Metamath Proof Explorer


Theorem aks6d1c6isolem3

Description: The preimage of a map sending a primitive root to its powers of zero is equal to the set of integers that divide R . (Contributed by metakunt, 15-May-2025)

Ref Expression
Hypotheses aks6d1c6isolem1.1 φ R CMnd
aks6d1c6isolem1.2 φ K
aks6d1c6isolem1.3 U = a Base R | i Base R i + R a = 0 R
aks6d1c6isolem1.4 F = x x R 𝑠 U M
aks6d1c6isolem1.5 φ M R PrimRoots K
aks6d1c6isolem3.1 S = RSpan ring
Assertion aks6d1c6isolem3 φ S K = F -1 0 R 𝑠 U

Proof

Step Hyp Ref Expression
1 aks6d1c6isolem1.1 φ R CMnd
2 aks6d1c6isolem1.2 φ K
3 aks6d1c6isolem1.3 U = a Base R | i Base R i + R a = 0 R
4 aks6d1c6isolem1.4 F = x x R 𝑠 U M
5 aks6d1c6isolem1.5 φ M R PrimRoots K
6 aks6d1c6isolem3.1 S = RSpan ring
7 zringring ring Ring
8 7 a1i φ ring Ring
9 2 nnzd φ K
10 zringbas = Base ring
11 dvdsrzring = r ring
12 10 6 11 rspsn ring Ring K S K = z | K z
13 8 9 12 syl2anc φ S K = z | K z
14 ovexd φ x x R 𝑠 U M V
15 14 4 fmptd φ F : V
16 15 ffnd φ F Fn
17 fniniseg2 F Fn F -1 0 R 𝑠 U = z | F z = 0 R 𝑠 U
18 16 17 syl φ F -1 0 R 𝑠 U = z | F z = 0 R 𝑠 U
19 4 a1i φ z F = x x R 𝑠 U M
20 simpr φ z x = z x = z
21 20 oveq1d φ z x = z x R 𝑠 U M = z R 𝑠 U M
22 simpr φ z z
23 ovexd φ z z R 𝑠 U M V
24 19 21 22 23 fvmptd φ z F z = z R 𝑠 U M
25 24 eqeq1d φ z F z = 0 R 𝑠 U z R 𝑠 U M = 0 R 𝑠 U
26 1 adantr φ z R CMnd
27 2 adantr φ z K
28 5 adantr φ z M R PrimRoots K
29 26 27 28 3 22 primrootspoweq0 φ z z R 𝑠 U M = 0 R 𝑠 U K z
30 25 29 bitrd φ z F z = 0 R 𝑠 U K z
31 30 rabbidva φ z | F z = 0 R 𝑠 U = z | K z
32 df-rab z | K z = z | z K z
33 32 a1i φ z | K z = z | z K z
34 simpr z K z K z
35 dvdszrcl K z K z
36 35 simprd K z z
37 36 ancri K z z K z
38 34 37 impbii z K z K z
39 38 a1i φ z K z K z
40 39 abbidv φ z | z K z = z | K z
41 33 40 eqtrd φ z | K z = z | K z
42 31 41 eqtrd φ z | F z = 0 R 𝑠 U = z | K z
43 18 42 eqtr2d φ z | K z = F -1 0 R 𝑠 U
44 13 43 eqtrd φ S K = F -1 0 R 𝑠 U