Metamath Proof Explorer


Theorem subrfld

Description: A subring of a field is an integral domain. (Contributed by Thierry Arnoux, 18-May-2025)

Ref Expression
Hypotheses subrfld.1 ( 𝜑𝑅 ∈ Field )
subrfld.2 ( 𝜑𝑆 ∈ ( SubRing ‘ 𝑅 ) )
Assertion subrfld ( 𝜑 → ( 𝑅s 𝑆 ) ∈ IDomn )

Proof

Step Hyp Ref Expression
1 subrfld.1 ( 𝜑𝑅 ∈ Field )
2 subrfld.2 ( 𝜑𝑆 ∈ ( SubRing ‘ 𝑅 ) )
3 fldidom ( 𝑅 ∈ Field → 𝑅 ∈ IDomn )
4 1 3 syl ( 𝜑𝑅 ∈ IDomn )
5 4 2 subridom ( 𝜑 → ( 𝑅s 𝑆 ) ∈ IDomn )