Metamath Proof Explorer


Theorem chpdmatlem0

Description: Lemma 0 for chpdmat . (Contributed by AV, 18-Aug-2019)

Ref Expression
Hypotheses chpdmat.c ⊢ C = N CharPlyMat R
chpdmat.p ⊢ P = Poly 1 ⁡ R
chpdmat.a ⊢ A = N Mat R
chpdmat.s ⊢ S = algSc ⁡ P
chpdmat.b ⊢ B = Base A
chpdmat.x ⊢ X = var 1 ⁡ R
chpdmat.0 ⊢ 0 ˙ = 0 R
chpdmat.g ⊢ G = mulGrp P
chpdmat.m ⊢ - ˙ = - P
chpdmatlem.q ⊢ Q = N Mat P
chpdmatlem.1 ⊢ 1 ˙ = 1 Q
chpdmatlem.m ⊢ · ˙ = ⋅ Q
Assertion chpdmatlem0 ⊢ N ∈ Fin ∧ R ∈ Ring → X · ˙ 1 ˙ ∈ Base Q

Proof

Step Hyp Ref Expression
1 chpdmat.c ⊢ C = N CharPlyMat R
2 chpdmat.p ⊢ P = Poly 1 ⁡ R
3 chpdmat.a ⊢ A = N Mat R
4 chpdmat.s ⊢ S = algSc ⁡ P
5 chpdmat.b ⊢ B = Base A
6 chpdmat.x ⊢ X = var 1 ⁡ R
7 chpdmat.0 ⊢ 0 ˙ = 0 R
8 chpdmat.g ⊢ G = mulGrp P
9 chpdmat.m ⊢ - ˙ = - P
10 chpdmatlem.q ⊢ Q = N Mat P
11 chpdmatlem.1 ⊢ 1 ˙ = 1 Q
12 chpdmatlem.m ⊢ · ˙ = ⋅ Q
13 2 10 pmatlmod ⊢ N ∈ Fin ∧ R ∈ Ring → Q ∈ LMod
14 eqid ⊢ Base P = Base P
15 6 2 14 vr1cl ⊢ R ∈ Ring → X ∈ Base P
16 15 adantl ⊢ N ∈ Fin ∧ R ∈ Ring → X ∈ Base P
17 2 ply1ring ⊢ R ∈ Ring → P ∈ Ring
18 10 matsca2 ⊢ N ∈ Fin ∧ P ∈ Ring → P = Scalar ⁡ Q
19 17 18 sylan2 ⊢ N ∈ Fin ∧ R ∈ Ring → P = Scalar ⁡ Q
20 19 eqcomd ⊢ N ∈ Fin ∧ R ∈ Ring → Scalar ⁡ Q = P
21 20 fveq2d ⊢ N ∈ Fin ∧ R ∈ Ring → Base Scalar ⁡ Q = Base P
22 16 21 eleqtrrd ⊢ N ∈ Fin ∧ R ∈ Ring → X ∈ Base Scalar ⁡ Q
23 2 10 pmatring ⊢ N ∈ Fin ∧ R ∈ Ring → Q ∈ Ring
24 eqid ⊢ Base Q = Base Q
25 24 11 ringidcl ⊢ Q ∈ Ring → 1 ˙ ∈ Base Q
26 23 25 syl ⊢ N ∈ Fin ∧ R ∈ Ring → 1 ˙ ∈ Base Q
27 eqid ⊢ Scalar ⁡ Q = Scalar ⁡ Q
28 eqid ⊢ Base Scalar ⁡ Q = Base Scalar ⁡ Q
29 24 27 12 28 lmodvscl ⊢ Q ∈ LMod ∧ X ∈ Base Scalar ⁡ Q ∧ 1 ˙ ∈ Base Q → X · ˙ 1 ˙ ∈ Base Q
30 13 22 26 29 syl3anc ⊢ N ∈ Fin ∧ R ∈ Ring → X · ˙ 1 ˙ ∈ Base Q