Metamath Proof Explorer


Theorem subrgascl

Description: The scalar injection function in a subring algebra is the same up to a restriction to the subring. (Contributed by Mario Carneiro, 4-Jul-2015)

Ref Expression
Hypotheses subrgascl.p ⊢ P = I mPoly R
subrgascl.a ⊢ A = algSc ⁡ P
subrgascl.h ⊢ H = R ↾ 𝑠 T
subrgascl.u ⊢ U = I mPoly H
subrgascl.i ⊢ φ → I ∈ W
subrgascl.r ⊢ φ → T ∈ SubRing ⁡ R
subrgascl.c ⊢ C = algSc ⁡ U
Assertion subrgascl ⊢ φ → C = A ↾ T

Proof

Step Hyp Ref Expression
1 subrgascl.p ⊢ P = I mPoly R
2 subrgascl.a ⊢ A = algSc ⁡ P
3 subrgascl.h ⊢ H = R ↾ 𝑠 T
4 subrgascl.u ⊢ U = I mPoly H
5 subrgascl.i ⊢ φ → I ∈ W
6 subrgascl.r ⊢ φ → T ∈ SubRing ⁡ R
7 subrgascl.c ⊢ C = algSc ⁡ U
8 eqid ⊢ Scalar ⁡ U = Scalar ⁡ U
9 eqid ⊢ Base Scalar ⁡ U = Base Scalar ⁡ U
10 7 8 9 asclfn ⊢ C Fn Base Scalar ⁡ U
11 3 subrgbas ⊢ T ∈ SubRing ⁡ R → T = Base H
12 6 11 syl ⊢ φ → T = Base H
13 3 ovexi ⊢ H ∈ V
14 13 a1i ⊢ φ → H ∈ V
15 4 5 14 mplsca ⊢ φ → H = Scalar ⁡ U
16 15 fveq2d ⊢ φ → Base H = Base Scalar ⁡ U
17 12 16 eqtrd ⊢ φ → T = Base Scalar ⁡ U
18 17 fneq2d ⊢ φ → C Fn T ↔ C Fn Base Scalar ⁡ U
19 10 18 mpbiri ⊢ φ → C Fn T
20 eqid ⊢ Scalar ⁡ P = Scalar ⁡ P
21 eqid ⊢ Base Scalar ⁡ P = Base Scalar ⁡ P
22 2 20 21 asclfn ⊢ A Fn Base Scalar ⁡ P
23 subrgrcl ⊢ T ∈ SubRing ⁡ R → R ∈ Ring
24 6 23 syl ⊢ φ → R ∈ Ring
25 1 5 24 mplsca ⊢ φ → R = Scalar ⁡ P
26 25 fveq2d ⊢ φ → Base R = Base Scalar ⁡ P
27 26 fneq2d ⊢ φ → A Fn Base R ↔ A Fn Base Scalar ⁡ P
28 22 27 mpbiri ⊢ φ → A Fn Base R
29 eqid ⊢ Base R = Base R
30 29 subrgss ⊢ T ∈ SubRing ⁡ R → T ⊆ Base R
31 6 30 syl ⊢ φ → T ⊆ Base R
32 fnssres ⊢ A Fn Base R ∧ T ⊆ Base R → A ↾ T Fn T
33 28 31 32 syl2anc ⊢ φ → A ↾ T Fn T
34 fvres ⊢ x ∈ T → A ↾ T ⁡ x = A ⁡ x
35 34 adantl ⊢ φ ∧ x ∈ T → A ↾ T ⁡ x = A ⁡ x
36 eqid ⊢ 0 R = 0 R
37 3 36 subrg0 ⊢ T ∈ SubRing ⁡ R → 0 R = 0 H
38 6 37 syl ⊢ φ → 0 R = 0 H
39 38 ifeq2d ⊢ φ → if y = I × 0 x 0 R = if y = I × 0 x 0 H
40 39 adantr ⊢ φ ∧ x ∈ T → if y = I × 0 x 0 R = if y = I × 0 x 0 H
41 40 mpteq2dv ⊢ φ ∧ x ∈ T → y ∈ f ∈ ℕ 0 I | f -1 ℕ ∈ Fin ⟼ if y = I × 0 x 0 R = y ∈ f ∈ ℕ 0 I | f -1 ℕ ∈ Fin ⟼ if y = I × 0 x 0 H
42 eqid ⊢ f ∈ ℕ 0 I | f -1 ℕ ∈ Fin = f ∈ ℕ 0 I | f -1 ℕ ∈ Fin
43 5 adantr ⊢ φ ∧ x ∈ T → I ∈ W
44 24 adantr ⊢ φ ∧ x ∈ T → R ∈ Ring
45 31 sselda ⊢ φ ∧ x ∈ T → x ∈ Base R
46 1 42 36 29 2 43 44 45 mplascl ⊢ φ ∧ x ∈ T → A ⁡ x = y ∈ f ∈ ℕ 0 I | f -1 ℕ ∈ Fin ⟼ if y = I × 0 x 0 R
47 eqid ⊢ 0 H = 0 H
48 eqid ⊢ Base H = Base H
49 3 subrgring ⊢ T ∈ SubRing ⁡ R → H ∈ Ring
50 6 49 syl ⊢ φ → H ∈ Ring
51 50 adantr ⊢ φ ∧ x ∈ T → H ∈ Ring
52 12 eleq2d ⊢ φ → x ∈ T ↔ x ∈ Base H
53 52 biimpa ⊢ φ ∧ x ∈ T → x ∈ Base H
54 4 42 47 48 7 43 51 53 mplascl ⊢ φ ∧ x ∈ T → C ⁡ x = y ∈ f ∈ ℕ 0 I | f -1 ℕ ∈ Fin ⟼ if y = I × 0 x 0 H
55 41 46 54 3eqtr4d ⊢ φ ∧ x ∈ T → A ⁡ x = C ⁡ x
56 35 55 eqtr2d ⊢ φ ∧ x ∈ T → C ⁡ x = A ↾ T ⁡ x
57 19 33 56 eqfnfvd ⊢ φ → C = A ↾ T