Metamath Proof Explorer


Theorem ply1sclrmsm

Description: The ring multiplication of a polynomial with a scalar polynomial is equal to the scalar multiplication of the polynomial with the corresponding scalar. (Contributed by AV, 14-Aug-2019)

Ref Expression
Hypotheses ply1sclrmsm.k ⊢ K = Base R
ply1sclrmsm.p ⊢ P = Poly 1 ⁡ R
ply1sclrmsm.b ⊢ E = Base P
ply1sclrmsm.x ⊢ X = var 1 ⁡ R
ply1sclrmsm.s ⊢ · ˙ = ⋅ P
ply1sclrmsm.m ⊢ × ˙ = ⋅ P
ply1sclrmsm.n ⊢ N = mulGrp P
ply1sclrmsm.e ⊢ × ˙ = ⋅ N
ply1sclrmsm.a ⊢ A = algSc ⁡ P
Assertion ply1sclrmsm ⊢ R ∈ Ring ∧ F ∈ K ∧ Z ∈ E → A ⁡ F × ˙ Z = F · ˙ Z

Proof

Step Hyp Ref Expression
1 ply1sclrmsm.k ⊢ K = Base R
2 ply1sclrmsm.p ⊢ P = Poly 1 ⁡ R
3 ply1sclrmsm.b ⊢ E = Base P
4 ply1sclrmsm.x ⊢ X = var 1 ⁡ R
5 ply1sclrmsm.s ⊢ · ˙ = ⋅ P
6 ply1sclrmsm.m ⊢ × ˙ = ⋅ P
7 ply1sclrmsm.n ⊢ N = mulGrp P
8 ply1sclrmsm.e ⊢ × ˙ = ⋅ N
9 ply1sclrmsm.a ⊢ A = algSc ⁡ P
10 2 ply1sca ⊢ R ∈ Ring → R = Scalar ⁡ P
11 10 fveq2d ⊢ R ∈ Ring → Base R = Base Scalar ⁡ P
12 1 11 eqtrid ⊢ R ∈ Ring → K = Base Scalar ⁡ P
13 12 eleq2d ⊢ R ∈ Ring → F ∈ K ↔ F ∈ Base Scalar ⁡ P
14 13 biimpa ⊢ R ∈ Ring ∧ F ∈ K → F ∈ Base Scalar ⁡ P
15 eqid ⊢ Scalar ⁡ P = Scalar ⁡ P
16 eqid ⊢ Base Scalar ⁡ P = Base Scalar ⁡ P
17 eqid ⊢ 1 P = 1 P
18 9 15 16 5 17 asclval ⊢ F ∈ Base Scalar ⁡ P → A ⁡ F = F · ˙ 1 P
19 14 18 syl ⊢ R ∈ Ring ∧ F ∈ K → A ⁡ F = F · ˙ 1 P
20 19 3adant3 ⊢ R ∈ Ring ∧ F ∈ K ∧ Z ∈ E → A ⁡ F = F · ˙ 1 P
21 20 oveq1d ⊢ R ∈ Ring ∧ F ∈ K ∧ Z ∈ E → A ⁡ F × ˙ Z = F · ˙ 1 P × ˙ Z
22 simp1 ⊢ R ∈ Ring ∧ F ∈ K ∧ Z ∈ E → R ∈ Ring
23 1 eleq2i ⊢ F ∈ K ↔ F ∈ Base R
24 23 biimpi ⊢ F ∈ K → F ∈ Base R
25 24 3ad2ant2 ⊢ R ∈ Ring ∧ F ∈ K ∧ Z ∈ E → F ∈ Base R
26 2 ply1ring ⊢ R ∈ Ring → P ∈ Ring
27 3 17 ringidcl ⊢ P ∈ Ring → 1 P ∈ E
28 26 27 syl ⊢ R ∈ Ring → 1 P ∈ E
29 28 3ad2ant1 ⊢ R ∈ Ring ∧ F ∈ K ∧ Z ∈ E → 1 P ∈ E
30 simp3 ⊢ R ∈ Ring ∧ F ∈ K ∧ Z ∈ E → Z ∈ E
31 eqid ⊢ Base R = Base R
32 2 6 3 31 5 ply1ass23l ⊢ R ∈ Ring ∧ F ∈ Base R ∧ 1 P ∈ E ∧ Z ∈ E → F · ˙ 1 P × ˙ Z = F · ˙ 1 P × ˙ Z
33 22 25 29 30 32 syl13anc ⊢ R ∈ Ring ∧ F ∈ K ∧ Z ∈ E → F · ˙ 1 P × ˙ Z = F · ˙ 1 P × ˙ Z
34 3 6 17 ringlidm ⊢ P ∈ Ring ∧ Z ∈ E → 1 P × ˙ Z = Z
35 26 34 sylan ⊢ R ∈ Ring ∧ Z ∈ E → 1 P × ˙ Z = Z
36 35 3adant2 ⊢ R ∈ Ring ∧ F ∈ K ∧ Z ∈ E → 1 P × ˙ Z = Z
37 36 oveq2d ⊢ R ∈ Ring ∧ F ∈ K ∧ Z ∈ E → F · ˙ 1 P × ˙ Z = F · ˙ Z
38 21 33 37 3eqtrd ⊢ R ∈ Ring ∧ F ∈ K ∧ Z ∈ E → A ⁡ F × ˙ Z = F · ˙ Z