Metamath Proof Explorer


Theorem isfieldidl

Description: Determine if a ring is a field based on its ideals. (Contributed by Jeff Madsen, 10-Jun-2010) (Revised by AV, 30-Jun-2026)

Ref Expression
Hypotheses isfieldidl.b ⊢ B = Base R
isfieldidl.0 ⊢ 0 ˙ = 0 R
isfieldidl.i ⊢ I = LIdeal ⁡ R
isfieldidl.1 ⊢ 1 ˙ = 1 R
Assertion isfieldidl ⊢ R ∈ Field ↔ R ∈ CRing ∧ 0 ˙ ≠ 1 ˙ ∧ I = 0 ˙ B

Proof

Step Hyp Ref Expression
1 isfieldidl.b ⊢ B = Base R
2 isfieldidl.0 ⊢ 0 ˙ = 0 R
3 isfieldidl.i ⊢ I = LIdeal ⁡ R
4 isfieldidl.1 ⊢ 1 ˙ = 1 R
5 isfld ⊢ R ∈ Field ↔ R ∈ DivRing ∧ R ∈ CRing
6 1 2 3 drngidl ⊢ R ∈ NzRing → R ∈ DivRing ↔ I = 0 ˙ B
7 6 adantr ⊢ R ∈ NzRing ∧ R ∈ CRing → R ∈ DivRing ↔ I = 0 ˙ B
8 4 2 nzrnz ⊢ R ∈ NzRing → 1 ˙ ≠ 0 ˙
9 8 necomd ⊢ R ∈ NzRing → 0 ˙ ≠ 1 ˙
10 9 a1d ⊢ R ∈ NzRing → I = 0 ˙ B → 0 ˙ ≠ 1 ˙
11 10 adantr ⊢ R ∈ NzRing ∧ R ∈ CRing → I = 0 ˙ B → 0 ˙ ≠ 1 ˙
12 11 pm4.71rd ⊢ R ∈ NzRing ∧ R ∈ CRing → I = 0 ˙ B ↔ 0 ˙ ≠ 1 ˙ ∧ I = 0 ˙ B
13 7 12 bitrd ⊢ R ∈ NzRing ∧ R ∈ CRing → R ∈ DivRing ↔ 0 ˙ ≠ 1 ˙ ∧ I = 0 ˙ B
14 drngnzr ⊢ R ∈ DivRing → R ∈ NzRing
15 14 con3i ⊢ ¬ R ∈ NzRing → ¬ R ∈ DivRing
16 15 adantr ⊢ ¬ R ∈ NzRing ∧ R ∈ CRing → ¬ R ∈ DivRing
17 ianor ⊢ ¬ R ∈ Ring ∧ 1 ˙ ≠ 0 ˙ ↔ ¬ R ∈ Ring ∨ ¬ 1 ˙ ≠ 0 ˙
18 4 2 isnzr ⊢ R ∈ NzRing ↔ R ∈ Ring ∧ 1 ˙ ≠ 0 ˙
19 17 18 xchnxbir ⊢ ¬ R ∈ NzRing ↔ ¬ R ∈ Ring ∨ ¬ 1 ˙ ≠ 0 ˙
20 pm2.24 ⊢ R ∈ Ring → ¬ R ∈ Ring → ¬ 0 ˙ ≠ 1 ˙ ∨ ¬ I = 0 ˙ B
21 crngring ⊢ R ∈ CRing → R ∈ Ring
22 20 21 syl11 ⊢ ¬ R ∈ Ring → R ∈ CRing → ¬ 0 ˙ ≠ 1 ˙ ∨ ¬ I = 0 ˙ B
23 id ⊢ 0 ˙ ≠ 1 ˙ → 0 ˙ ≠ 1 ˙
24 23 necomd ⊢ 0 ˙ ≠ 1 ˙ → 1 ˙ ≠ 0 ˙
25 24 con3i ⊢ ¬ 1 ˙ ≠ 0 ˙ → ¬ 0 ˙ ≠ 1 ˙
26 25 orcd ⊢ ¬ 1 ˙ ≠ 0 ˙ → ¬ 0 ˙ ≠ 1 ˙ ∨ ¬ I = 0 ˙ B
27 26 a1d ⊢ ¬ 1 ˙ ≠ 0 ˙ → R ∈ CRing → ¬ 0 ˙ ≠ 1 ˙ ∨ ¬ I = 0 ˙ B
28 22 27 jaoi ⊢ ¬ R ∈ Ring ∨ ¬ 1 ˙ ≠ 0 ˙ → R ∈ CRing → ¬ 0 ˙ ≠ 1 ˙ ∨ ¬ I = 0 ˙ B
29 19 28 sylbi ⊢ ¬ R ∈ NzRing → R ∈ CRing → ¬ 0 ˙ ≠ 1 ˙ ∨ ¬ I = 0 ˙ B
30 29 imp ⊢ ¬ R ∈ NzRing ∧ R ∈ CRing → ¬ 0 ˙ ≠ 1 ˙ ∨ ¬ I = 0 ˙ B
31 ianor ⊢ ¬ 0 ˙ ≠ 1 ˙ ∧ I = 0 ˙ B ↔ ¬ 0 ˙ ≠ 1 ˙ ∨ ¬ I = 0 ˙ B
32 30 31 sylibr ⊢ ¬ R ∈ NzRing ∧ R ∈ CRing → ¬ 0 ˙ ≠ 1 ˙ ∧ I = 0 ˙ B
33 16 32 2falsed ⊢ ¬ R ∈ NzRing ∧ R ∈ CRing → R ∈ DivRing ↔ 0 ˙ ≠ 1 ˙ ∧ I = 0 ˙ B
34 13 33 pm2.61ian ⊢ R ∈ CRing → R ∈ DivRing ↔ 0 ˙ ≠ 1 ˙ ∧ I = 0 ˙ B
35 34 pm5.32i ⊢ R ∈ CRing ∧ R ∈ DivRing ↔ R ∈ CRing ∧ 0 ˙ ≠ 1 ˙ ∧ I = 0 ˙ B
36 ancom ⊢ R ∈ DivRing ∧ R ∈ CRing ↔ R ∈ CRing ∧ R ∈ DivRing
37 3anass ⊢ R ∈ CRing ∧ 0 ˙ ≠ 1 ˙ ∧ I = 0 ˙ B ↔ R ∈ CRing ∧ 0 ˙ ≠ 1 ˙ ∧ I = 0 ˙ B
38 35 36 37 3bitr4i ⊢ R ∈ DivRing ∧ R ∈ CRing ↔ R ∈ CRing ∧ 0 ˙ ≠ 1 ˙ ∧ I = 0 ˙ B
39 5 38 bitri ⊢ R ∈ Field ↔ R ∈ CRing ∧ 0 ˙ ≠ 1 ˙ ∧ I = 0 ˙ B