Metamath Proof Explorer


Theorem aks5lem1

Description: Section 5 of https://www3.nd.edu/%7eandyp/notes/AKS.pdf. Construction of a ring homomorphism out of Zn X to K. (Contributed by metakunt, 7-Jun-2025)

Ref Expression
Hypotheses aks5lem1.1 ⊢ φ → K ∈ Field
aks5lem1.2 ⊢ P = chr ⁡ K
aks5lem1.3 ⊢ φ → P ∈ ℙ ∧ N ∈ ℕ ∧ P ∥ N
aks5lem1.4 ⊢ F = p ∈ Base Poly 1 ⁡ ℤ/Nℤ ⟼ G ∘ p
aks5lem1.5 ⊢ G = q ∈ Base ℤ/Nℤ ⟼ ⋃ ℤRHom ⁡ K q
aks5lem1.6 ⊢ H = r ∈ Base Poly 1 ⁡ K ⟼ eval 1 ⁡ K ⁡ r ⁡ M
aks5lem1.7 ⊢ φ → M ∈ Base K
Assertion aks5lem1 ⊢ φ → H ∘ F ∈ Poly 1 ⁡ ℤ/Nℤ RingHom K

Proof

Step Hyp Ref Expression
1 aks5lem1.1 ⊢ φ → K ∈ Field
2 aks5lem1.2 ⊢ P = chr ⁡ K
3 aks5lem1.3 ⊢ φ → P ∈ ℙ ∧ N ∈ ℕ ∧ P ∥ N
4 aks5lem1.4 ⊢ F = p ∈ Base Poly 1 ⁡ ℤ/Nℤ ⟼ G ∘ p
5 aks5lem1.5 ⊢ G = q ∈ Base ℤ/Nℤ ⟼ ⋃ ℤRHom ⁡ K q
6 aks5lem1.6 ⊢ H = r ∈ Base Poly 1 ⁡ K ⟼ eval 1 ⁡ K ⁡ r ⁡ M
7 aks5lem1.7 ⊢ φ → M ∈ Base K
8 eqid ⊢ eval 1 ⁡ K = eval 1 ⁡ K
9 eqid ⊢ Poly 1 ⁡ K = Poly 1 ⁡ K
10 eqid ⊢ Base K = Base K
11 eqid ⊢ Base Poly 1 ⁡ K = Base Poly 1 ⁡ K
12 1 fldcrngd ⊢ φ → K ∈ CRing
13 8 9 10 11 12 7 6 evl1maprhm ⊢ φ → H ∈ Poly 1 ⁡ K RingHom K
14 eqid ⊢ Poly 1 ⁡ ℤ/Nℤ = Poly 1 ⁡ ℤ/Nℤ
15 eqid ⊢ Base Poly 1 ⁡ ℤ/Nℤ = Base Poly 1 ⁡ ℤ/Nℤ
16 crngring ⊢ K ∈ CRing → K ∈ Ring
17 12 16 syl ⊢ φ → K ∈ Ring
18 3 simp2d ⊢ φ → N ∈ ℕ
19 2 eqcomi ⊢ chr ⁡ K = P
20 3 simp1d ⊢ φ → P ∈ ℙ
21 prmnn ⊢ P ∈ ℙ → P ∈ ℕ
22 20 21 syl ⊢ φ → P ∈ ℕ
23 22 nnzd ⊢ φ → P ∈ ℤ
24 19 23 eqeltrid ⊢ φ → chr ⁡ K ∈ ℤ
25 3 simp3d ⊢ φ → P ∥ N
26 19 25 eqbrtrid ⊢ φ → chr ⁡ K ∥ N
27 eqid ⊢ ℤ/Nℤ = ℤ/Nℤ
28 17 18 24 26 27 5 zndvdchrrhm ⊢ φ → G ∈ ℤ/Nℤ RingHom K
29 14 9 15 4 28 rhmply1 ⊢ φ → F ∈ Poly 1 ⁡ ℤ/Nℤ RingHom Poly 1 ⁡ K
30 rhmco ⊢ H ∈ Poly 1 ⁡ K RingHom K ∧ F ∈ Poly 1 ⁡ ℤ/Nℤ RingHom Poly 1 ⁡ K → H ∘ F ∈ Poly 1 ⁡ ℤ/Nℤ RingHom K
31 13 29 30 syl2anc ⊢ φ → H ∘ F ∈ Poly 1 ⁡ ℤ/Nℤ RingHom K