Metamath Proof Explorer


Theorem rlmdim

Description: The left vector space induced by a ring over itself has dimension 1. (Contributed by Thierry Arnoux, 5-Aug-2023) Generalize to division rings. (Revised by SN, 22-Mar-2025)

Ref Expression
Hypothesis rlmdim.1 V = ringLMod F
Assertion rlmdim F DivRing dim V = 1

Proof

Step Hyp Ref Expression
1 rlmdim.1 V = ringLMod F
2 rlmlvec F DivRing ringLMod F LVec
3 1 2 eqeltrid F DivRing V LVec
4 ssid Base F Base F
5 rlmval ringLMod F = subringAlg F Base F
6 1 5 eqtri V = subringAlg F Base F
7 eqid Base F = Base F
8 6 7 sradrng F DivRing Base F Base F V DivRing
9 4 8 mpan2 F DivRing V DivRing
10 9 drngringd F DivRing V Ring
11 eqid Base V = Base V
12 eqid 1 V = 1 V
13 11 12 ringidcl V Ring 1 V Base V
14 10 13 syl F DivRing 1 V Base V
15 eqid 0 V = 0 V
16 15 12 drngunz V DivRing 1 V 0 V
17 9 16 syl F DivRing 1 V 0 V
18 11 15 lindssn V LVec 1 V Base V 1 V 0 V 1 V LIndS V
19 3 14 17 18 syl3anc F DivRing 1 V LIndS V
20 drngring F DivRing F Ring
21 1 fveq2i LSpan V = LSpan ringLMod F
22 rspval RSpan F = LSpan ringLMod F
23 21 22 eqtr4i LSpan V = RSpan F
24 eqid 1 F = 1 F
25 23 7 24 rsp1 F Ring LSpan V 1 F = Base F
26 20 25 syl F DivRing LSpan V 1 F = Base F
27 6 a1i F DivRing V = subringAlg F Base F
28 eqidd F DivRing 1 F = 1 F
29 ssidd F DivRing Base F Base F
30 27 28 29 sra1r F DivRing 1 F = 1 V
31 30 sneqd F DivRing 1 F = 1 V
32 31 fveq2d F DivRing LSpan V 1 F = LSpan V 1 V
33 27 29 srabase F DivRing Base F = Base V
34 26 32 33 3eqtr3d F DivRing LSpan V 1 V = Base V
35 eqid LBasis V = LBasis V
36 eqid LSpan V = LSpan V
37 11 35 36 islbs4 1 V LBasis V 1 V LIndS V LSpan V 1 V = Base V
38 19 34 37 sylanbrc F DivRing 1 V LBasis V
39 35 dimval V LVec 1 V LBasis V dim V = 1 V
40 3 38 39 syl2anc F DivRing dim V = 1 V
41 fvex 1 V V
42 hashsng 1 V V 1 V = 1
43 41 42 ax-mp 1 V = 1
44 40 43 eqtrdi F DivRing dim V = 1