Metamath Proof Explorer


Theorem erng0g

Description: The division ring zero of an endomorphism ring. (Contributed by NM, 5-Nov-2013) (Revised by Mario Carneiro, 23-Jun-2014)

Ref Expression
Hypotheses erng0g.b 𝐵 = ( Base ‘ 𝐾 )
erng0g.h 𝐻 = ( LHyp ‘ 𝐾 )
erng0g.t 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 )
erng0g.d 𝐷 = ( ( EDRing ‘ 𝐾 ) ‘ 𝑊 )
erng0g.o 𝑂 = ( 𝑓𝑇 ↦ ( I ↾ 𝐵 ) )
erng0g.z 0 = ( 0g𝐷 )
Assertion erng0g ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → 0 = 𝑂 )

Proof

Step Hyp Ref Expression
1 erng0g.b 𝐵 = ( Base ‘ 𝐾 )
2 erng0g.h 𝐻 = ( LHyp ‘ 𝐾 )
3 erng0g.t 𝑇 = ( ( LTrn ‘ 𝐾 ) ‘ 𝑊 )
4 erng0g.d 𝐷 = ( ( EDRing ‘ 𝐾 ) ‘ 𝑊 )
5 erng0g.o 𝑂 = ( 𝑓𝑇 ↦ ( I ↾ 𝐵 ) )
6 erng0g.z 0 = ( 0g𝐷 )
7 eqid ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) = ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 )
8 eqid ( +g𝐷 ) = ( +g𝐷 )
9 2 3 7 4 8 erngfplus ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → ( +g𝐷 ) = ( 𝑠 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) , 𝑡 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) ↦ ( 𝑓𝑇 ↦ ( ( 𝑠𝑓 ) ∘ ( 𝑡𝑓 ) ) ) ) )
10 9 oveqd ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → ( 𝑂 ( +g𝐷 ) 𝑂 ) = ( 𝑂 ( 𝑠 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) , 𝑡 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) ↦ ( 𝑓𝑇 ↦ ( ( 𝑠𝑓 ) ∘ ( 𝑡𝑓 ) ) ) ) 𝑂 ) )
11 1 2 3 7 5 tendo0cl ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → 𝑂 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) )
12 eqid ( 𝑠 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) , 𝑡 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) ↦ ( 𝑓𝑇 ↦ ( ( 𝑠𝑓 ) ∘ ( 𝑡𝑓 ) ) ) ) = ( 𝑠 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) , 𝑡 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) ↦ ( 𝑓𝑇 ↦ ( ( 𝑠𝑓 ) ∘ ( 𝑡𝑓 ) ) ) )
13 1 2 3 7 5 12 tendo0pl ( ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) ∧ 𝑂 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) ) → ( 𝑂 ( 𝑠 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) , 𝑡 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) ↦ ( 𝑓𝑇 ↦ ( ( 𝑠𝑓 ) ∘ ( 𝑡𝑓 ) ) ) ) 𝑂 ) = 𝑂 )
14 11 13 mpdan ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → ( 𝑂 ( 𝑠 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) , 𝑡 ∈ ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) ↦ ( 𝑓𝑇 ↦ ( ( 𝑠𝑓 ) ∘ ( 𝑡𝑓 ) ) ) ) 𝑂 ) = 𝑂 )
15 10 14 eqtrd ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → ( 𝑂 ( +g𝐷 ) 𝑂 ) = 𝑂 )
16 2 4 eringring ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → 𝐷 ∈ Ring )
17 ringgrp ( 𝐷 ∈ Ring → 𝐷 ∈ Grp )
18 16 17 syl ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → 𝐷 ∈ Grp )
19 eqid ( Base ‘ 𝐷 ) = ( Base ‘ 𝐷 )
20 2 3 7 4 19 erngbase ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → ( Base ‘ 𝐷 ) = ( ( TEndo ‘ 𝐾 ) ‘ 𝑊 ) )
21 11 20 eleqtrrd ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → 𝑂 ∈ ( Base ‘ 𝐷 ) )
22 19 8 6 grpid ( ( 𝐷 ∈ Grp ∧ 𝑂 ∈ ( Base ‘ 𝐷 ) ) → ( ( 𝑂 ( +g𝐷 ) 𝑂 ) = 𝑂0 = 𝑂 ) )
23 18 21 22 syl2anc ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → ( ( 𝑂 ( +g𝐷 ) 𝑂 ) = 𝑂0 = 𝑂 ) )
24 15 23 mpbid ( ( 𝐾 ∈ HL ∧ 𝑊𝐻 ) → 0 = 𝑂 )