Metamath Proof Explorer


Theorem drngmxidl

Description: The zero ideal is the only ideal of a division ring. (Contributed by Thierry Arnoux, 16-Mar-2025)

Ref Expression
Hypothesis drngmxidl.1 0 = ( 0g𝑅 )
Assertion drngmxidl ( 𝑅 ∈ DivRing → ( MaxIdeal ‘ 𝑅 ) = { { 0 } } )

Proof

Step Hyp Ref Expression
1 drngmxidl.1 0 = ( 0g𝑅 )
2 drngring ( 𝑅 ∈ DivRing → 𝑅 ∈ Ring )
3 eqid ( Base ‘ 𝑅 ) = ( Base ‘ 𝑅 )
4 3 mxidlidl ( ( 𝑅 ∈ Ring ∧ 𝑖 ∈ ( MaxIdeal ‘ 𝑅 ) ) → 𝑖 ∈ ( LIdeal ‘ 𝑅 ) )
5 4 ex ( 𝑅 ∈ Ring → ( 𝑖 ∈ ( MaxIdeal ‘ 𝑅 ) → 𝑖 ∈ ( LIdeal ‘ 𝑅 ) ) )
6 5 ssrdv ( 𝑅 ∈ Ring → ( MaxIdeal ‘ 𝑅 ) ⊆ ( LIdeal ‘ 𝑅 ) )
7 2 6 syl ( 𝑅 ∈ DivRing → ( MaxIdeal ‘ 𝑅 ) ⊆ ( LIdeal ‘ 𝑅 ) )
8 eqid ( LIdeal ‘ 𝑅 ) = ( LIdeal ‘ 𝑅 )
9 3 1 8 drngnidl ( 𝑅 ∈ DivRing → ( LIdeal ‘ 𝑅 ) = { { 0 } , ( Base ‘ 𝑅 ) } )
10 7 9 sseqtrd ( 𝑅 ∈ DivRing → ( MaxIdeal ‘ 𝑅 ) ⊆ { { 0 } , ( Base ‘ 𝑅 ) } )
11 3 mxidlnr ( ( 𝑅 ∈ Ring ∧ 𝑖 ∈ ( MaxIdeal ‘ 𝑅 ) ) → 𝑖 ≠ ( Base ‘ 𝑅 ) )
12 2 11 sylan ( ( 𝑅 ∈ DivRing ∧ 𝑖 ∈ ( MaxIdeal ‘ 𝑅 ) ) → 𝑖 ≠ ( Base ‘ 𝑅 ) )
13 12 nelrdva ( 𝑅 ∈ DivRing → ¬ ( Base ‘ 𝑅 ) ∈ ( MaxIdeal ‘ 𝑅 ) )
14 ssdifsn ( ( MaxIdeal ‘ 𝑅 ) ⊆ ( { { 0 } , ( Base ‘ 𝑅 ) } ∖ { ( Base ‘ 𝑅 ) } ) ↔ ( ( MaxIdeal ‘ 𝑅 ) ⊆ { { 0 } , ( Base ‘ 𝑅 ) } ∧ ¬ ( Base ‘ 𝑅 ) ∈ ( MaxIdeal ‘ 𝑅 ) ) )
15 10 13 14 sylanbrc ( 𝑅 ∈ DivRing → ( MaxIdeal ‘ 𝑅 ) ⊆ ( { { 0 } , ( Base ‘ 𝑅 ) } ∖ { ( Base ‘ 𝑅 ) } ) )
16 drngnzr ( 𝑅 ∈ DivRing → 𝑅 ∈ NzRing )
17 1 3 drnglidl1ne0 ( 𝑅 ∈ NzRing → ( Base ‘ 𝑅 ) ≠ { 0 } )
18 17 necomd ( 𝑅 ∈ NzRing → { 0 } ≠ ( Base ‘ 𝑅 ) )
19 difprsn2 ( { 0 } ≠ ( Base ‘ 𝑅 ) → ( { { 0 } , ( Base ‘ 𝑅 ) } ∖ { ( Base ‘ 𝑅 ) } ) = { { 0 } } )
20 16 18 19 3syl ( 𝑅 ∈ DivRing → ( { { 0 } , ( Base ‘ 𝑅 ) } ∖ { ( Base ‘ 𝑅 ) } ) = { { 0 } } )
21 15 20 sseqtrd ( 𝑅 ∈ DivRing → ( MaxIdeal ‘ 𝑅 ) ⊆ { { 0 } } )
22 1 drng0mxidl ( 𝑅 ∈ DivRing → { 0 } ∈ ( MaxIdeal ‘ 𝑅 ) )
23 22 snssd ( 𝑅 ∈ DivRing → { { 0 } } ⊆ ( MaxIdeal ‘ 𝑅 ) )
24 21 23 eqssd ( 𝑅 ∈ DivRing → ( MaxIdeal ‘ 𝑅 ) = { { 0 } } )