Metamath Proof Explorer


Theorem isclmi

Description: Reverse direction of isclm . (Contributed by Mario Carneiro, 30-Oct-2015)

Ref Expression
Hypothesis clm0.f 𝐹 = ( Scalar ‘ 𝑊 )
Assertion isclmi ( ( 𝑊 ∈ LMod ∧ 𝐹 = ( ℂflds 𝐾 ) ∧ 𝐾 ∈ ( SubRing ‘ ℂfld ) ) → 𝑊 ∈ ℂMod )

Proof

Step Hyp Ref Expression
1 clm0.f 𝐹 = ( Scalar ‘ 𝑊 )
2 simp1 ( ( 𝑊 ∈ LMod ∧ 𝐹 = ( ℂflds 𝐾 ) ∧ 𝐾 ∈ ( SubRing ‘ ℂfld ) ) → 𝑊 ∈ LMod )
3 simp2 ( ( 𝑊 ∈ LMod ∧ 𝐹 = ( ℂflds 𝐾 ) ∧ 𝐾 ∈ ( SubRing ‘ ℂfld ) ) → 𝐹 = ( ℂflds 𝐾 ) )
4 eqid ( ℂflds 𝐾 ) = ( ℂflds 𝐾 )
5 4 subrgbas ( 𝐾 ∈ ( SubRing ‘ ℂfld ) → 𝐾 = ( Base ‘ ( ℂflds 𝐾 ) ) )
6 5 3ad2ant3 ( ( 𝑊 ∈ LMod ∧ 𝐹 = ( ℂflds 𝐾 ) ∧ 𝐾 ∈ ( SubRing ‘ ℂfld ) ) → 𝐾 = ( Base ‘ ( ℂflds 𝐾 ) ) )
7 3 fveq2d ( ( 𝑊 ∈ LMod ∧ 𝐹 = ( ℂflds 𝐾 ) ∧ 𝐾 ∈ ( SubRing ‘ ℂfld ) ) → ( Base ‘ 𝐹 ) = ( Base ‘ ( ℂflds 𝐾 ) ) )
8 6 7 eqtr4d ( ( 𝑊 ∈ LMod ∧ 𝐹 = ( ℂflds 𝐾 ) ∧ 𝐾 ∈ ( SubRing ‘ ℂfld ) ) → 𝐾 = ( Base ‘ 𝐹 ) )
9 8 oveq2d ( ( 𝑊 ∈ LMod ∧ 𝐹 = ( ℂflds 𝐾 ) ∧ 𝐾 ∈ ( SubRing ‘ ℂfld ) ) → ( ℂflds 𝐾 ) = ( ℂflds ( Base ‘ 𝐹 ) ) )
10 3 9 eqtrd ( ( 𝑊 ∈ LMod ∧ 𝐹 = ( ℂflds 𝐾 ) ∧ 𝐾 ∈ ( SubRing ‘ ℂfld ) ) → 𝐹 = ( ℂflds ( Base ‘ 𝐹 ) ) )
11 simp3 ( ( 𝑊 ∈ LMod ∧ 𝐹 = ( ℂflds 𝐾 ) ∧ 𝐾 ∈ ( SubRing ‘ ℂfld ) ) → 𝐾 ∈ ( SubRing ‘ ℂfld ) )
12 8 11 eqeltrrd ( ( 𝑊 ∈ LMod ∧ 𝐹 = ( ℂflds 𝐾 ) ∧ 𝐾 ∈ ( SubRing ‘ ℂfld ) ) → ( Base ‘ 𝐹 ) ∈ ( SubRing ‘ ℂfld ) )
13 eqid ( Base ‘ 𝐹 ) = ( Base ‘ 𝐹 )
14 1 13 isclm ( 𝑊 ∈ ℂMod ↔ ( 𝑊 ∈ LMod ∧ 𝐹 = ( ℂflds ( Base ‘ 𝐹 ) ) ∧ ( Base ‘ 𝐹 ) ∈ ( SubRing ‘ ℂfld ) ) )
15 2 10 12 14 syl3anbrc ( ( 𝑊 ∈ LMod ∧ 𝐹 = ( ℂflds 𝐾 ) ∧ 𝐾 ∈ ( SubRing ‘ ℂfld ) ) → 𝑊 ∈ ℂMod )