Metamath Proof Explorer


Theorem qusrhm

Description: If S is a two-sided ideal in R , then the "natural map" from elements to their cosets is a ring homomorphism from R to R / S . (Contributed by Mario Carneiro, 15-Jun-2015)

Ref Expression
Hypotheses qusring.u U=R/𝑠R~QGS
qusring.i I=2IdealR
qusrhm.x X=BaseR
qusrhm.f F=xXxR~QGS
Assertion qusrhm RRingSIFRRingHomU

Proof

Step Hyp Ref Expression
1 qusring.u U=R/𝑠R~QGS
2 qusring.i I=2IdealR
3 qusrhm.x X=BaseR
4 qusrhm.f F=xXxR~QGS
5 eqid 1R=1R
6 eqid 1U=1U
7 eqid R=R
8 eqid U=U
9 simpl RRingSIRRing
10 1 2 qusring RRingSIURing
11 eqid LIdealR=LIdealR
12 eqid opprR=opprR
13 eqid LIdealopprR=LIdealopprR
14 11 12 13 2 2idlval I=LIdealRLIdealopprR
15 14 elin2 SISLIdealRSLIdealopprR
16 15 simplbi SISLIdealR
17 11 lidlsubg RRingSLIdealRSSubGrpR
18 16 17 sylan2 RRingSISSubGrpR
19 eqid R~QGS=R~QGS
20 3 19 eqger SSubGrpRR~QGSErX
21 18 20 syl RRingSIR~QGSErX
22 3 fvexi XV
23 22 a1i RRingSIXV
24 21 23 4 divsfval RRingSIF1R=1RR~QGS
25 1 2 5 qus1 RRingSIURing1RR~QGS=1U
26 25 simprd RRingSI1RR~QGS=1U
27 24 26 eqtrd RRingSIF1R=1U
28 1 a1i RRingSIU=R/𝑠R~QGS
29 3 a1i RRingSIX=BaseR
30 3 19 2 7 2idlcpbl RRingSIaR~QGScbR~QGSdaRbR~QGScRd
31 3 7 ringcl RRingyXzXyRzX
32 31 3expb RRingyXzXyRzX
33 32 adantlr RRingSIyXzXyRzX
34 33 caovclg RRingSIcXdXcRdX
35 28 29 21 9 30 34 7 8 qusmulval RRingSIyXzXyR~QGSUzR~QGS=yRzR~QGS
36 35 3expb RRingSIyXzXyR~QGSUzR~QGS=yRzR~QGS
37 21 adantr RRingSIyXzXR~QGSErX
38 22 a1i RRingSIyXzXXV
39 37 38 4 divsfval RRingSIyXzXFy=yR~QGS
40 37 38 4 divsfval RRingSIyXzXFz=zR~QGS
41 39 40 oveq12d RRingSIyXzXFyUFz=yR~QGSUzR~QGS
42 37 38 4 divsfval RRingSIyXzXFyRz=yRzR~QGS
43 36 41 42 3eqtr4rd RRingSIyXzXFyRz=FyUFz
44 ringabl RRingRAbel
45 44 adantr RRingSIRAbel
46 ablnsg RAbelNrmSGrpR=SubGrpR
47 45 46 syl RRingSINrmSGrpR=SubGrpR
48 18 47 eleqtrrd RRingSISNrmSGrpR
49 3 1 4 qusghm SNrmSGrpRFRGrpHomU
50 48 49 syl RRingSIFRGrpHomU
51 3 5 6 7 8 9 10 27 43 50 isrhm2d RRingSIFRRingHomU