Metamath Proof Explorer


Theorem ply1scl1

Description: The one scalar is the unit polynomial. (Contributed by Stefan O'Rear, 1-Apr-2015) (Proof shortened by SN, 12-Mar-2025)

Ref Expression
Hypotheses ply1scl.p ⊢ P = Poly 1 ⁡ R
ply1scl.a ⊢ A = algSc ⁡ P
ply1scl1.o ⊢ 1 ˙ = 1 R
ply1scl1.n ⊢ N = 1 P
Assertion ply1scl1 ⊢ R ∈ Ring → A ⁡ 1 ˙ = N

Proof

Step Hyp Ref Expression
1 ply1scl.p ⊢ P = Poly 1 ⁡ R
2 ply1scl.a ⊢ A = algSc ⁡ P
3 ply1scl1.o ⊢ 1 ˙ = 1 R
4 ply1scl1.n ⊢ N = 1 P
5 1 ply1sca ⊢ R ∈ Ring → R = Scalar ⁡ P
6 5 fveq2d ⊢ R ∈ Ring → 1 R = 1 Scalar ⁡ P
7 3 6 eqtrid ⊢ R ∈ Ring → 1 ˙ = 1 Scalar ⁡ P
8 7 fveq2d ⊢ R ∈ Ring → A ⁡ 1 ˙ = A ⁡ 1 Scalar ⁡ P
9 eqid ⊢ Scalar ⁡ P = Scalar ⁡ P
10 1 ply1lmod ⊢ R ∈ Ring → P ∈ LMod
11 1 ply1ring ⊢ R ∈ Ring → P ∈ Ring
12 2 9 10 11 ascl1 ⊢ R ∈ Ring → A ⁡ 1 Scalar ⁡ P = 1 P
13 8 12 eqtrd ⊢ R ∈ Ring → A ⁡ 1 ˙ = 1 P
14 13 4 eqtr4di ⊢ R ∈ Ring → A ⁡ 1 ˙ = N