Metamath Proof Explorer


Theorem assafld

Description: If an algebra A of finite degree over a division ring K is an integral domain, then it is a field. Corollary of Proposition 2. in Chapter 5. of BourbakiAlg2 p. 113. (Contributed by Thierry Arnoux, 3-Aug-2025)

Ref Expression
Hypotheses assafld.k ⊢ K = Scalar ⁡ A
assafld.a ⊢ φ → A ∈ AssAlg
assafld.1 ⊢ φ → A ∈ IDomn
assafld.2 ⊢ φ → K ∈ DivRing
assafld.3 ⊢ φ → dim ⁡ A ∈ ℕ 0
Assertion assafld ⊢ φ → A ∈ Field

Proof

Step Hyp Ref Expression
1 assafld.k ⊢ K = Scalar ⁡ A
2 assafld.a ⊢ φ → A ∈ AssAlg
3 assafld.1 ⊢ φ → A ∈ IDomn
4 assafld.2 ⊢ φ → K ∈ DivRing
5 assafld.3 ⊢ φ → dim ⁡ A ∈ ℕ 0
6 3 idomringd ⊢ φ → A ∈ Ring
7 eqid ⊢ Base A = Base A
8 eqid ⊢ Unit ⁡ A = Unit ⁡ A
9 7 8 unitss ⊢ Unit ⁡ A ⊆ Base A
10 9 a1i ⊢ φ → Unit ⁡ A ⊆ Base A
11 eqid ⊢ 0 A = 0 A
12 3 idomdomd ⊢ φ → A ∈ Domn
13 domnnzr ⊢ A ∈ Domn → A ∈ NzRing
14 12 13 syl ⊢ φ → A ∈ NzRing
15 14 adantr ⊢ φ ∧ 0 A ∈ Unit ⁡ A → A ∈ NzRing
16 simpr ⊢ φ ∧ 0 A ∈ Unit ⁡ A → 0 A ∈ Unit ⁡ A
17 8 11 15 16 unitnz ⊢ φ ∧ 0 A ∈ Unit ⁡ A → 0 A ≠ 0 A
18 neirr ⊢ ¬ 0 A ≠ 0 A
19 18 a1i ⊢ φ ∧ 0 A ∈ Unit ⁡ A → ¬ 0 A ≠ 0 A
20 17 19 pm2.65da ⊢ φ → ¬ 0 A ∈ Unit ⁡ A
21 ssdifsn ⊢ Unit ⁡ A ⊆ Base A ∖ 0 A ↔ Unit ⁡ A ⊆ Base A ∧ ¬ 0 A ∈ Unit ⁡ A
22 10 20 21 sylanbrc ⊢ φ → Unit ⁡ A ⊆ Base A ∖ 0 A
23 eqid ⊢ RLReg ⁡ A = RLReg ⁡ A
24 2 adantr ⊢ φ ∧ x ∈ Base A ∖ 0 A → A ∈ AssAlg
25 4 adantr ⊢ φ ∧ x ∈ Base A ∖ 0 A → K ∈ DivRing
26 5 adantr ⊢ φ ∧ x ∈ Base A ∖ 0 A → dim ⁡ A ∈ ℕ 0
27 12 adantr ⊢ φ ∧ x ∈ Base A ∖ 0 A → A ∈ Domn
28 simpr ⊢ φ ∧ x ∈ Base A ∖ 0 A → x ∈ Base A ∖ 0 A
29 28 eldifad ⊢ φ ∧ x ∈ Base A ∖ 0 A → x ∈ Base A
30 eldifsni ⊢ x ∈ Base A ∖ 0 A → x ≠ 0 A
31 28 30 syl ⊢ φ ∧ x ∈ Base A ∖ 0 A → x ≠ 0 A
32 7 23 11 domnrrg ⊢ A ∈ Domn ∧ x ∈ Base A ∧ x ≠ 0 A → x ∈ RLReg ⁡ A
33 27 29 31 32 syl3anc ⊢ φ ∧ x ∈ Base A ∖ 0 A → x ∈ RLReg ⁡ A
34 23 8 1 24 25 26 33 assarrginv ⊢ φ ∧ x ∈ Base A ∖ 0 A → x ∈ Unit ⁡ A
35 22 34 eqelssd ⊢ φ → Unit ⁡ A = Base A ∖ 0 A
36 7 8 11 isdrng ⊢ A ∈ DivRing ↔ A ∈ Ring ∧ Unit ⁡ A = Base A ∖ 0 A
37 6 35 36 sylanbrc ⊢ φ → A ∈ DivRing
38 3 idomcringd ⊢ φ → A ∈ CRing
39 isfld ⊢ A ∈ Field ↔ A ∈ DivRing ∧ A ∈ CRing
40 37 38 39 sylanbrc ⊢ φ → A ∈ Field