Metamath Proof Explorer


Theorem ascl0

Description: The scalar 0 embedded into a left module corresponds to the 0 of the left module if the left module is also a ring. (Contributed by AV, 31-Jul-2019)

Ref Expression
Hypotheses ascl0.a
|- A = ( algSc ` W )
ascl0.f
|- F = ( Scalar ` W )
ascl0.l
|- ( ph -> W e. LMod )
ascl0.r
|- ( ph -> W e. Ring )
Assertion ascl0
|- ( ph -> ( A ` ( 0g ` F ) ) = ( 0g ` W ) )

Proof

Step Hyp Ref Expression
1 ascl0.a
 |-  A = ( algSc ` W )
2 ascl0.f
 |-  F = ( Scalar ` W )
3 ascl0.l
 |-  ( ph -> W e. LMod )
4 ascl0.r
 |-  ( ph -> W e. Ring )
5 2 lmodfgrp
 |-  ( W e. LMod -> F e. Grp )
6 3 5 syl
 |-  ( ph -> F e. Grp )
7 eqid
 |-  ( Base ` F ) = ( Base ` F )
8 eqid
 |-  ( 0g ` F ) = ( 0g ` F )
9 7 8 grpidcl
 |-  ( F e. Grp -> ( 0g ` F ) e. ( Base ` F ) )
10 6 9 syl
 |-  ( ph -> ( 0g ` F ) e. ( Base ` F ) )
11 eqid
 |-  ( .s ` W ) = ( .s ` W )
12 eqid
 |-  ( 1r ` W ) = ( 1r ` W )
13 1 2 7 11 12 asclval
 |-  ( ( 0g ` F ) e. ( Base ` F ) -> ( A ` ( 0g ` F ) ) = ( ( 0g ` F ) ( .s ` W ) ( 1r ` W ) ) )
14 10 13 syl
 |-  ( ph -> ( A ` ( 0g ` F ) ) = ( ( 0g ` F ) ( .s ` W ) ( 1r ` W ) ) )
15 eqid
 |-  ( Base ` W ) = ( Base ` W )
16 15 12 ringidcl
 |-  ( W e. Ring -> ( 1r ` W ) e. ( Base ` W ) )
17 4 16 syl
 |-  ( ph -> ( 1r ` W ) e. ( Base ` W ) )
18 eqid
 |-  ( 0g ` W ) = ( 0g ` W )
19 15 2 11 8 18 lmod0vs
 |-  ( ( W e. LMod /\ ( 1r ` W ) e. ( Base ` W ) ) -> ( ( 0g ` F ) ( .s ` W ) ( 1r ` W ) ) = ( 0g ` W ) )
20 3 17 19 syl2anc
 |-  ( ph -> ( ( 0g ` F ) ( .s ` W ) ( 1r ` W ) ) = ( 0g ` W ) )
21 14 20 eqtrd
 |-  ( ph -> ( A ` ( 0g ` F ) ) = ( 0g ` W ) )