Metamath Proof Explorer


Theorem sdrgdvcl

Description: A sub-division-ring is closed under the ring division operation. (Contributed by Thierry Arnoux, 15-Jan-2025)

Ref Expression
Hypotheses sdrgdvcl.i ⊢ × ˙ = / r ⁡ R
sdrgdvcl.0 ⊢ 0 ˙ = 0 R
sdrgdvcl.a ⊢ φ → A ∈ SubDRing ⁡ R
sdrgdvcl.x ⊢ φ → X ∈ A
sdrgdvcl.y ⊢ φ → Y ∈ A
sdrgdvcl.1 ⊢ φ → Y ≠ 0 ˙
Assertion sdrgdvcl ⊢ φ → X × ˙ Y ∈ A

Proof

Step Hyp Ref Expression
1 sdrgdvcl.i ⊢ × ˙ = / r ⁡ R
2 sdrgdvcl.0 ⊢ 0 ˙ = 0 R
3 sdrgdvcl.a ⊢ φ → A ∈ SubDRing ⁡ R
4 sdrgdvcl.x ⊢ φ → X ∈ A
5 sdrgdvcl.y ⊢ φ → Y ∈ A
6 sdrgdvcl.1 ⊢ φ → Y ≠ 0 ˙
7 issdrg ⊢ A ∈ SubDRing ⁡ R ↔ R ∈ DivRing ∧ A ∈ SubRing ⁡ R ∧ R ↾ 𝑠 A ∈ DivRing
8 3 7 sylib ⊢ φ → R ∈ DivRing ∧ A ∈ SubRing ⁡ R ∧ R ↾ 𝑠 A ∈ DivRing
9 8 simp3d ⊢ φ → R ↾ 𝑠 A ∈ DivRing
10 9 drngringd ⊢ φ → R ↾ 𝑠 A ∈ Ring
11 8 simp2d ⊢ φ → A ∈ SubRing ⁡ R
12 eqid ⊢ R ↾ 𝑠 A = R ↾ 𝑠 A
13 12 subrgbas ⊢ A ∈ SubRing ⁡ R → A = Base R ↾ 𝑠 A
14 11 13 syl ⊢ φ → A = Base R ↾ 𝑠 A
15 4 14 eleqtrd ⊢ φ → X ∈ Base R ↾ 𝑠 A
16 5 14 eleqtrd ⊢ φ → Y ∈ Base R ↾ 𝑠 A
17 12 2 subrg0 ⊢ A ∈ SubRing ⁡ R → 0 ˙ = 0 R ↾ 𝑠 A
18 11 17 syl ⊢ φ → 0 ˙ = 0 R ↾ 𝑠 A
19 6 18 neeqtrd ⊢ φ → Y ≠ 0 R ↾ 𝑠 A
20 eqid ⊢ Base R ↾ 𝑠 A = Base R ↾ 𝑠 A
21 eqid ⊢ Unit ⁡ R ↾ 𝑠 A = Unit ⁡ R ↾ 𝑠 A
22 eqid ⊢ 0 R ↾ 𝑠 A = 0 R ↾ 𝑠 A
23 20 21 22 drngunit ⊢ R ↾ 𝑠 A ∈ DivRing → Y ∈ Unit ⁡ R ↾ 𝑠 A ↔ Y ∈ Base R ↾ 𝑠 A ∧ Y ≠ 0 R ↾ 𝑠 A
24 23 biimpar ⊢ R ↾ 𝑠 A ∈ DivRing ∧ Y ∈ Base R ↾ 𝑠 A ∧ Y ≠ 0 R ↾ 𝑠 A → Y ∈ Unit ⁡ R ↾ 𝑠 A
25 9 16 19 24 syl12anc ⊢ φ → Y ∈ Unit ⁡ R ↾ 𝑠 A
26 eqid ⊢ / r ⁡ R ↾ 𝑠 A = / r ⁡ R ↾ 𝑠 A
27 20 21 26 dvrcl ⊢ R ↾ 𝑠 A ∈ Ring ∧ X ∈ Base R ↾ 𝑠 A ∧ Y ∈ Unit ⁡ R ↾ 𝑠 A → X / r ⁡ R ↾ 𝑠 A Y ∈ Base R ↾ 𝑠 A
28 10 15 25 27 syl3anc ⊢ φ → X / r ⁡ R ↾ 𝑠 A Y ∈ Base R ↾ 𝑠 A
29 12 1 21 26 subrgdv ⊢ A ∈ SubRing ⁡ R ∧ X ∈ A ∧ Y ∈ Unit ⁡ R ↾ 𝑠 A → X × ˙ Y = X / r ⁡ R ↾ 𝑠 A Y
30 11 4 25 29 syl3anc ⊢ φ → X × ˙ Y = X / r ⁡ R ↾ 𝑠 A Y
31 28 30 14 3eltr4d ⊢ φ → X × ˙ Y ∈ A