Metamath Proof Explorer


Theorem minplyelirng

Description: If the minimal polynomial F of an element X of a field R has nonnegative degree, then X is integral. (Contributed by Thierry Arnoux, 26-Oct-2025)

Ref Expression
Hypotheses minplyelirng.b ⊢ B = Base R
minplyelirng.m ⊢ M = R minPoly S
minplyelirng.d ⊢ D = deg 1 ⁡ R ↾ 𝑠 S
minplyelirng.r ⊢ φ → R ∈ Field
minplyelirng.s ⊢ φ → S ∈ SubDRing ⁡ R
minplyelirng.a ⊢ φ → A ∈ B
minplyelirng.1 ⊢ φ → D ⁡ M ⁡ A ∈ ℕ 0
Assertion minplyelirng ⊢ φ → A ∈ R IntgRing S

Proof

Step Hyp Ref Expression
1 minplyelirng.b ⊢ B = Base R
2 minplyelirng.m ⊢ M = R minPoly S
3 minplyelirng.d ⊢ D = deg 1 ⁡ R ↾ 𝑠 S
4 minplyelirng.r ⊢ φ → R ∈ Field
5 minplyelirng.s ⊢ φ → S ∈ SubDRing ⁡ R
6 minplyelirng.a ⊢ φ → A ∈ B
7 minplyelirng.1 ⊢ φ → D ⁡ M ⁡ A ∈ ℕ 0
8 fveq2 ⊢ m = M ⁡ A → R evalSub 1 S ⁡ m = R evalSub 1 S ⁡ M ⁡ A
9 8 fveq1d ⊢ m = M ⁡ A → R evalSub 1 S ⁡ m ⁡ A = R evalSub 1 S ⁡ M ⁡ A ⁡ A
10 9 eqeq1d ⊢ m = M ⁡ A → R evalSub 1 S ⁡ m ⁡ A = 0 R ↔ R evalSub 1 S ⁡ M ⁡ A ⁡ A = 0 R
11 eqid ⊢ 0 Poly 1 ⁡ R = 0 Poly 1 ⁡ R
12 sdrgsubrg ⊢ S ∈ SubDRing ⁡ R → S ∈ SubRing ⁡ R
13 5 12 syl ⊢ φ → S ∈ SubRing ⁡ R
14 eqid ⊢ R ↾ 𝑠 S = R ↾ 𝑠 S
15 14 subrgring ⊢ S ∈ SubRing ⁡ R → R ↾ 𝑠 S ∈ Ring
16 13 15 syl ⊢ φ → R ↾ 𝑠 S ∈ Ring
17 eqid ⊢ R evalSub 1 S = R evalSub 1 S
18 eqid ⊢ Poly 1 ⁡ R ↾ 𝑠 S = Poly 1 ⁡ R ↾ 𝑠 S
19 eqid ⊢ 0 R = 0 R
20 eqid ⊢ q ∈ dom ⁡ R evalSub 1 S | R evalSub 1 S ⁡ q ⁡ A = 0 R = q ∈ dom ⁡ R evalSub 1 S | R evalSub 1 S ⁡ q ⁡ A = 0 R
21 eqid ⊢ RSpan ⁡ Poly 1 ⁡ R ↾ 𝑠 S = RSpan ⁡ Poly 1 ⁡ R ↾ 𝑠 S
22 eqid ⊢ idlGen 1p ⁡ R ↾ 𝑠 S = idlGen 1p ⁡ R ↾ 𝑠 S
23 17 18 1 4 5 6 19 20 21 22 2 minplycl ⊢ φ → M ⁡ A ∈ Base Poly 1 ⁡ R ↾ 𝑠 S
24 eqid ⊢ 0 Poly 1 ⁡ R ↾ 𝑠 S = 0 Poly 1 ⁡ R ↾ 𝑠 S
25 eqid ⊢ Base Poly 1 ⁡ R ↾ 𝑠 S = Base Poly 1 ⁡ R ↾ 𝑠 S
26 3 18 24 25 deg1nn0clb ⊢ R ↾ 𝑠 S ∈ Ring ∧ M ⁡ A ∈ Base Poly 1 ⁡ R ↾ 𝑠 S → M ⁡ A ≠ 0 Poly 1 ⁡ R ↾ 𝑠 S ↔ D ⁡ M ⁡ A ∈ ℕ 0
27 26 biimpar ⊢ R ↾ 𝑠 S ∈ Ring ∧ M ⁡ A ∈ Base Poly 1 ⁡ R ↾ 𝑠 S ∧ D ⁡ M ⁡ A ∈ ℕ 0 → M ⁡ A ≠ 0 Poly 1 ⁡ R ↾ 𝑠 S
28 16 23 7 27 syl21anc ⊢ φ → M ⁡ A ≠ 0 Poly 1 ⁡ R ↾ 𝑠 S
29 eqid ⊢ Poly 1 ⁡ R = Poly 1 ⁡ R
30 29 14 18 25 13 11 ressply10g ⊢ φ → 0 Poly 1 ⁡ R = 0 Poly 1 ⁡ R ↾ 𝑠 S
31 28 30 neeqtrrd ⊢ φ → M ⁡ A ≠ 0 Poly 1 ⁡ R
32 eqid ⊢ Monic 1p ⁡ R ↾ 𝑠 S = Monic 1p ⁡ R ↾ 𝑠 S
33 1 11 4 5 2 6 31 32 minplynzm1p ⊢ φ → M ⁡ A ∈ Monic 1p ⁡ R ↾ 𝑠 S
34 17 18 1 4 5 6 19 2 minplyann ⊢ φ → R evalSub 1 S ⁡ M ⁡ A ⁡ A = 0 R
35 10 33 34 rspcedvdw ⊢ φ → ∃ m ∈ Monic 1p ⁡ R ↾ 𝑠 S R evalSub 1 S ⁡ m ⁡ A = 0 R
36 4 fldcrngd ⊢ φ → R ∈ CRing
37 17 14 1 19 36 13 elirng ⊢ φ → A ∈ R IntgRing S ↔ A ∈ B ∧ ∃ m ∈ Monic 1p ⁡ R ↾ 𝑠 S R evalSub 1 S ⁡ m ⁡ A = 0 R
38 6 35 37 mpbir2and ⊢ φ → A ∈ R IntgRing S