Metamath Proof Explorer


Theorem dchrzrhmul

Description: A Dirichlet character is completely multiplicative. (Contributed by Mario Carneiro, 4-May-2016)

Ref Expression
Hypotheses dchrmhm.g ⊢ G = DChr ⁡ N
dchrmhm.z ⊢ Z = ℤ/Nℤ
dchrmhm.b ⊢ D = Base G
dchrelbas4.l ⊢ L = ℤRHom ⁡ Z
dchrzrh1.x ⊢ φ → X ∈ D
dchrzrh1.a ⊢ φ → A ∈ ℤ
dchrzrh1.c ⊢ φ → C ∈ ℤ
Assertion dchrzrhmul ⊢ φ → X ⁡ L ⁡ A ⁢ C = X ⁡ L ⁡ A ⁢ X ⁡ L ⁡ C

Proof

Step Hyp Ref Expression
1 dchrmhm.g ⊢ G = DChr ⁡ N
2 dchrmhm.z ⊢ Z = ℤ/Nℤ
3 dchrmhm.b ⊢ D = Base G
4 dchrelbas4.l ⊢ L = ℤRHom ⁡ Z
5 dchrzrh1.x ⊢ φ → X ∈ D
6 dchrzrh1.a ⊢ φ → A ∈ ℤ
7 dchrzrh1.c ⊢ φ → C ∈ ℤ
8 1 3 dchrrcl ⊢ X ∈ D → N ∈ ℕ
9 5 8 syl ⊢ φ → N ∈ ℕ
10 9 nnnn0d ⊢ φ → N ∈ ℕ 0
11 2 zncrng ⊢ N ∈ ℕ 0 → Z ∈ CRing
12 10 11 syl ⊢ φ → Z ∈ CRing
13 crngring ⊢ Z ∈ CRing → Z ∈ Ring
14 12 13 syl ⊢ φ → Z ∈ Ring
15 4 zrhrhm ⊢ Z ∈ Ring → L ∈ ℤ ring RingHom Z
16 14 15 syl ⊢ φ → L ∈ ℤ ring RingHom Z
17 zringbas ⊢ ℤ = Base ℤ ring
18 zringmulr ⊢ × = ⋅ ℤ ring
19 eqid ⊢ ⋅ Z = ⋅ Z
20 17 18 19 rhmmul ⊢ L ∈ ℤ ring RingHom Z ∧ A ∈ ℤ ∧ C ∈ ℤ → L ⁡ A ⁢ C = L ⁡ A ⋅ Z L ⁡ C
21 16 6 7 20 syl3anc ⊢ φ → L ⁡ A ⁢ C = L ⁡ A ⋅ Z L ⁡ C
22 21 fveq2d ⊢ φ → X ⁡ L ⁡ A ⁢ C = X ⁡ L ⁡ A ⋅ Z L ⁡ C
23 1 2 3 dchrmhm ⊢ D ⊆ mulGrp Z MndHom mulGrp ℂ fld
24 23 5 sselid ⊢ φ → X ∈ mulGrp Z MndHom mulGrp ℂ fld
25 eqid ⊢ Base Z = Base Z
26 17 25 rhmf ⊢ L ∈ ℤ ring RingHom Z → L : ℤ ⟶ Base Z
27 16 26 syl ⊢ φ → L : ℤ ⟶ Base Z
28 27 6 ffvelcdmd ⊢ φ → L ⁡ A ∈ Base Z
29 27 7 ffvelcdmd ⊢ φ → L ⁡ C ∈ Base Z
30 eqid ⊢ mulGrp Z = mulGrp Z
31 30 25 mgpbas ⊢ Base Z = Base mulGrp Z
32 30 19 mgpplusg ⊢ ⋅ Z = + mulGrp Z
33 eqid ⊢ mulGrp ℂ fld = mulGrp ℂ fld
34 cnfldmul ⊢ × = ⋅ ℂ fld
35 33 34 mgpplusg ⊢ × = + mulGrp ℂ fld
36 31 32 35 mhmlin ⊢ X ∈ mulGrp Z MndHom mulGrp ℂ fld ∧ L ⁡ A ∈ Base Z ∧ L ⁡ C ∈ Base Z → X ⁡ L ⁡ A ⋅ Z L ⁡ C = X ⁡ L ⁡ A ⁢ X ⁡ L ⁡ C
37 24 28 29 36 syl3anc ⊢ φ → X ⁡ L ⁡ A ⋅ Z L ⁡ C = X ⁡ L ⁡ A ⁢ X ⁡ L ⁡ C
38 22 37 eqtrd ⊢ φ → X ⁡ L ⁡ A ⁢ C = X ⁡ L ⁡ A ⁢ X ⁡ L ⁡ C