Metamath Proof Explorer


Theorem sradrng

Description: Condition for a subring algebra to be a division ring. (Contributed by Thierry Arnoux, 29-Jul-2023)

Ref Expression
Hypotheses sradrng.1 ⊢ A = subringAlg ⁡ R ⁡ V
sradrng.2 ⊢ B = Base R
Assertion sradrng ⊢ R ∈ DivRing ∧ V ⊆ B → A ∈ DivRing

Proof

Step Hyp Ref Expression
1 sradrng.1 ⊢ A = subringAlg ⁡ R ⁡ V
2 sradrng.2 ⊢ B = Base R
3 drngring ⊢ R ∈ DivRing → R ∈ Ring
4 1 2 sraring ⊢ R ∈ Ring ∧ V ⊆ B → A ∈ Ring
5 3 4 sylan ⊢ R ∈ DivRing ∧ V ⊆ B → A ∈ Ring
6 eqid ⊢ Base R = Base R
7 eqid ⊢ Unit ⁡ R = Unit ⁡ R
8 eqid ⊢ 0 R = 0 R
9 6 7 8 isdrng ⊢ R ∈ DivRing ↔ R ∈ Ring ∧ Unit ⁡ R = Base R ∖ 0 R
10 9 simprbi ⊢ R ∈ DivRing → Unit ⁡ R = Base R ∖ 0 R
11 10 adantr ⊢ R ∈ DivRing ∧ V ⊆ B → Unit ⁡ R = Base R ∖ 0 R
12 eqidd ⊢ R ∈ DivRing ∧ V ⊆ B → Base R = Base R
13 1 a1i ⊢ R ∈ DivRing ∧ V ⊆ B → A = subringAlg ⁡ R ⁡ V
14 simpr ⊢ R ∈ DivRing ∧ V ⊆ B → V ⊆ B
15 14 2 sseqtrdi ⊢ R ∈ DivRing ∧ V ⊆ B → V ⊆ Base R
16 13 15 srabase ⊢ R ∈ DivRing ∧ V ⊆ B → Base R = Base A
17 13 15 sramulr ⊢ R ∈ DivRing ∧ V ⊆ B → ⋅ R = ⋅ A
18 17 oveqdr ⊢ R ∈ DivRing ∧ V ⊆ B ∧ x ∈ Base R ∧ y ∈ Base R → x ⋅ R y = x ⋅ A y
19 12 16 18 unitpropd ⊢ R ∈ DivRing ∧ V ⊆ B → Unit ⁡ R = Unit ⁡ A
20 eqidd ⊢ R ∈ DivRing ∧ V ⊆ B → 0 R = 0 R
21 13 20 15 sralmod0 ⊢ R ∈ DivRing ∧ V ⊆ B → 0 R = 0 A
22 21 sneqd ⊢ R ∈ DivRing ∧ V ⊆ B → 0 R = 0 A
23 16 22 difeq12d ⊢ R ∈ DivRing ∧ V ⊆ B → Base R ∖ 0 R = Base A ∖ 0 A
24 11 19 23 3eqtr3d ⊢ R ∈ DivRing ∧ V ⊆ B → Unit ⁡ A = Base A ∖ 0 A
25 eqid ⊢ Base A = Base A
26 eqid ⊢ Unit ⁡ A = Unit ⁡ A
27 eqid ⊢ 0 A = 0 A
28 25 26 27 isdrng ⊢ A ∈ DivRing ↔ A ∈ Ring ∧ Unit ⁡ A = Base A ∖ 0 A
29 5 24 28 sylanbrc ⊢ R ∈ DivRing ∧ V ⊆ B → A ∈ DivRing