Metamath Proof Explorer


Theorem lpi0

Description: The zero ideal is always principal. (Contributed by Stefan O'Rear, 3-Jan-2015)

Ref Expression
Hypotheses lpival.p ⊢ P = LPIdeal ⁡ R
lpi0.z ⊢ 0 ˙ = 0 R
Assertion lpi0 ⊢ R ∈ Ring → 0 ˙ ∈ P

Proof

Step Hyp Ref Expression
1 lpival.p ⊢ P = LPIdeal ⁡ R
2 lpi0.z ⊢ 0 ˙ = 0 R
3 eqid ⊢ Base R = Base R
4 3 2 ring0cl ⊢ R ∈ Ring → 0 ˙ ∈ Base R
5 eqid ⊢ RSpan ⁡ R = RSpan ⁡ R
6 5 2 rsp0 ⊢ R ∈ Ring → RSpan ⁡ R ⁡ 0 ˙ = 0 ˙
7 6 eqcomd ⊢ R ∈ Ring → 0 ˙ = RSpan ⁡ R ⁡ 0 ˙
8 sneq ⊢ g = 0 ˙ → g = 0 ˙
9 8 fveq2d ⊢ g = 0 ˙ → RSpan ⁡ R ⁡ g = RSpan ⁡ R ⁡ 0 ˙
10 9 rspceeqv ⊢ 0 ˙ ∈ Base R ∧ 0 ˙ = RSpan ⁡ R ⁡ 0 ˙ → ∃ g ∈ Base R 0 ˙ = RSpan ⁡ R ⁡ g
11 4 7 10 syl2anc ⊢ R ∈ Ring → ∃ g ∈ Base R 0 ˙ = RSpan ⁡ R ⁡ g
12 1 5 3 islpidl ⊢ R ∈ Ring → 0 ˙ ∈ P ↔ ∃ g ∈ Base R 0 ˙ = RSpan ⁡ R ⁡ g
13 11 12 mpbird ⊢ R ∈ Ring → 0 ˙ ∈ P