Metamath Proof Explorer


Theorem ressascl

Description: The lifting of scalars is invariant between subalgebras and superalgebras. (Contributed by Mario Carneiro, 9-Mar-2015)

Ref Expression
Hypotheses ressascl.a ⊢ A = algSc ⁡ W
ressascl.x ⊢ X = W ↾ 𝑠 S
Assertion ressascl ⊢ S ∈ SubRing ⁡ W → A = algSc ⁡ X

Proof

Step Hyp Ref Expression
1 ressascl.a ⊢ A = algSc ⁡ W
2 ressascl.x ⊢ X = W ↾ 𝑠 S
3 eqid ⊢ Scalar ⁡ W = Scalar ⁡ W
4 2 3 resssca ⊢ S ∈ SubRing ⁡ W → Scalar ⁡ W = Scalar ⁡ X
5 4 fveq2d ⊢ S ∈ SubRing ⁡ W → Base Scalar ⁡ W = Base Scalar ⁡ X
6 eqid ⊢ ⋅ W = ⋅ W
7 2 6 ressvsca ⊢ S ∈ SubRing ⁡ W → ⋅ W = ⋅ X
8 eqidd ⊢ S ∈ SubRing ⁡ W → x = x
9 eqid ⊢ 1 W = 1 W
10 2 9 subrg1 ⊢ S ∈ SubRing ⁡ W → 1 W = 1 X
11 7 8 10 oveq123d ⊢ S ∈ SubRing ⁡ W → x ⋅ W 1 W = x ⋅ X 1 X
12 5 11 mpteq12dv ⊢ S ∈ SubRing ⁡ W → x ∈ Base Scalar ⁡ W ⟼ x ⋅ W 1 W = x ∈ Base Scalar ⁡ X ⟼ x ⋅ X 1 X
13 eqid ⊢ Base Scalar ⁡ W = Base Scalar ⁡ W
14 1 3 13 6 9 asclfval ⊢ A = x ∈ Base Scalar ⁡ W ⟼ x ⋅ W 1 W
15 eqid ⊢ algSc ⁡ X = algSc ⁡ X
16 eqid ⊢ Scalar ⁡ X = Scalar ⁡ X
17 eqid ⊢ Base Scalar ⁡ X = Base Scalar ⁡ X
18 eqid ⊢ ⋅ X = ⋅ X
19 eqid ⊢ 1 X = 1 X
20 15 16 17 18 19 asclfval ⊢ algSc ⁡ X = x ∈ Base Scalar ⁡ X ⟼ x ⋅ X 1 X
21 12 14 20 3eqtr4g ⊢ S ∈ SubRing ⁡ W → A = algSc ⁡ X