Metamath Proof Explorer


Theorem lidlrsppropd

Description: The left ideals and ring span of a ring depend only on the ring components. Here W is expected to be either B (when closure is available) or _V (when strong equality is available). (Contributed by Mario Carneiro, 14-Jun-2015)

Ref Expression
Hypotheses lidlpropd.1 φB=BaseK
lidlpropd.2 φB=BaseL
lidlpropd.3 φBW
lidlpropd.4 φxWyWx+Ky=x+Ly
lidlpropd.5 φxByBxKyW
lidlpropd.6 φxByBxKy=xLy
Assertion lidlrsppropd φLIdealK=LIdealLRSpanK=RSpanL

Proof

Step Hyp Ref Expression
1 lidlpropd.1 φB=BaseK
2 lidlpropd.2 φB=BaseL
3 lidlpropd.3 φBW
4 lidlpropd.4 φxWyWx+Ky=x+Ly
5 lidlpropd.5 φxByBxKyW
6 lidlpropd.6 φxByBxKy=xLy
7 rlmbas BaseK=BaseringLModK
8 1 7 eqtrdi φB=BaseringLModK
9 rlmbas BaseL=BaseringLModL
10 2 9 eqtrdi φB=BaseringLModL
11 rlmplusg +K=+ringLModK
12 11 oveqi x+Ky=x+ringLModKy
13 rlmplusg +L=+ringLModL
14 13 oveqi x+Ly=x+ringLModLy
15 4 12 14 3eqtr3g φxWyWx+ringLModKy=x+ringLModLy
16 rlmvsca K=ringLModK
17 16 oveqi xKy=xringLModKy
18 17 5 eqeltrrid φxByBxringLModKyW
19 rlmvsca L=ringLModL
20 19 oveqi xLy=xringLModLy
21 6 17 20 3eqtr3g φxByBxringLModKy=xringLModLy
22 baseid Base=SlotBasendx
23 eqid BaseK=BaseK
24 22 23 strfvi BaseK=BaseIK
25 rlmsca2 IK=ScalarringLModK
26 25 fveq2i BaseIK=BaseScalarringLModK
27 24 26 eqtri BaseK=BaseScalarringLModK
28 1 27 eqtrdi φB=BaseScalarringLModK
29 eqid BaseL=BaseL
30 22 29 strfvi BaseL=BaseIL
31 rlmsca2 IL=ScalarringLModL
32 31 fveq2i BaseIL=BaseScalarringLModL
33 30 32 eqtri BaseL=BaseScalarringLModL
34 2 33 eqtrdi φB=BaseScalarringLModL
35 8 10 3 15 18 21 28 34 lsspropd φLSubSpringLModK=LSubSpringLModL
36 lidlval LIdealK=LSubSpringLModK
37 lidlval LIdealL=LSubSpringLModL
38 35 36 37 3eqtr4g φLIdealK=LIdealL
39 fvexd φringLModKV
40 fvexd φringLModLV
41 8 10 3 15 18 21 28 34 39 40 lsppropd φLSpanringLModK=LSpanringLModL
42 rspval RSpanK=LSpanringLModK
43 rspval RSpanL=LSpanringLModL
44 41 42 43 3eqtr4g φRSpanK=RSpanL
45 38 44 jca φLIdealK=LIdealLRSpanK=RSpanL