Metamath Proof Explorer


Theorem chpdmatlem3

Description: Lemma 3 for chpdmat . (Contributed by AV, 18-Aug-2019)

Ref Expression
Hypotheses chpdmat.c C=NCharPlyMatR
chpdmat.p P=Poly1R
chpdmat.a A=NMatR
chpdmat.s S=algScP
chpdmat.b B=BaseA
chpdmat.x X=var1R
chpdmat.0 0˙=0R
chpdmat.g G=mulGrpP
chpdmat.m -˙=-P
chpdmatlem.q Q=NMatP
chpdmatlem.1 1˙=1Q
chpdmatlem.m ·˙=Q
chpdmatlem.z Z=-Q
chpdmatlem.t T=NmatToPolyMatR
Assertion chpdmatlem3 NFinRRingMBKNKX·˙1˙ZTMK=X-˙SKMK

Proof

Step Hyp Ref Expression
1 chpdmat.c C=NCharPlyMatR
2 chpdmat.p P=Poly1R
3 chpdmat.a A=NMatR
4 chpdmat.s S=algScP
5 chpdmat.b B=BaseA
6 chpdmat.x X=var1R
7 chpdmat.0 0˙=0R
8 chpdmat.g G=mulGrpP
9 chpdmat.m -˙=-P
10 chpdmatlem.q Q=NMatP
11 chpdmatlem.1 1˙=1Q
12 chpdmatlem.m ·˙=Q
13 chpdmatlem.z Z=-Q
14 chpdmatlem.t T=NmatToPolyMatR
15 2 ply1ring RRingPRing
16 15 3ad2ant2 NFinRRingMBPRing
17 16 adantr NFinRRingMBKNPRing
18 1 2 3 4 5 6 7 8 9 10 11 12 chpdmatlem0 NFinRRingX·˙1˙BaseQ
19 18 3adant3 NFinRRingMBX·˙1˙BaseQ
20 14 3 5 2 10 mat2pmatbas NFinRRingMBTMBaseQ
21 19 20 jca NFinRRingMBX·˙1˙BaseQTMBaseQ
22 21 adantr NFinRRingMBKNX·˙1˙BaseQTMBaseQ
23 simpr NFinRRingMBKNKN
24 eqid BaseQ=BaseQ
25 10 24 13 9 matsubgcell PRingX·˙1˙BaseQTMBaseQKNKNKX·˙1˙ZTMK=KX·˙1˙K-˙KTMK
26 17 22 23 23 25 syl112anc NFinRRingMBKNKX·˙1˙ZTMK=KX·˙1˙K-˙KTMK
27 eqid BaseP=BaseP
28 6 2 27 vr1cl RRingXBaseP
29 28 adantl NFinRRingXBaseP
30 2 10 pmatring NFinRRingQRing
31 24 11 ringidcl QRing1˙BaseQ
32 30 31 syl NFinRRing1˙BaseQ
33 29 32 jca NFinRRingXBaseP1˙BaseQ
34 33 3adant3 NFinRRingMBXBaseP1˙BaseQ
35 34 adantr NFinRRingMBKNXBaseP1˙BaseQ
36 eqid P=P
37 10 24 27 12 36 matvscacell PRingXBaseP1˙BaseQKNKNKX·˙1˙K=XPK1˙K
38 17 35 23 23 37 syl112anc NFinRRingMBKNKX·˙1˙K=XPK1˙K
39 eqid 1P=1P
40 eqid 0P=0P
41 simpl1 NFinRRingMBKNNFin
42 10 39 40 41 17 23 23 11 mat1ov NFinRRingMBKNK1˙K=ifK=K1P0P
43 eqid K=K
44 43 iftruei ifK=K1P0P=1P
45 42 44 eqtrdi NFinRRingMBKNK1˙K=1P
46 45 oveq2d NFinRRingMBKNXPK1˙K=XP1P
47 15 28 jca RRingPRingXBaseP
48 47 3ad2ant2 NFinRRingMBPRingXBaseP
49 27 36 39 ringridm PRingXBasePXP1P=X
50 48 49 syl NFinRRingMBXP1P=X
51 50 adantr NFinRRingMBKNXP1P=X
52 38 46 51 3eqtrd NFinRRingMBKNKX·˙1˙K=X
53 14 3 5 2 4 mat2pmatvalel NFinRRingMBKNKNKTMK=SKMK
54 53 anabsan2 NFinRRingMBKNKTMK=SKMK
55 52 54 oveq12d NFinRRingMBKNKX·˙1˙K-˙KTMK=X-˙SKMK
56 26 55 eqtrd NFinRRingMBKNKX·˙1˙ZTMK=X-˙SKMK