Metamath Proof Explorer


Theorem zringidom

Description: The ring of integers is an integral domain. (Contributed by Thierry Arnoux, 4-May-2025)

Ref Expression
Assertion zringidom ring ∈ IDomn

Proof

Step Hyp Ref Expression
1 zringcrng ring ∈ CRing
2 zringnzr ring ∈ NzRing
3 eldifi ( 𝑥 ∈ ( ℤ ∖ { 0 } ) → 𝑥 ∈ ℤ )
4 3 ad2antrr ( ( ( 𝑥 ∈ ( ℤ ∖ { 0 } ) ∧ 𝑦 ∈ ℤ ) ∧ ( 𝑥 · 𝑦 ) = 0 ) → 𝑥 ∈ ℤ )
5 4 zcnd ( ( ( 𝑥 ∈ ( ℤ ∖ { 0 } ) ∧ 𝑦 ∈ ℤ ) ∧ ( 𝑥 · 𝑦 ) = 0 ) → 𝑥 ∈ ℂ )
6 simplr ( ( ( 𝑥 ∈ ( ℤ ∖ { 0 } ) ∧ 𝑦 ∈ ℤ ) ∧ ( 𝑥 · 𝑦 ) = 0 ) → 𝑦 ∈ ℤ )
7 6 zcnd ( ( ( 𝑥 ∈ ( ℤ ∖ { 0 } ) ∧ 𝑦 ∈ ℤ ) ∧ ( 𝑥 · 𝑦 ) = 0 ) → 𝑦 ∈ ℂ )
8 simpr ( ( ( 𝑥 ∈ ( ℤ ∖ { 0 } ) ∧ 𝑦 ∈ ℤ ) ∧ ( 𝑥 · 𝑦 ) = 0 ) → ( 𝑥 · 𝑦 ) = 0 )
9 mul0or ( ( 𝑥 ∈ ℂ ∧ 𝑦 ∈ ℂ ) → ( ( 𝑥 · 𝑦 ) = 0 ↔ ( 𝑥 = 0 ∨ 𝑦 = 0 ) ) )
10 9 biimpa ( ( ( 𝑥 ∈ ℂ ∧ 𝑦 ∈ ℂ ) ∧ ( 𝑥 · 𝑦 ) = 0 ) → ( 𝑥 = 0 ∨ 𝑦 = 0 ) )
11 5 7 8 10 syl21anc ( ( ( 𝑥 ∈ ( ℤ ∖ { 0 } ) ∧ 𝑦 ∈ ℤ ) ∧ ( 𝑥 · 𝑦 ) = 0 ) → ( 𝑥 = 0 ∨ 𝑦 = 0 ) )
12 eldifsni ( 𝑥 ∈ ( ℤ ∖ { 0 } ) → 𝑥 ≠ 0 )
13 12 ad2antrr ( ( ( 𝑥 ∈ ( ℤ ∖ { 0 } ) ∧ 𝑦 ∈ ℤ ) ∧ ( 𝑥 · 𝑦 ) = 0 ) → 𝑥 ≠ 0 )
14 13 neneqd ( ( ( 𝑥 ∈ ( ℤ ∖ { 0 } ) ∧ 𝑦 ∈ ℤ ) ∧ ( 𝑥 · 𝑦 ) = 0 ) → ¬ 𝑥 = 0 )
15 11 14 orcnd ( ( ( 𝑥 ∈ ( ℤ ∖ { 0 } ) ∧ 𝑦 ∈ ℤ ) ∧ ( 𝑥 · 𝑦 ) = 0 ) → 𝑦 = 0 )
16 15 ex ( ( 𝑥 ∈ ( ℤ ∖ { 0 } ) ∧ 𝑦 ∈ ℤ ) → ( ( 𝑥 · 𝑦 ) = 0 → 𝑦 = 0 ) )
17 16 ralrimiva ( 𝑥 ∈ ( ℤ ∖ { 0 } ) → ∀ 𝑦 ∈ ℤ ( ( 𝑥 · 𝑦 ) = 0 → 𝑦 = 0 ) )
18 eqid ( RLReg ‘ ℤring ) = ( RLReg ‘ ℤring )
19 zringbas ℤ = ( Base ‘ ℤring )
20 zringmulr · = ( .r ‘ ℤring )
21 zring0 0 = ( 0g ‘ ℤring )
22 18 19 20 21 isrrg ( 𝑥 ∈ ( RLReg ‘ ℤring ) ↔ ( 𝑥 ∈ ℤ ∧ ∀ 𝑦 ∈ ℤ ( ( 𝑥 · 𝑦 ) = 0 → 𝑦 = 0 ) ) )
23 3 17 22 sylanbrc ( 𝑥 ∈ ( ℤ ∖ { 0 } ) → 𝑥 ∈ ( RLReg ‘ ℤring ) )
24 23 ssriv ( ℤ ∖ { 0 } ) ⊆ ( RLReg ‘ ℤring )
25 19 18 21 isdomn2 ( ℤring ∈ Domn ↔ ( ℤring ∈ NzRing ∧ ( ℤ ∖ { 0 } ) ⊆ ( RLReg ‘ ℤring ) ) )
26 2 24 25 mpbir2an ring ∈ Domn
27 isidom ( ℤring ∈ IDomn ↔ ( ℤring ∈ CRing ∧ ℤring ∈ Domn ) )
28 1 26 27 mpbir2an ring ∈ IDomn