Metamath Proof Explorer


Theorem asclply1subcl

Description: Closure of the algebra scalar injection function in a polynomial on a subring. (Contributed by Thierry Arnoux, 5-Feb-2025)

Ref Expression
Hypotheses asclply1subcl.1 A = algSc V
asclply1subcl.2 U = R 𝑠 S
asclply1subcl.3 V = Poly 1 R
asclply1subcl.4 W = Poly 1 U
asclply1subcl.5 P = Base W
asclply1subcl.6 φ S SubRing R
asclply1subcl.7 φ Z S
Assertion asclply1subcl φ A Z P

Proof

Step Hyp Ref Expression
1 asclply1subcl.1 A = algSc V
2 asclply1subcl.2 U = R 𝑠 S
3 asclply1subcl.3 V = Poly 1 R
4 asclply1subcl.4 W = Poly 1 U
5 asclply1subcl.5 P = Base W
6 asclply1subcl.6 φ S SubRing R
7 asclply1subcl.7 φ Z S
8 eqid Base R = Base R
9 8 subrgss S SubRing R S Base R
10 6 9 syl φ S Base R
11 10 7 sseldd φ Z Base R
12 subrgrcl S SubRing R R Ring
13 3 ply1sca R Ring R = Scalar V
14 6 12 13 3syl φ R = Scalar V
15 14 fveq2d φ Base R = Base Scalar V
16 11 15 eleqtrd φ Z Base Scalar V
17 eqid Scalar V = Scalar V
18 eqid Base Scalar V = Base Scalar V
19 eqid V = V
20 eqid 1 V = 1 V
21 1 17 18 19 20 asclval Z Base Scalar V A Z = Z V 1 V
22 16 21 syl φ A Z = Z V 1 V
23 3 2 4 5 subrgply1 S SubRing R P SubRing V
24 eqid V 𝑠 P = V 𝑠 P
25 24 19 ressvsca P SubRing V V = V 𝑠 P
26 6 23 25 3syl φ V = V 𝑠 P
27 26 oveqd φ Z V 1 V = Z V 𝑠 P 1 V
28 id φ φ
29 20 subrg1cl P SubRing V 1 V P
30 6 23 29 3syl φ 1 V P
31 3 2 4 5 6 24 ressply1vsca φ Z S 1 V P Z W 1 V = Z V 𝑠 P 1 V
32 28 7 30 31 syl12anc φ Z W 1 V = Z V 𝑠 P 1 V
33 27 32 eqtr4d φ Z V 1 V = Z W 1 V
34 2 subrgring S SubRing R U Ring
35 4 ply1lmod U Ring W LMod
36 6 34 35 3syl φ W LMod
37 2 8 ressbas2 S Base R S = Base U
38 6 9 37 3syl φ S = Base U
39 7 38 eleqtrd φ Z Base U
40 2 ovexi U V
41 4 ply1sca U V U = Scalar W
42 40 41 ax-mp U = Scalar W
43 eqid W = W
44 eqid Base U = Base U
45 5 42 43 44 lmodvscl W LMod Z Base U 1 V P Z W 1 V P
46 36 39 30 45 syl3anc φ Z W 1 V P
47 33 46 eqeltrd φ Z V 1 V P
48 22 47 eqeltrd φ A Z P