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 ring
znzrh2.r ˙ = ring ~ QG S N
znzrh2.y Y = /N
znzrh2.2 L = ℤRHom Y
Assertion znzrh2 N 0 L = x x ˙

Proof

Step Hyp Ref Expression
1 znzrh2.s S = RSpan ring
2 znzrh2.r ˙ = ring ~ QG S N
3 znzrh2.y Y = /N
4 znzrh2.2 L = ℤRHom Y
5 zringring ring Ring
6 nn0z N 0 N
7 1 znlidl N S N LIdeal ring
8 6 7 syl N 0 S N LIdeal ring
9 2 oveq2i ring / 𝑠 ˙ = ring / 𝑠 ring ~ QG S N
10 zringcrng ring CRing
11 eqid LIdeal ring = LIdeal ring
12 11 crng2idl ring CRing LIdeal ring = 2Ideal ring
13 10 12 ax-mp LIdeal ring = 2Ideal ring
14 zringbas = Base ring
15 eceq2 ˙ = ring ~ QG S N x ˙ = x ring ~ QG S N
16 2 15 ax-mp x ˙ = x ring ~ QG S N
17 16 mpteq2i x x ˙ = x x ring ~ QG S N
18 9 13 14 17 qusrhm ring Ring S N LIdeal ring x x ˙ ring RingHom ring / 𝑠 ˙
19 5 8 18 sylancr N 0 x x ˙ ring RingHom ring / 𝑠 ˙
20 1 9 zncrng2 N ring / 𝑠 ˙ CRing
21 crngring ring / 𝑠 ˙ CRing ring / 𝑠 ˙ Ring
22 eqid ℤRHom ring / 𝑠 ˙ = ℤRHom ring / 𝑠 ˙
23 22 zrhrhmb ring / 𝑠 ˙ Ring x x ˙ ring RingHom ring / 𝑠 ˙ x x ˙ = ℤRHom ring / 𝑠 ˙
24 6 20 21 23 4syl N 0 x x ˙ ring RingHom ring / 𝑠 ˙ x x ˙ = ℤRHom ring / 𝑠 ˙
25 19 24 mpbid N 0 x x ˙ = ℤRHom ring / 𝑠 ˙
26 1 9 3 znzrh N 0 ℤRHom ring / 𝑠 ˙ = ℤRHom Y
27 25 26 eqtr2d N 0 ℤRHom Y = x x ˙
28 4 27 syl5eq N 0 L = x x ˙