Metamath Proof Explorer


Theorem ascl1

Description: The scalar 1 embedded into a left module corresponds to the 1 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 ( 𝜑𝑊 ∈ LMod )
ascl0.r ( 𝜑𝑊 ∈ Ring )
Assertion ascl1 ( 𝜑 → ( 𝐴 ‘ ( 1r𝐹 ) ) = ( 1r𝑊 ) )

Proof

Step Hyp Ref Expression
1 ascl0.a 𝐴 = ( algSc ‘ 𝑊 )
2 ascl0.f 𝐹 = ( Scalar ‘ 𝑊 )
3 ascl0.l ( 𝜑𝑊 ∈ LMod )
4 ascl0.r ( 𝜑𝑊 ∈ Ring )
5 2 lmodring ( 𝑊 ∈ LMod → 𝐹 ∈ Ring )
6 eqid ( Base ‘ 𝐹 ) = ( Base ‘ 𝐹 )
7 eqid ( 1r𝐹 ) = ( 1r𝐹 )
8 6 7 ringidcl ( 𝐹 ∈ Ring → ( 1r𝐹 ) ∈ ( Base ‘ 𝐹 ) )
9 eqid ( ·𝑠𝑊 ) = ( ·𝑠𝑊 )
10 eqid ( 1r𝑊 ) = ( 1r𝑊 )
11 1 2 6 9 10 asclval ( ( 1r𝐹 ) ∈ ( Base ‘ 𝐹 ) → ( 𝐴 ‘ ( 1r𝐹 ) ) = ( ( 1r𝐹 ) ( ·𝑠𝑊 ) ( 1r𝑊 ) ) )
12 3 5 8 11 4syl ( 𝜑 → ( 𝐴 ‘ ( 1r𝐹 ) ) = ( ( 1r𝐹 ) ( ·𝑠𝑊 ) ( 1r𝑊 ) ) )
13 eqid ( Base ‘ 𝑊 ) = ( Base ‘ 𝑊 )
14 13 10 ringidcl ( 𝑊 ∈ Ring → ( 1r𝑊 ) ∈ ( Base ‘ 𝑊 ) )
15 4 14 syl ( 𝜑 → ( 1r𝑊 ) ∈ ( Base ‘ 𝑊 ) )
16 13 2 9 7 lmodvs1 ( ( 𝑊 ∈ LMod ∧ ( 1r𝑊 ) ∈ ( Base ‘ 𝑊 ) ) → ( ( 1r𝐹 ) ( ·𝑠𝑊 ) ( 1r𝑊 ) ) = ( 1r𝑊 ) )
17 3 15 16 syl2anc ( 𝜑 → ( ( 1r𝐹 ) ( ·𝑠𝑊 ) ( 1r𝑊 ) ) = ( 1r𝑊 ) )
18 12 17 eqtrd ( 𝜑 → ( 𝐴 ‘ ( 1r𝐹 ) ) = ( 1r𝑊 ) )