Metamath Proof Explorer


Theorem chpdmatlem1

Description: Lemma 1 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 chpdmatlem1 NFinRRingMBX·˙1˙ZTMBaseQ

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 10 pmatring NFinRRingQRing
16 15 3adant3 NFinRRingMBQRing
17 ringgrp QRingQGrp
18 16 17 syl NFinRRingMBQGrp
19 1 2 3 4 5 6 7 8 9 10 11 12 chpdmatlem0 NFinRRingX·˙1˙BaseQ
20 19 3adant3 NFinRRingMBX·˙1˙BaseQ
21 14 3 5 2 10 mat2pmatbas NFinRRingMBTMBaseQ
22 eqid BaseQ=BaseQ
23 22 13 grpsubcl QGrpX·˙1˙BaseQTMBaseQX·˙1˙ZTMBaseQ
24 18 20 21 23 syl3anc NFinRRingMBX·˙1˙ZTMBaseQ