Metamath Proof Explorer


Theorem mulgrhm

Description: The powers of the element 1 give a ring homomorphism from ZZ to a ring. (Contributed by Mario Carneiro, 14-Jun-2015) (Revised by AV, 12-Jun-2019)

Ref Expression
Hypotheses mulgghm2.m ·˙=R
mulgghm2.f F=nn·˙1˙
mulgrhm.1 1˙=1R
Assertion mulgrhm RRingFringRingHomR

Proof

Step Hyp Ref Expression
1 mulgghm2.m ·˙=R
2 mulgghm2.f F=nn·˙1˙
3 mulgrhm.1 1˙=1R
4 zringbas =Basering
5 zring1 1=1ring
6 zringmulr ×=ring
7 eqid R=R
8 zringring ringRing
9 8 a1i RRingringRing
10 id RRingRRing
11 1z 1
12 oveq1 n=1n·˙1˙=1·˙1˙
13 ovex 1·˙1˙V
14 12 2 13 fvmpt 1F1=1·˙1˙
15 11 14 ax-mp F1=1·˙1˙
16 eqid BaseR=BaseR
17 16 3 ringidcl RRing1˙BaseR
18 16 1 mulg1 1˙BaseR1·˙1˙=1˙
19 17 18 syl RRing1·˙1˙=1˙
20 15 19 eqtrid RRingF1=1˙
21 ringgrp RRingRGrp
22 21 adantr RRingxyRGrp
23 simprr RRingxyy
24 17 adantr RRingxy1˙BaseR
25 16 1 mulgcl RGrpy1˙BaseRy·˙1˙BaseR
26 22 23 24 25 syl3anc RRingxyy·˙1˙BaseR
27 16 7 3 ringlidm RRingy·˙1˙BaseR1˙Ry·˙1˙=y·˙1˙
28 26 27 syldan RRingxy1˙Ry·˙1˙=y·˙1˙
29 28 oveq2d RRingxyx·˙1˙Ry·˙1˙=x·˙y·˙1˙
30 simpl RRingxyRRing
31 simprl RRingxyx
32 16 1 7 mulgass2 RRingx1˙BaseRy·˙1˙BaseRx·˙1˙Ry·˙1˙=x·˙1˙Ry·˙1˙
33 30 31 24 26 32 syl13anc RRingxyx·˙1˙Ry·˙1˙=x·˙1˙Ry·˙1˙
34 16 1 mulgass RGrpxy1˙BaseRxy·˙1˙=x·˙y·˙1˙
35 22 31 23 24 34 syl13anc RRingxyxy·˙1˙=x·˙y·˙1˙
36 29 33 35 3eqtr4rd RRingxyxy·˙1˙=x·˙1˙Ry·˙1˙
37 zmulcl xyxy
38 37 adantl RRingxyxy
39 oveq1 n=xyn·˙1˙=xy·˙1˙
40 ovex xy·˙1˙V
41 39 2 40 fvmpt xyFxy=xy·˙1˙
42 38 41 syl RRingxyFxy=xy·˙1˙
43 oveq1 n=xn·˙1˙=x·˙1˙
44 ovex x·˙1˙V
45 43 2 44 fvmpt xFx=x·˙1˙
46 oveq1 n=yn·˙1˙=y·˙1˙
47 ovex y·˙1˙V
48 46 2 47 fvmpt yFy=y·˙1˙
49 45 48 oveqan12d xyFxRFy=x·˙1˙Ry·˙1˙
50 49 adantl RRingxyFxRFy=x·˙1˙Ry·˙1˙
51 36 42 50 3eqtr4d RRingxyFxy=FxRFy
52 1 2 16 mulgghm2 RGrp1˙BaseRFringGrpHomR
53 21 17 52 syl2anc RRingFringGrpHomR
54 4 5 3 6 7 9 10 20 51 53 isrhm2d RRingFringRingHomR