Metamath Proof Explorer


Theorem sraidom

Description: Condition for a subring algebra to be an integral domain. (Contributed by Thierry Arnoux, 13-Oct-2025)

Ref Expression
Hypotheses sraidom.1 𝐴 = ( ( subringAlg ‘ 𝑅 ) ‘ 𝑉 )
sraidom.2 𝐵 = ( Base ‘ 𝑅 )
sraidom.3 ( 𝜑𝑅 ∈ IDomn )
sraidom.4 ( 𝜑𝑉𝐵 )
Assertion sraidom ( 𝜑𝐴 ∈ IDomn )

Proof

Step Hyp Ref Expression
1 sraidom.1 𝐴 = ( ( subringAlg ‘ 𝑅 ) ‘ 𝑉 )
2 sraidom.2 𝐵 = ( Base ‘ 𝑅 )
3 sraidom.3 ( 𝜑𝑅 ∈ IDomn )
4 sraidom.4 ( 𝜑𝑉𝐵 )
5 eqidd ( 𝜑 → ( Base ‘ 𝑅 ) = ( Base ‘ 𝑅 ) )
6 1 a1i ( 𝜑𝐴 = ( ( subringAlg ‘ 𝑅 ) ‘ 𝑉 ) )
7 4 2 sseqtrdi ( 𝜑𝑉 ⊆ ( Base ‘ 𝑅 ) )
8 6 7 srabase ( 𝜑 → ( Base ‘ 𝑅 ) = ( Base ‘ 𝐴 ) )
9 6 7 sraaddg ( 𝜑 → ( +g𝑅 ) = ( +g𝐴 ) )
10 9 oveqdr ( ( 𝜑 ∧ ( 𝑥 ∈ ( Base ‘ 𝑅 ) ∧ 𝑦 ∈ ( Base ‘ 𝑅 ) ) ) → ( 𝑥 ( +g𝑅 ) 𝑦 ) = ( 𝑥 ( +g𝐴 ) 𝑦 ) )
11 6 7 sramulr ( 𝜑 → ( .r𝑅 ) = ( .r𝐴 ) )
12 11 oveqdr ( ( 𝜑 ∧ ( 𝑥 ∈ ( Base ‘ 𝑅 ) ∧ 𝑦 ∈ ( Base ‘ 𝑅 ) ) ) → ( 𝑥 ( .r𝑅 ) 𝑦 ) = ( 𝑥 ( .r𝐴 ) 𝑦 ) )
13 5 8 10 12 idompropd ( 𝜑 → ( 𝑅 ∈ IDomn ↔ 𝐴 ∈ IDomn ) )
14 3 13 mpbid ( 𝜑𝐴 ∈ IDomn )