Metamath Proof Explorer


Theorem clmsub

Description: Subtraction in the scalar ring of a subcomplex module. (Contributed by Mario Carneiro, 16-Oct-2015)

Ref Expression
Hypotheses clm0.f 𝐹 = ( Scalar ‘ 𝑊 )
clmsub.k 𝐾 = ( Base ‘ 𝐹 )
Assertion clmsub ( ( 𝑊 ∈ ℂMod ∧ 𝐴𝐾𝐵𝐾 ) → ( 𝐴𝐵 ) = ( 𝐴 ( -g𝐹 ) 𝐵 ) )

Proof

Step Hyp Ref Expression
1 clm0.f 𝐹 = ( Scalar ‘ 𝑊 )
2 clmsub.k 𝐾 = ( Base ‘ 𝐹 )
3 1 2 clmsubrg ( 𝑊 ∈ ℂMod → 𝐾 ∈ ( SubRing ‘ ℂfld ) )
4 subrgsubg ( 𝐾 ∈ ( SubRing ‘ ℂfld ) → 𝐾 ∈ ( SubGrp ‘ ℂfld ) )
5 3 4 syl ( 𝑊 ∈ ℂMod → 𝐾 ∈ ( SubGrp ‘ ℂfld ) )
6 cnfldsub − = ( -g ‘ ℂfld )
7 eqid ( ℂflds 𝐾 ) = ( ℂflds 𝐾 )
8 eqid ( -g ‘ ( ℂflds 𝐾 ) ) = ( -g ‘ ( ℂflds 𝐾 ) )
9 6 7 8 subgsub ( ( 𝐾 ∈ ( SubGrp ‘ ℂfld ) ∧ 𝐴𝐾𝐵𝐾 ) → ( 𝐴𝐵 ) = ( 𝐴 ( -g ‘ ( ℂflds 𝐾 ) ) 𝐵 ) )
10 5 9 syl3an1 ( ( 𝑊 ∈ ℂMod ∧ 𝐴𝐾𝐵𝐾 ) → ( 𝐴𝐵 ) = ( 𝐴 ( -g ‘ ( ℂflds 𝐾 ) ) 𝐵 ) )
11 1 2 clmsca ( 𝑊 ∈ ℂMod → 𝐹 = ( ℂflds 𝐾 ) )
12 11 fveq2d ( 𝑊 ∈ ℂMod → ( -g𝐹 ) = ( -g ‘ ( ℂflds 𝐾 ) ) )
13 12 3ad2ant1 ( ( 𝑊 ∈ ℂMod ∧ 𝐴𝐾𝐵𝐾 ) → ( -g𝐹 ) = ( -g ‘ ( ℂflds 𝐾 ) ) )
14 13 oveqd ( ( 𝑊 ∈ ℂMod ∧ 𝐴𝐾𝐵𝐾 ) → ( 𝐴 ( -g𝐹 ) 𝐵 ) = ( 𝐴 ( -g ‘ ( ℂflds 𝐾 ) ) 𝐵 ) )
15 10 14 eqtr4d ( ( 𝑊 ∈ ℂMod ∧ 𝐴𝐾𝐵𝐾 ) → ( 𝐴𝐵 ) = ( 𝐴 ( -g𝐹 ) 𝐵 ) )