Metamath Proof Explorer


Theorem pm2mpghmlem2

Description: Lemma 2 for pm2mpghm . (Contributed by AV, 15-Oct-2019) (Revised by AV, 4-Dec-2019)

Ref Expression
Hypotheses pm2mpfo.p P = Poly 1 R
pm2mpfo.c C = N Mat P
pm2mpfo.b B = Base C
pm2mpfo.m ˙ = Q
pm2mpfo.e × ˙ = mulGrp Q
pm2mpfo.x X = var 1 A
pm2mpfo.a A = N Mat R
pm2mpfo.q Q = Poly 1 A
Assertion pm2mpghmlem2 N Fin R Ring M B finSupp 0 Q k 0 M decompPMat k ˙ k × ˙ X

Proof

Step Hyp Ref Expression
1 pm2mpfo.p P = Poly 1 R
2 pm2mpfo.c C = N Mat P
3 pm2mpfo.b B = Base C
4 pm2mpfo.m ˙ = Q
5 pm2mpfo.e × ˙ = mulGrp Q
6 pm2mpfo.x X = var 1 A
7 pm2mpfo.a A = N Mat R
8 pm2mpfo.q Q = Poly 1 A
9 nn0ex 0 V
10 9 a1i N Fin R Ring M B 0 V
11 7 matring N Fin R Ring A Ring
12 11 3adant3 N Fin R Ring M B A Ring
13 8 ply1lmod A Ring Q LMod
14 12 13 syl N Fin R Ring M B Q LMod
15 8 ply1sca A Ring A = Scalar Q
16 12 15 syl N Fin R Ring M B A = Scalar Q
17 eqid Base Q = Base Q
18 simpl2 N Fin R Ring M B k 0 R Ring
19 simpl3 N Fin R Ring M B k 0 M B
20 simpr N Fin R Ring M B k 0 k 0
21 eqid Base A = Base A
22 1 2 3 7 21 decpmatcl R Ring M B k 0 M decompPMat k Base A
23 18 19 20 22 syl3anc N Fin R Ring M B k 0 M decompPMat k Base A
24 eqid mulGrp Q = mulGrp Q
25 8 6 24 5 17 ply1moncl A Ring k 0 k × ˙ X Base Q
26 12 25 sylan N Fin R Ring M B k 0 k × ˙ X Base Q
27 eqid 0 Q = 0 Q
28 eqid 0 A = 0 A
29 1 2 3 7 28 decpmatfsupp R Ring M B finSupp 0 A k 0 M decompPMat k
30 29 3adant1 N Fin R Ring M B finSupp 0 A k 0 M decompPMat k
31 10 14 16 17 23 26 27 28 4 30 mptscmfsupp0 N Fin R Ring M B finSupp 0 Q k 0 M decompPMat k ˙ k × ˙ X