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 biimtrid ⊢ 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 biimtrid ⊢ 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