Metamath Proof Explorer


Theorem gsumply1subr

Description: Evaluate a group sum in a polynomial ring over a subring. (Contributed by AV, 22-Sep-2019) (Proof shortened by AV, 31-Jan-2020)

Ref Expression
Hypotheses subrgply1.s S = Poly 1 R
subrgply1.h H = R 𝑠 T
subrgply1.u U = Poly 1 H
subrgply1.b B = Base U
gsumply1subr.s φ T SubRing R
gsumply1subr.a φ A V
gsumply1subr.f φ F : A B
Assertion gsumply1subr φ S F = U F

Proof

Step Hyp Ref Expression
1 subrgply1.s S = Poly 1 R
2 subrgply1.h H = R 𝑠 T
3 subrgply1.u U = Poly 1 H
4 subrgply1.b B = Base U
5 gsumply1subr.s φ T SubRing R
6 gsumply1subr.a φ A V
7 gsumply1subr.f φ F : A B
8 1 2 3 4 subrgply1 T SubRing R B SubRing S
9 subrgsubg B SubRing S B SubGrp S
10 subgsubm B SubGrp S B SubMnd S
11 9 10 syl B SubRing S B SubMnd S
12 5 8 11 3syl φ B SubMnd S
13 eqid S 𝑠 B = S 𝑠 B
14 6 12 7 13 gsumsubm φ S F = S 𝑠 B F
15 7 6 fexd φ F V
16 ovexd φ S 𝑠 B V
17 3 fvexi U V
18 17 a1i φ U V
19 eqid Base U = Base U
20 4 oveq2i S 𝑠 B = S 𝑠 Base U
21 1 2 3 19 5 20 ressply1bas φ Base U = Base S 𝑠 B
22 21 eqcomd φ Base S 𝑠 B = Base U
23 13 subrgring B SubRing S S 𝑠 B Ring
24 8 23 syl T SubRing R S 𝑠 B Ring
25 ringmgm S 𝑠 B Ring S 𝑠 B Mgm
26 5 24 25 3syl φ S 𝑠 B Mgm
27 simpl φ s Base S 𝑠 B t Base S 𝑠 B φ
28 1 2 3 4 5 13 ressply1bas φ B = Base S 𝑠 B
29 28 eqcomd φ Base S 𝑠 B = B
30 29 eleq2d φ s Base S 𝑠 B s B
31 30 biimpcd s Base S 𝑠 B φ s B
32 31 adantr s Base S 𝑠 B t Base S 𝑠 B φ s B
33 32 impcom φ s Base S 𝑠 B t Base S 𝑠 B s B
34 29 eleq2d φ t Base S 𝑠 B t B
35 34 biimpcd t Base S 𝑠 B φ t B
36 35 adantl s Base S 𝑠 B t Base S 𝑠 B φ t B
37 36 impcom φ s Base S 𝑠 B t Base S 𝑠 B t B
38 1 2 3 4 5 13 ressply1add φ s B t B s + U t = s + S 𝑠 B t
39 27 33 37 38 syl12anc φ s Base S 𝑠 B t Base S 𝑠 B s + U t = s + S 𝑠 B t
40 39 eqcomd φ s Base S 𝑠 B t Base S 𝑠 B s + S 𝑠 B t = s + U t
41 7 ffund φ Fun F
42 7 frnd φ ran F B
43 42 28 sseqtrd φ ran F Base S 𝑠 B
44 15 16 18 22 26 40 41 43 gsummgmpropd φ S 𝑠 B F = U F
45 14 44 eqtrd φ S F = U F