Metamath Proof Explorer


Theorem znzrh2

Description: The ZZ ring homomorphism maps elements to their equivalence classes. (Contributed by Mario Carneiro, 15-Jun-2015) (Revised by AV, 13-Jun-2019)

Ref Expression
Hypotheses znzrh2.s ⊢ S = RSpan ⁡ ℤ ring
znzrh2.r ⊢ ∼ ˙ = ℤ ring ~ QG S ⁡ N
znzrh2.y ⊢ Y = ℤ/Nℤ
znzrh2.2 ⊢ L = ℤRHom ⁡ Y
Assertion znzrh2 ⊢ N ∈ ℕ 0 → L = x ∈ ℤ ⟼ x ∼ ˙

Proof

Step Hyp Ref Expression
1 znzrh2.s ⊢ S = RSpan ⁡ ℤ ring
2 znzrh2.r ⊢ ∼ ˙ = ℤ ring ~ QG S ⁡ N
3 znzrh2.y ⊢ Y = ℤ/Nℤ
4 znzrh2.2 ⊢ L = ℤRHom ⁡ Y
5 zringring ⊢ ℤ ring ∈ Ring
6 nn0z ⊢ N ∈ ℕ 0 → N ∈ ℤ
7 1 znlidl ⊢ N ∈ ℤ → S ⁡ N ∈ LIdeal ⁡ ℤ ring
8 6 7 syl ⊢ N ∈ ℕ 0 → S ⁡ N ∈ LIdeal ⁡ ℤ ring
9 2 oveq2i ⊢ ℤ ring / 𝑠 ∼ ˙ = ℤ ring / 𝑠 ℤ ring ~ QG S ⁡ N
10 zringcrng ⊢ ℤ ring ∈ CRing
11 eqid ⊢ LIdeal ⁡ ℤ ring = LIdeal ⁡ ℤ ring
12 11 crng2idl ⊢ ℤ ring ∈ CRing → LIdeal ⁡ ℤ ring = 2Ideal ⁡ ℤ ring
13 10 12 ax-mp ⊢ LIdeal ⁡ ℤ ring = 2Ideal ⁡ ℤ ring
14 zringbas ⊢ ℤ = Base ℤ ring
15 eceq2 ⊢ ∼ ˙ = ℤ ring ~ QG S ⁡ N → x ∼ ˙ = x ℤ ring ~ QG S ⁡ N
16 2 15 ax-mp ⊢ x ∼ ˙ = x ℤ ring ~ QG S ⁡ N
17 16 mpteq2i ⊢ x ∈ ℤ ⟼ x ∼ ˙ = x ∈ ℤ ⟼ x ℤ ring ~ QG S ⁡ N
18 9 13 14 17 qusrhm ⊢ ℤ ring ∈ Ring ∧ S ⁡ N ∈ LIdeal ⁡ ℤ ring → x ∈ ℤ ⟼ x ∼ ˙ ∈ ℤ ring RingHom ℤ ring / 𝑠 ∼ ˙
19 5 8 18 sylancr ⊢ N ∈ ℕ 0 → x ∈ ℤ ⟼ x ∼ ˙ ∈ ℤ ring RingHom ℤ ring / 𝑠 ∼ ˙
20 1 9 zncrng2 ⊢ N ∈ ℤ → ℤ ring / 𝑠 ∼ ˙ ∈ CRing
21 crngring ⊢ ℤ ring / 𝑠 ∼ ˙ ∈ CRing → ℤ ring / 𝑠 ∼ ˙ ∈ Ring
22 eqid ⊢ ℤRHom ⁡ ℤ ring / 𝑠 ∼ ˙ = ℤRHom ⁡ ℤ ring / 𝑠 ∼ ˙
23 22 zrhrhmb ⊢ ℤ ring / 𝑠 ∼ ˙ ∈ Ring → x ∈ ℤ ⟼ x ∼ ˙ ∈ ℤ ring RingHom ℤ ring / 𝑠 ∼ ˙ ↔ x ∈ ℤ ⟼ x ∼ ˙ = ℤRHom ⁡ ℤ ring / 𝑠 ∼ ˙
24 6 20 21 23 4syl ⊢ N ∈ ℕ 0 → x ∈ ℤ ⟼ x ∼ ˙ ∈ ℤ ring RingHom ℤ ring / 𝑠 ∼ ˙ ↔ x ∈ ℤ ⟼ x ∼ ˙ = ℤRHom ⁡ ℤ ring / 𝑠 ∼ ˙
25 19 24 mpbid ⊢ N ∈ ℕ 0 → x ∈ ℤ ⟼ x ∼ ˙ = ℤRHom ⁡ ℤ ring / 𝑠 ∼ ˙
26 1 9 3 znzrh ⊢ N ∈ ℕ 0 → ℤRHom ⁡ ℤ ring / 𝑠 ∼ ˙ = ℤRHom ⁡ Y
27 25 26 eqtr2d ⊢ N ∈ ℕ 0 → ℤRHom ⁡ Y = x ∈ ℤ ⟼ x ∼ ˙
28 4 27 eqtrid ⊢ N ∈ ℕ 0 → L = x ∈ ℤ ⟼ x ∼ ˙