Metamath Proof Explorer


Theorem irrednzr

Description: A ring with an irreducible element cannot be the zero ring. (Contributed by Thierry Arnoux, 18-May-2025)

Ref Expression
Hypotheses irrednzr.1 ⊢ 𝐼 = ( Irred ‘ 𝑅 )
irrednzr.2 ⊢ ( 𝜑 → 𝑅 ∈ Ring )
irrednzr.3 ⊢ ( 𝜑 → 𝑋 ∈ 𝐼 )
Assertion irrednzr ( 𝜑 → 𝑅 ∈ NzRing )

Proof

Step Hyp Ref Expression
1 irrednzr.1 ⊢ 𝐼 = ( Irred ‘ 𝑅 )
2 irrednzr.2 ⊢ ( 𝜑 → 𝑅 ∈ Ring )
3 irrednzr.3 ⊢ ( 𝜑 → 𝑋 ∈ 𝐼 )
4 eqid ⊢ ( Base ‘ 𝑅 ) = ( Base ‘ 𝑅 )
5 1 4 irredcl ⊢ ( 𝑋 ∈ 𝐼 → 𝑋 ∈ ( Base ‘ 𝑅 ) )
6 3 5 syl ⊢ ( 𝜑 → 𝑋 ∈ ( Base ‘ 𝑅 ) )
7 eqid ⊢ ( 0g ‘ 𝑅 ) = ( 0g ‘ 𝑅 )
8 1 7 irredn0 ⊢ ( ( 𝑅 ∈ Ring ∧ 𝑋 ∈ 𝐼 ) → 𝑋 ≠ ( 0g ‘ 𝑅 ) )
9 2 3 8 syl2anc ⊢ ( 𝜑 → 𝑋 ≠ ( 0g ‘ 𝑅 ) )
10 6 9 eldifsnd ⊢ ( 𝜑 → 𝑋 ∈ ( ( Base ‘ 𝑅 ) ∖ { ( 0g ‘ 𝑅 ) } ) )
11 7 4 ringelnzr ⊢ ( ( 𝑅 ∈ Ring ∧ 𝑋 ∈ ( ( Base ‘ 𝑅 ) ∖ { ( 0g ‘ 𝑅 ) } ) ) → 𝑅 ∈ NzRing )
12 2 10 11 syl2anc ⊢ ( 𝜑 → 𝑅 ∈ NzRing )