Metamath Proof Explorer


Theorem m1pmeq

Description: If two monic polynomials I and J differ by a unit factor K , then they are equal. (Contributed by Thierry Arnoux, 27-Apr-2025)

Ref Expression
Hypotheses m1pmeq.p P = Poly 1 F
m1pmeq.m M = Monic 1p F
m1pmeq.u U = Unit P
m1pmeq.t · ˙ = P
m1pmeq.r φ F Field
m1pmeq.f φ I M
m1pmeq.g φ J M
m1pmeq.h φ K U
m1pmeq.1 φ I = K · ˙ J
Assertion m1pmeq φ I = J

Proof

Step Hyp Ref Expression
1 m1pmeq.p P = Poly 1 F
2 m1pmeq.m M = Monic 1p F
3 m1pmeq.u U = Unit P
4 m1pmeq.t · ˙ = P
5 m1pmeq.r φ F Field
6 m1pmeq.f φ I M
7 m1pmeq.g φ J M
8 m1pmeq.h φ K U
9 m1pmeq.1 φ I = K · ˙ J
10 5 flddrngd φ F DivRing
11 10 drngringd φ F Ring
12 eqid Base P = Base P
13 12 3 unitcl K U K Base P
14 8 13 syl φ K Base P
15 8 3 eleqtrdi φ K Unit P
16 eqid algSc P = algSc P
17 eqid Base F = Base F
18 eqid 0 F = 0 F
19 eqid deg 1 F = deg 1 F
20 1 16 17 18 5 19 14 ply1unit φ K Unit P deg 1 F K = 0
21 15 20 mpbid φ deg 1 F K = 0
22 0le0 0 0
23 21 22 eqbrtrdi φ deg 1 F K 0
24 19 1 12 16 deg1le0 F Ring K Base P deg 1 F K 0 K = algSc P coe 1 K 0
25 24 biimpa F Ring K Base P deg 1 F K 0 K = algSc P coe 1 K 0
26 11 14 23 25 syl21anc φ K = algSc P coe 1 K 0
27 eqid F = F
28 eqid 1 F = 1 F
29 21 fveq2d φ coe 1 K deg 1 F K = coe 1 K 0
30 0nn0 0 0
31 21 30 eqeltrdi φ deg 1 F K 0
32 eqid coe 1 K = coe 1 K
33 32 12 1 17 coe1fvalcl K Base P deg 1 F K 0 coe 1 K deg 1 F K Base F
34 14 31 33 syl2anc φ coe 1 K deg 1 F K Base F
35 29 34 eqeltrrd φ coe 1 K 0 Base F
36 17 27 28 11 35 ringridmd φ coe 1 K 0 F 1 F = coe 1 K 0
37 9 fveq2d φ coe 1 I = coe 1 K · ˙ J
38 9 fveq2d φ deg 1 F I = deg 1 F K · ˙ J
39 eqid RLReg F = RLReg F
40 eqid 0 P = 0 P
41 drngnzr F DivRing F NzRing
42 10 41 syl φ F NzRing
43 1 ply1nz F NzRing P NzRing
44 42 43 syl φ P NzRing
45 3 40 44 8 unitnz φ K 0 P
46 fldidom F Field F IDomn
47 5 46 syl φ F IDomn
48 47 idomdomd φ F Domn
49 19 1 18 12 40 11 14 23 deg1le0eq0 φ K = 0 P coe 1 K 0 = 0 F
50 49 necon3bid φ K 0 P coe 1 K 0 0 F
51 45 50 mpbid φ coe 1 K 0 0 F
52 29 51 eqnetrd φ coe 1 K deg 1 F K 0 F
53 17 39 18 domnrrg F Domn coe 1 K deg 1 F K Base F coe 1 K deg 1 F K 0 F coe 1 K deg 1 F K RLReg F
54 48 34 52 53 syl3anc φ coe 1 K deg 1 F K RLReg F
55 1 12 2 mon1pcl J M J Base P
56 7 55 syl φ J Base P
57 1 40 2 mon1pn0 J M J 0 P
58 7 57 syl φ J 0 P
59 19 1 39 12 4 40 11 14 45 54 56 58 deg1mul2 φ deg 1 F K · ˙ J = deg 1 F K + deg 1 F J
60 38 59 eqtrd φ deg 1 F I = deg 1 F K + deg 1 F J
61 37 60 fveq12d φ coe 1 I deg 1 F I = coe 1 K · ˙ J deg 1 F K + deg 1 F J
62 19 28 2 mon1pldg I M coe 1 I deg 1 F I = 1 F
63 6 62 syl φ coe 1 I deg 1 F I = 1 F
64 1 4 27 12 19 40 11 14 45 56 58 coe1mul4 φ coe 1 K · ˙ J deg 1 F K + deg 1 F J = coe 1 K deg 1 F K F coe 1 J deg 1 F J
65 19 28 2 mon1pldg J M coe 1 J deg 1 F J = 1 F
66 7 65 syl φ coe 1 J deg 1 F J = 1 F
67 29 66 oveq12d φ coe 1 K deg 1 F K F coe 1 J deg 1 F J = coe 1 K 0 F 1 F
68 64 67 eqtrd φ coe 1 K · ˙ J deg 1 F K + deg 1 F J = coe 1 K 0 F 1 F
69 61 63 68 3eqtr3rd φ coe 1 K 0 F 1 F = 1 F
70 36 69 eqtr3d φ coe 1 K 0 = 1 F
71 70 fveq2d φ algSc P coe 1 K 0 = algSc P 1 F
72 eqid 1 P = 1 P
73 1 16 28 72 11 ply1ascl1 φ algSc P 1 F = 1 P
74 26 71 73 3eqtrd φ K = 1 P
75 74 oveq1d φ K · ˙ J = 1 P · ˙ J
76 1 ply1ring F Ring P Ring
77 11 76 syl φ P Ring
78 12 4 72 77 56 ringlidmd φ 1 P · ˙ J = J
79 9 75 78 3eqtrd φ I = J