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
|- ( ph -> R e. Field )
subrfld.2
|- ( ph -> S e. ( SubRing ` R ) )
Assertion subrfld
|- ( ph -> ( R |`s S ) e. IDomn )

Proof

Step Hyp Ref Expression
1 subrfld.1
 |-  ( ph -> R e. Field )
2 subrfld.2
 |-  ( ph -> S e. ( SubRing ` R ) )
3 fldidom
 |-  ( R e. Field -> R e. IDomn )
4 1 3 syl
 |-  ( ph -> R e. IDomn )
5 4 2 subridom
 |-  ( ph -> ( R |`s S ) e. IDomn )