Metamath Proof Explorer


Theorem chpdmatlem0

Description: Lemma 0 for chpdmat . (Contributed by AV, 18-Aug-2019)

Ref Expression
Hypotheses chpdmat.c C = N CharPlyMat R
chpdmat.p P = Poly 1 R
chpdmat.a A = N Mat R
chpdmat.s S = algSc P
chpdmat.b B = Base A
chpdmat.x X = var 1 R
chpdmat.0 0 ˙ = 0 R
chpdmat.g G = mulGrp P
chpdmat.m - ˙ = - P
chpdmatlem.q Q = N Mat P
chpdmatlem.1 1 ˙ = 1 Q
chpdmatlem.m · ˙ = Q
Assertion chpdmatlem0 N Fin R Ring X · ˙ 1 ˙ Base Q

Proof

Step Hyp Ref Expression
1 chpdmat.c C = N CharPlyMat R
2 chpdmat.p P = Poly 1 R
3 chpdmat.a A = N Mat R
4 chpdmat.s S = algSc P
5 chpdmat.b B = Base A
6 chpdmat.x X = var 1 R
7 chpdmat.0 0 ˙ = 0 R
8 chpdmat.g G = mulGrp P
9 chpdmat.m - ˙ = - P
10 chpdmatlem.q Q = N Mat P
11 chpdmatlem.1 1 ˙ = 1 Q
12 chpdmatlem.m · ˙ = Q
13 2 10 pmatlmod N Fin R Ring Q LMod
14 eqid Base P = Base P
15 6 2 14 vr1cl R Ring X Base P
16 15 adantl N Fin R Ring X Base P
17 2 ply1ring R Ring P Ring
18 10 matsca2 N Fin P Ring P = Scalar Q
19 17 18 sylan2 N Fin R Ring P = Scalar Q
20 19 eqcomd N Fin R Ring Scalar Q = P
21 20 fveq2d N Fin R Ring Base Scalar Q = Base P
22 16 21 eleqtrrd N Fin R Ring X Base Scalar Q
23 2 10 pmatring N Fin R Ring Q Ring
24 eqid Base Q = Base Q
25 24 11 ringidcl Q Ring 1 ˙ Base Q
26 23 25 syl N Fin R Ring 1 ˙ Base Q
27 eqid Scalar Q = Scalar Q
28 eqid Base Scalar Q = Base Scalar Q
29 24 27 12 28 lmodvscl Q LMod X Base Scalar Q 1 ˙ Base Q X · ˙ 1 ˙ Base Q
30 13 22 26 29 syl3anc N Fin R Ring X · ˙ 1 ˙ Base Q