Metamath Proof Explorer


Theorem mplascl

Description: Value of the scalar injection into the polynomial algebra. (Contributed by Stefan O'Rear, 9-Mar-2015)

Ref Expression
Hypotheses mplascl.p ⊢ P = I mPoly R
mplascl.d ⊢ D = f ∈ ℕ 0 I | f -1 ℕ ∈ Fin
mplascl.z ⊢ 0 ˙ = 0 R
mplascl.b ⊢ B = Base R
mplascl.a ⊢ A = algSc ⁡ P
mplascl.i ⊢ φ → I ∈ W
mplascl.r ⊢ φ → R ∈ Ring
mplascl.x ⊢ φ → X ∈ B
Assertion mplascl ⊢ φ → A ⁡ X = y ∈ D ⟼ if y = I × 0 X 0 ˙

Proof

Step Hyp Ref Expression
1 mplascl.p ⊢ P = I mPoly R
2 mplascl.d ⊢ D = f ∈ ℕ 0 I | f -1 ℕ ∈ Fin
3 mplascl.z ⊢ 0 ˙ = 0 R
4 mplascl.b ⊢ B = Base R
5 mplascl.a ⊢ A = algSc ⁡ P
6 mplascl.i ⊢ φ → I ∈ W
7 mplascl.r ⊢ φ → R ∈ Ring
8 mplascl.x ⊢ φ → X ∈ B
9 1 6 7 mplsca ⊢ φ → R = Scalar ⁡ P
10 9 fveq2d ⊢ φ → Base R = Base Scalar ⁡ P
11 4 10 eqtrid ⊢ φ → B = Base Scalar ⁡ P
12 8 11 eleqtrd ⊢ φ → X ∈ Base Scalar ⁡ P
13 eqid ⊢ Scalar ⁡ P = Scalar ⁡ P
14 eqid ⊢ Base Scalar ⁡ P = Base Scalar ⁡ P
15 eqid ⊢ ⋅ P = ⋅ P
16 eqid ⊢ 1 P = 1 P
17 5 13 14 15 16 asclval ⊢ X ∈ Base Scalar ⁡ P → A ⁡ X = X ⋅ P 1 P
18 12 17 syl ⊢ φ → A ⁡ X = X ⋅ P 1 P
19 eqid ⊢ 1 R = 1 R
20 1 2 3 19 16 6 7 mpl1 ⊢ φ → 1 P = y ∈ D ⟼ if y = I × 0 1 R 0 ˙
21 20 oveq2d ⊢ φ → X ⋅ P 1 P = X ⋅ P y ∈ D ⟼ if y = I × 0 1 R 0 ˙
22 2 psrbag0 ⊢ I ∈ W → I × 0 ∈ D
23 6 22 syl ⊢ φ → I × 0 ∈ D
24 1 15 2 19 3 4 6 7 23 8 mplmon2 ⊢ φ → X ⋅ P y ∈ D ⟼ if y = I × 0 1 R 0 ˙ = y ∈ D ⟼ if y = I × 0 X 0 ˙
25 18 21 24 3eqtrd ⊢ φ → A ⁡ X = y ∈ D ⟼ if y = I × 0 X 0 ˙