Metamath Proof Explorer


Theorem chpdmatlem0

Description: Lemma 0 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
Assertion chpdmatlem0 NFinRRingX·˙1˙BaseQ

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 2 10 pmatlmod NFinRRingQLMod
14 eqid BaseP=BaseP
15 6 2 14 vr1cl RRingXBaseP
16 15 adantl NFinRRingXBaseP
17 2 ply1ring RRingPRing
18 10 matsca2 NFinPRingP=ScalarQ
19 17 18 sylan2 NFinRRingP=ScalarQ
20 19 eqcomd NFinRRingScalarQ=P
21 20 fveq2d NFinRRingBaseScalarQ=BaseP
22 16 21 eleqtrrd NFinRRingXBaseScalarQ
23 2 10 pmatring NFinRRingQRing
24 eqid BaseQ=BaseQ
25 24 11 ringidcl QRing1˙BaseQ
26 23 25 syl NFinRRing1˙BaseQ
27 eqid ScalarQ=ScalarQ
28 eqid BaseScalarQ=BaseScalarQ
29 24 27 12 28 lmodvscl QLModXBaseScalarQ1˙BaseQX·˙1˙BaseQ
30 13 22 26 29 syl3anc NFinRRingX·˙1˙BaseQ