Metamath Proof Explorer

Theorem subrgchr

Description: If A is a subring of R , then they have the same characteristic. (Contributed by Thierry Arnoux, 24-Feb-2018)

Ref Expression
Assertion subrgchr 
|- ( A e. ( SubRing ` R ) -> ( chr ` ( R |`s A ) ) = ( chr ` R ) )

Proof

Step Hyp Ref Expression
1 subrgsubg 
 |-  ( A e. ( SubRing ` R ) -> A e. ( SubGrp ` R ) )
2 eqid 
 |-  ( 1r ` R ) = ( 1r ` R )
3 2 subrg1cl 
 |-  ( A e. ( SubRing ` R ) -> ( 1r ` R ) e. A )
4 eqid 
 |-  ( R |`s A ) = ( R |`s A )
5 eqid 
 |-  ( od ` R ) = ( od ` R )
6 eqid 
 |-  ( od ` ( R |`s A ) ) = ( od ` ( R |`s A ) )
7 4 5 6 subgod 
 |-  ( ( A e. ( SubGrp ` R ) /\ ( 1r ` R ) e. A ) -> ( ( od ` R ) ` ( 1r ` R ) ) = ( ( od ` ( R |`s A ) ) ` ( 1r ` R ) ) )
8 1 3 7 syl2anc 
 |-  ( A e. ( SubRing ` R ) -> ( ( od ` R ) ` ( 1r ` R ) ) = ( ( od ` ( R |`s A ) ) ` ( 1r ` R ) ) )
9 4 2 subrg1 
 |-  ( A e. ( SubRing ` R ) -> ( 1r ` R ) = ( 1r ` ( R |`s A ) ) )
10 9 fveq2d 
 |-  ( A e. ( SubRing ` R ) -> ( ( od ` ( R |`s A ) ) ` ( 1r ` R ) ) = ( ( od ` ( R |`s A ) ) ` ( 1r ` ( R |`s A ) ) ) )
11 8 10 eqtr2d 
 |-  ( A e. ( SubRing ` R ) -> ( ( od ` ( R |`s A ) ) ` ( 1r ` ( R |`s A ) ) ) = ( ( od ` R ) ` ( 1r ` R ) ) )
12 eqid 
 |-  ( 1r ` ( R |`s A ) ) = ( 1r ` ( R |`s A ) )
13 eqid 
 |-  ( chr ` ( R |`s A ) ) = ( chr ` ( R |`s A ) )
14 6 12 13 chrval 
 |-  ( ( od ` ( R |`s A ) ) ` ( 1r ` ( R |`s A ) ) ) = ( chr ` ( R |`s A ) )
15 eqid 
 |-  ( chr ` R ) = ( chr ` R )
16 5 2 15 chrval 
 |-  ( ( od ` R ) ` ( 1r ` R ) ) = ( chr ` R )
17 11 14 16 3eqtr3g 
 |-  ( A e. ( SubRing ` R ) -> ( chr ` ( R |`s A ) ) = ( chr ` R ) )