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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 = ( 0g ‘ 𝑅 )
Assertion lpi0 ( 𝑅 ∈ Ring → { 0 } ∈ 𝑃 )

Proof

Step Hyp Ref Expression
1 lpival.p ⊢ 𝑃 = ( LPIdeal ‘ 𝑅 )
2 lpi0.z ⊢ 0 = ( 0g ‘ 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 )