Metamath Proof Explorer


Theorem subrg1ascl

Description: The scalar injection function in a subring algebra is the same up to a restriction to the subring. (Contributed by Mario Carneiro, 4-Jul-2015)

Ref Expression
Hypotheses subrg1ascl.p 𝑃 = ( Poly1𝑅 )
subrg1ascl.a 𝐴 = ( algSc ‘ 𝑃 )
subrg1ascl.h 𝐻 = ( 𝑅s 𝑇 )
subrg1ascl.u 𝑈 = ( Poly1𝐻 )
subrg1ascl.r ( 𝜑𝑇 ∈ ( SubRing ‘ 𝑅 ) )
subrg1ascl.c 𝐶 = ( algSc ‘ 𝑈 )
Assertion subrg1ascl ( 𝜑𝐶 = ( 𝐴𝑇 ) )

Proof

Step Hyp Ref Expression
1 subrg1ascl.p 𝑃 = ( Poly1𝑅 )
2 subrg1ascl.a 𝐴 = ( algSc ‘ 𝑃 )
3 subrg1ascl.h 𝐻 = ( 𝑅s 𝑇 )
4 subrg1ascl.u 𝑈 = ( Poly1𝐻 )
5 subrg1ascl.r ( 𝜑𝑇 ∈ ( SubRing ‘ 𝑅 ) )
6 subrg1ascl.c 𝐶 = ( algSc ‘ 𝑈 )
7 eqid ( 1o mPoly 𝑅 ) = ( 1o mPoly 𝑅 )
8 1 2 ply1ascl 𝐴 = ( algSc ‘ ( 1o mPoly 𝑅 ) )
9 eqid ( 1o mPoly 𝐻 ) = ( 1o mPoly 𝐻 )
10 1on 1o ∈ On
11 10 a1i ( 𝜑 → 1o ∈ On )
12 4 6 ply1ascl 𝐶 = ( algSc ‘ ( 1o mPoly 𝐻 ) )
13 7 8 3 9 11 5 12 subrgascl ( 𝜑𝐶 = ( 𝐴𝑇 ) )