Metamath Proof Explorer


Theorem dvply2g

Description: The derivative of a polynomial with coefficients in a subring is a polynomial with coefficients in the same ring. (Contributed by Mario Carneiro, 1-Jan-2017)

Ref Expression
Assertion dvply2g S SubRing fld F Poly S D F Poly S

Proof

Step Hyp Ref Expression
1 plyf F Poly S F :
2 1 adantl S SubRing fld F Poly S F :
3 2 feqmptd S SubRing fld F Poly S F = a F a
4 simplr S SubRing fld F Poly S a F Poly S
5 dgrcl F Poly S deg F 0
6 5 adantl S SubRing fld F Poly S deg F 0
7 6 nn0zd S SubRing fld F Poly S deg F
8 7 adantr S SubRing fld F Poly S a deg F
9 uzid deg F deg F deg F
10 peano2uz deg F deg F deg F + 1 deg F
11 8 9 10 3syl S SubRing fld F Poly S a deg F + 1 deg F
12 simpr S SubRing fld F Poly S a a
13 eqid coeff F = coeff F
14 eqid deg F = deg F
15 13 14 coeid3 F Poly S deg F + 1 deg F a F a = b = 0 deg F + 1 coeff F b a b
16 4 11 12 15 syl3anc S SubRing fld F Poly S a F a = b = 0 deg F + 1 coeff F b a b
17 16 mpteq2dva S SubRing fld F Poly S a F a = a b = 0 deg F + 1 coeff F b a b
18 3 17 eqtrd S SubRing fld F Poly S F = a b = 0 deg F + 1 coeff F b a b
19 6 nn0cnd S SubRing fld F Poly S deg F
20 ax-1cn 1
21 pncan deg F 1 deg F + 1 - 1 = deg F
22 19 20 21 sylancl S SubRing fld F Poly S deg F + 1 - 1 = deg F
23 22 eqcomd S SubRing fld F Poly S deg F = deg F + 1 - 1
24 23 oveq2d S SubRing fld F Poly S 0 deg F = 0 deg F + 1 - 1
25 24 sumeq1d S SubRing fld F Poly S b = 0 deg F c 0 c + 1 coeff F c + 1 b a b = b = 0 deg F + 1 - 1 c 0 c + 1 coeff F c + 1 b a b
26 25 mpteq2dv S SubRing fld F Poly S a b = 0 deg F c 0 c + 1 coeff F c + 1 b a b = a b = 0 deg F + 1 - 1 c 0 c + 1 coeff F c + 1 b a b
27 13 coef3 F Poly S coeff F : 0
28 27 adantl S SubRing fld F Poly S coeff F : 0
29 oveq1 c = b c + 1 = b + 1
30 fvoveq1 c = b coeff F c + 1 = coeff F b + 1
31 29 30 oveq12d c = b c + 1 coeff F c + 1 = b + 1 coeff F b + 1
32 31 cbvmptv c 0 c + 1 coeff F c + 1 = b 0 b + 1 coeff F b + 1
33 peano2nn0 deg F 0 deg F + 1 0
34 6 33 syl S SubRing fld F Poly S deg F + 1 0
35 18 26 28 32 34 dvply1 S SubRing fld F Poly S D F = a b = 0 deg F c 0 c + 1 coeff F c + 1 b a b
36 cnfldbas = Base fld
37 36 subrgss S SubRing fld S
38 37 adantr S SubRing fld F Poly S S
39 elfznn0 b 0 deg F b 0
40 simpll S SubRing fld F Poly S c 0 S SubRing fld
41 zsssubrg S SubRing fld S
42 41 ad2antrr S SubRing fld F Poly S c 0 S
43 peano2nn0 c 0 c + 1 0
44 43 adantl S SubRing fld F Poly S c 0 c + 1 0
45 44 nn0zd S SubRing fld F Poly S c 0 c + 1
46 42 45 sseldd S SubRing fld F Poly S c 0 c + 1 S
47 simplr S SubRing fld F Poly S c 0 F Poly S
48 subrgsubg S SubRing fld S SubGrp fld
49 cnfld0 0 = 0 fld
50 49 subg0cl S SubGrp fld 0 S
51 48 50 syl S SubRing fld 0 S
52 51 ad2antrr S SubRing fld F Poly S c 0 0 S
53 13 coef2 F Poly S 0 S coeff F : 0 S
54 47 52 53 syl2anc S SubRing fld F Poly S c 0 coeff F : 0 S
55 54 44 ffvelrnd S SubRing fld F Poly S c 0 coeff F c + 1 S
56 cnfldmul × = fld
57 56 subrgmcl S SubRing fld c + 1 S coeff F c + 1 S c + 1 coeff F c + 1 S
58 40 46 55 57 syl3anc S SubRing fld F Poly S c 0 c + 1 coeff F c + 1 S
59 58 fmpttd S SubRing fld F Poly S c 0 c + 1 coeff F c + 1 : 0 S
60 59 ffvelrnda S SubRing fld F Poly S b 0 c 0 c + 1 coeff F c + 1 b S
61 39 60 sylan2 S SubRing fld F Poly S b 0 deg F c 0 c + 1 coeff F c + 1 b S
62 38 6 61 elplyd S SubRing fld F Poly S a b = 0 deg F c 0 c + 1 coeff F c + 1 b a b Poly S
63 35 62 eqeltrd S SubRing fld F Poly S D F Poly S