137 136 adantr φ B < C P B C P C B
138 131 137 mtbird φ B < C ¬ P B C
139 94 138 syl φ ¬ B = C ¬ C < B ¬ P B C
140 84 139 pm2.61dan φ ¬ B = C ¬ P B C
141 140 ex φ ¬ B = C ¬ P B C
142 141 con4d φ P B C B = C
143 40 142 impbid φ B = C P B C
144 1 fldcrngd φ K CRing
145 crngring K CRing K Ring
146 144 145 syl φ K Ring
147 eqid ℤRHom K = ℤRHom K
148 eqid 0 K = 0 K
149 3 147 148 chrdvds K Ring B C P B C ℤRHom K B C = 0 K
150 146 47 149 syl2anc φ P B C ℤRHom K B C = 0 K
151 143 150 bitr2d φ ℤRHom K B C = 0 K B = C
152 26 151 bitrd φ ℤRHom K 0 C + ring B = 0 K B = C
153 152 bicomd φ B = C ℤRHom K 0 C + ring B = 0 K
154 147 zrhrhm K Ring ℤRHom K ring RingHom K
155 rhmghm ℤRHom K ring RingHom K ℤRHom K ring GrpHom K
156 154 155 syl K Ring ℤRHom K ring GrpHom K
157 146 156 syl φ ℤRHom K ring GrpHom K
158 0zd φ 0
159 158 16 zsubcld φ 0 C
160 zringbas = Base ring
161 159 160 eleqtrdi φ 0 C Base ring
162 18 160 eleqtrdi φ B Base ring
163 eqid Base ring = Base ring
164 eqid + ring = + ring
165 eqid + K = + K
166 163 164 165 ghmlin ℤRHom K ring GrpHom K 0 C Base ring B Base ring ℤRHom K 0 C + ring B = ℤRHom K 0 C + K ℤRHom K B
167 157 161 162 166 syl3anc φ ℤRHom K 0 C + ring B = ℤRHom K 0 C + K ℤRHom K B
168 167 eqeq1d φ ℤRHom K 0 C + ring B = 0 K ℤRHom K 0 C + K ℤRHom K B = 0 K
169 153 168 bitrd φ B = C ℤRHom K 0 C + K ℤRHom K B = 0 K
170 eqid eval 1 K = eval 1 K
171 eqid Poly 1 K = Poly 1 K
172 eqid Base K = Base K
173 eqid Base Poly 1 K = Base Poly 1 K
174 160 172 ghmf ℤRHom K ring GrpHom K ℤRHom K : Base K
175 157 174 syl φ ℤRHom K : Base K
176 175 159 ffvelcdmd φ ℤRHom K 0 C Base K
177 170 6 172 171 173 144 176 evl1vard φ X Base Poly 1 K eval 1 K X ℤRHom K 0 C = ℤRHom K 0 C
178 eqid algSc Poly 1 K = algSc Poly 1 K
179 175 18 ffvelcdmd φ ℤRHom K B Base K
180 170 171 172 178 173 144 179 176 evl1scad φ algSc Poly 1 K ℤRHom K B Base Poly 1 K eval 1 K algSc Poly 1 K ℤRHom K B ℤRHom K 0 C = ℤRHom K B
181 eqid + Poly 1 K = + Poly 1 K
182 170 171 172 173 144 176 177 180 181 165 evl1addd φ X + Poly 1 K algSc Poly 1 K ℤRHom K B Base Poly 1 K eval 1 K X + Poly 1 K algSc Poly 1 K ℤRHom K B ℤRHom K 0 C = ℤRHom K 0 C + K ℤRHom K B
183 182 simprd φ eval 1 K X + Poly 1 K algSc Poly 1 K ℤRHom K B ℤRHom K 0 C = ℤRHom K 0 C + K ℤRHom K B
184 183 eqcomd φ ℤRHom K 0 C + K ℤRHom K B = eval 1 K X + Poly 1 K algSc Poly 1 K ℤRHom K B ℤRHom K 0 C
185 184 eqeq1d φ ℤRHom K 0 C + K ℤRHom K B = 0 K eval 1 K X + Poly 1 K algSc Poly 1 K ℤRHom K B ℤRHom K 0 C = 0 K
186 169 185 bitrd φ B = C eval 1 K X + Poly 1 K algSc Poly 1 K ℤRHom K B ℤRHom K 0 C = 0 K