Metamath Proof Explorer


Theorem zzngim

Description: The ZZ ring homomorphism is an isomorphism for N = 0 . (We only show group isomorphism here, but ring isomorphism follows, since it is a bijective ring homomorphism.) (Contributed by Mario Carneiro, 21-Apr-2016) (Revised by AV, 13-Jun-2019)

Ref Expression
Hypotheses zzngim.y ⊢ 𝑌 = ( ℤ/nℤ ‘ 0 )
zzngim.2 ⊢ 𝐿 = ( ℤRHom ‘ 𝑌 )
Assertion zzngim 𝐿 ∈ ( ℤring GrpIso 𝑌 )

Proof

Step Hyp Ref Expression
1 zzngim.y ⊢ 𝑌 = ( ℤ/nℤ ‘ 0 )
2 zzngim.2 ⊢ 𝐿 = ( ℤRHom ‘ 𝑌 )
3 0nn0 ⊢ 0 ∈ ℕ0
4 1 zncrng ⊢ ( 0 ∈ ℕ0 → 𝑌 ∈ CRing )
5 crngring ⊢ ( 𝑌 ∈ CRing → 𝑌 ∈ Ring )
6 3 4 5 mp2b ⊢ 𝑌 ∈ Ring
7 2 zrhrhm ⊢ ( 𝑌 ∈ Ring → 𝐿 ∈ ( ℤring RingHom 𝑌 ) )
8 rhmghm ⊢ ( 𝐿 ∈ ( ℤring RingHom 𝑌 ) → 𝐿 ∈ ( ℤring GrpHom 𝑌 ) )
9 6 7 8 mp2b ⊢ 𝐿 ∈ ( ℤring GrpHom 𝑌 )
10 eqid ⊢ ( Base ‘ 𝑌 ) = ( Base ‘ 𝑌 )
11 1 10 2 znzrhfo ⊢ ( 0 ∈ ℕ0 → 𝐿 : ℤ –onto→ ( Base ‘ 𝑌 ) )
12 3 11 ax-mp ⊢ 𝐿 : ℤ –onto→ ( Base ‘ 𝑌 )
13 fofn ⊢ ( 𝐿 : ℤ –onto→ ( Base ‘ 𝑌 ) → 𝐿 Fn ℤ )
14 fnresdm ⊢ ( 𝐿 Fn ℤ → ( 𝐿 ↾ ℤ ) = 𝐿 )
15 12 13 14 mp2b ⊢ ( 𝐿 ↾ ℤ ) = 𝐿
16 2 reseq1i ⊢ ( 𝐿 ↾ ℤ ) = ( ( ℤRHom ‘ 𝑌 ) ↾ ℤ )
17 15 16 eqtr3i ⊢ 𝐿 = ( ( ℤRHom ‘ 𝑌 ) ↾ ℤ )
18 eqid ⊢ 0 = 0
19 18 iftruei ⊢ if ( 0 = 0 , ℤ , ( 0 ..^ 0 ) ) = ℤ
20 19 eqcomi ⊢ ℤ = if ( 0 = 0 , ℤ , ( 0 ..^ 0 ) )
21 1 10 17 20 znf1o ⊢ ( 0 ∈ ℕ0 → 𝐿 : ℤ –1-1-onto→ ( Base ‘ 𝑌 ) )
22 3 21 ax-mp ⊢ 𝐿 : ℤ –1-1-onto→ ( Base ‘ 𝑌 )
23 zringbas ⊢ ℤ = ( Base ‘ ℤring )
24 23 10 isgim ⊢ ( 𝐿 ∈ ( ℤring GrpIso 𝑌 ) ↔ ( 𝐿 ∈ ( ℤring GrpHom 𝑌 ) ∧ 𝐿 : ℤ –1-1-onto→ ( Base ‘ 𝑌 ) ) )
25 9 22 24 mpbir2an ⊢ 𝐿 ∈ ( ℤring GrpIso 𝑌 )