Metamath Proof Explorer


Theorem imasrhm

Description: Given a function F with homomorphic properties, build the image of a ring. (Contributed by Thierry Arnoux, 2-Apr-2025)

Ref Expression
Hypotheses imasmhm.b B = Base W
imasmhm.f φ F : B C
imasmhm.1 + ˙ = + W
imasmhm.2 φ a B b B p B q B F a = F p F b = F q F a + ˙ b = F p + ˙ q
imasrhm.3 · ˙ = W
imasrhm.4 φ a B b B p B q B F a = F p F b = F q F a · ˙ b = F p · ˙ q
imasrhm.w φ W Ring
Assertion imasrhm φ F 𝑠 W Ring F W RingHom F 𝑠 W

Proof

Step Hyp Ref Expression
1 imasmhm.b B = Base W
2 imasmhm.f φ F : B C
3 imasmhm.1 + ˙ = + W
4 imasmhm.2 φ a B b B p B q B F a = F p F b = F q F a + ˙ b = F p + ˙ q
5 imasrhm.3 · ˙ = W
6 imasrhm.4 φ a B b B p B q B F a = F p F b = F q F a · ˙ b = F p · ˙ q
7 imasrhm.w φ W Ring
8 eqidd φ F 𝑠 W = F 𝑠 W
9 1 a1i φ B = Base W
10 eqid 1 W = 1 W
11 fimadmfo F : B C F : B onto F B
12 2 11 syl φ F : B onto F B
13 8 9 3 5 10 12 4 6 7 imasring φ F 𝑠 W Ring F 1 W = 1 F 𝑠 W
14 13 simpld φ F 𝑠 W Ring
15 eqid 1 F 𝑠 W = 1 F 𝑠 W
16 eqid F 𝑠 W = F 𝑠 W
17 13 simprd φ F 1 W = 1 F 𝑠 W
18 12 6 8 9 7 5 16 imasmulval φ x B y B F x F 𝑠 W F y = F x · ˙ y
19 18 3expb φ x B y B F x F 𝑠 W F y = F x · ˙ y
20 19 eqcomd φ x B y B F x · ˙ y = F x F 𝑠 W F y
21 eqid Base F 𝑠 W = Base F 𝑠 W
22 eqid + F 𝑠 W = + F 𝑠 W
23 fof F : B onto F B F : B F B
24 12 23 syl φ F : B F B
25 8 9 12 7 imasbas φ F B = Base F 𝑠 W
26 25 feq3d φ F : B F B F : B Base F 𝑠 W
27 24 26 mpbid φ F : B Base F 𝑠 W
28 12 4 8 9 7 3 22 imasaddval φ x B y B F x + F 𝑠 W F y = F x + ˙ y
29 28 3expb φ x B y B F x + F 𝑠 W F y = F x + ˙ y
30 29 eqcomd φ x B y B F x + ˙ y = F x + F 𝑠 W F y
31 1 10 15 5 16 7 14 17 20 21 3 22 27 30 isrhmd φ F W RingHom F 𝑠 W
32 14 31 jca φ F 𝑠 W Ring F W RingHom F 𝑠 W