Metamath Proof Explorer


Theorem lindsrng01

Description: Any subset of a module is always linearly independent if the underlying ring has at most one element. Since the underlying ring cannot be the empty set (see lmodsn0 ), this means that the underlying ring has only one element, so it is a zero ring. (Contributed by AV, 14-Apr-2019) (Revised by AV, 27-Apr-2019)

Ref Expression
Hypotheses lindsrng01.b B = Base M
lindsrng01.r R = Scalar M
lindsrng01.e E = Base R
Assertion lindsrng01 M LMod E = 0 E = 1 S 𝒫 B S linIndS M

Proof

Step Hyp Ref Expression
1 lindsrng01.b B = Base M
2 lindsrng01.r R = Scalar M
3 lindsrng01.e E = Base R
4 2 3 lmodsn0 M LMod E
5 3 fvexi E V
6 hasheq0 E V E = 0 E =
7 5 6 ax-mp E = 0 E =
8 eqneqall E = E S linIndS M
9 8 com12 E E = S linIndS M
10 7 9 syl5bi E E = 0 S linIndS M
11 4 10 syl M LMod E = 0 S linIndS M
12 11 adantr M LMod S 𝒫 B E = 0 S linIndS M
13 12 com12 E = 0 M LMod S 𝒫 B S linIndS M
14 2 lmodring M LMod R Ring
15 14 adantr M LMod S 𝒫 B R Ring
16 eqid 0 R = 0 R
17 3 16 0ring R Ring E = 1 E = 0 R
18 15 17 sylan M LMod S 𝒫 B E = 1 E = 0 R
19 simpr M LMod S 𝒫 B S 𝒫 B
20 19 adantr M LMod S 𝒫 B E = 1 S 𝒫 B
21 20 adantl E = 0 R M LMod S 𝒫 B E = 1 S 𝒫 B
22 snex 0 R V
23 20 22 jctil M LMod S 𝒫 B E = 1 0 R V S 𝒫 B
24 23 adantl E = 0 R M LMod S 𝒫 B E = 1 0 R V S 𝒫 B
25 elmapg 0 R V S 𝒫 B f 0 R S f : S 0 R
26 24 25 syl E = 0 R M LMod S 𝒫 B E = 1 f 0 R S f : S 0 R
27 fvex 0 R V
28 27 fconst2 f : S 0 R f = S × 0 R
29 fconstmpt S × 0 R = x S 0 R
30 29 eqeq2i f = S × 0 R f = x S 0 R
31 28 30 bitri f : S 0 R f = x S 0 R
32 eqidd E = 0 R M LMod S 𝒫 B E = 1 v S x S 0 R = x S 0 R
33 eqidd E = 0 R M LMod S 𝒫 B E = 1 v S x = v 0 R = 0 R
34 simpr E = 0 R M LMod S 𝒫 B E = 1 v S v S
35 fvexd E = 0 R M LMod S 𝒫 B E = 1 v S 0 R V
36 32 33 34 35 fvmptd E = 0 R M LMod S 𝒫 B E = 1 v S x S 0 R v = 0 R
37 36 ralrimiva E = 0 R M LMod S 𝒫 B E = 1 v S x S 0 R v = 0 R
38 37 a1d E = 0 R M LMod S 𝒫 B E = 1 finSupp 0 R x S 0 R x S 0 R linC M S = 0 M v S x S 0 R v = 0 R
39 breq1 f = x S 0 R finSupp 0 R f finSupp 0 R x S 0 R
40 oveq1 f = x S 0 R f linC M S = x S 0 R linC M S
41 40 eqeq1d f = x S 0 R f linC M S = 0 M x S 0 R linC M S = 0 M
42 39 41 anbi12d f = x S 0 R finSupp 0 R f f linC M S = 0 M finSupp 0 R x S 0 R x S 0 R linC M S = 0 M
43 fveq1 f = x S 0 R f v = x S 0 R v
44 43 eqeq1d f = x S 0 R f v = 0 R x S 0 R v = 0 R
45 44 ralbidv f = x S 0 R v S f v = 0 R v S x S 0 R v = 0 R
46 42 45 imbi12d f = x S 0 R finSupp 0 R f f linC M S = 0 M v S f v = 0 R finSupp 0 R x S 0 R x S 0 R linC M S = 0 M v S x S 0 R v = 0 R
47 38 46 syl5ibrcom E = 0 R M LMod S 𝒫 B E = 1 f = x S 0 R finSupp 0 R f f linC M S = 0 M v S f v = 0 R
48 31 47 syl5bi E = 0 R M LMod S 𝒫 B E = 1 f : S 0 R finSupp 0 R f f linC M S = 0 M v S f v = 0 R
49 26 48 sylbid E = 0 R M LMod S 𝒫 B E = 1 f 0 R S finSupp 0 R f f linC M S = 0 M v S f v = 0 R
50 49 ralrimiv E = 0 R M LMod S 𝒫 B E = 1 f 0 R S finSupp 0 R f f linC M S = 0 M v S f v = 0 R
51 oveq1 E = 0 R E S = 0 R S
52 51 raleqdv E = 0 R f E S finSupp 0 R f f linC M S = 0 M v S f v = 0 R f 0 R S finSupp 0 R f f linC M S = 0 M v S f v = 0 R
53 52 adantr E = 0 R M LMod S 𝒫 B E = 1 f E S finSupp 0 R f f linC M S = 0 M v S f v = 0 R f 0 R S finSupp 0 R f f linC M S = 0 M v S f v = 0 R
54 50 53 mpbird E = 0 R M LMod S 𝒫 B E = 1 f E S finSupp 0 R f f linC M S = 0 M v S f v = 0 R
55 simpl M LMod S 𝒫 B E = 1 M LMod S 𝒫 B
56 55 ancomd M LMod S 𝒫 B E = 1 S 𝒫 B M LMod
57 56 adantl E = 0 R M LMod S 𝒫 B E = 1 S 𝒫 B M LMod
58 eqid 0 M = 0 M
59 1 58 2 3 16 islininds S 𝒫 B M LMod S linIndS M S 𝒫 B f E S finSupp 0 R f f linC M S = 0 M v S f v = 0 R
60 57 59 syl E = 0 R M LMod S 𝒫 B E = 1 S linIndS M S 𝒫 B f E S finSupp 0 R f f linC M S = 0 M v S f v = 0 R
61 21 54 60 mpbir2and E = 0 R M LMod S 𝒫 B E = 1 S linIndS M
62 18 61 mpancom M LMod S 𝒫 B E = 1 S linIndS M
63 62 expcom E = 1 M LMod S 𝒫 B S linIndS M
64 13 63 jaoi E = 0 E = 1 M LMod S 𝒫 B S linIndS M
65 64 expd E = 0 E = 1 M LMod S 𝒫 B S linIndS M
66 65 com12 M LMod E = 0 E = 1 S 𝒫 B S linIndS M
67 66 3imp M LMod E = 0 E = 1 S 𝒫 B S linIndS M