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=algScV
asclply1subcl.2 U=R𝑠S
asclply1subcl.3 V=Poly1R
asclply1subcl.4 W=Poly1U
asclply1subcl.5 P=BaseW
asclply1subcl.6 φSSubRingR
asclply1subcl.7 φZS
Assertion asclply1subcl φAZP

Proof

Step Hyp Ref Expression
1 asclply1subcl.1 A=algScV
2 asclply1subcl.2 U=R𝑠S
3 asclply1subcl.3 V=Poly1R
4 asclply1subcl.4 W=Poly1U
5 asclply1subcl.5 P=BaseW
6 asclply1subcl.6 φSSubRingR
7 asclply1subcl.7 φZS
8 eqid BaseR=BaseR
9 8 subrgss SSubRingRSBaseR
10 6 9 syl φSBaseR
11 10 7 sseldd φZBaseR
12 subrgrcl SSubRingRRRing
13 3 ply1sca RRingR=ScalarV
14 6 12 13 3syl φR=ScalarV
15 14 fveq2d φBaseR=BaseScalarV
16 11 15 eleqtrd φZBaseScalarV
17 eqid ScalarV=ScalarV
18 eqid BaseScalarV=BaseScalarV
19 eqid V=V
20 eqid 1V=1V
21 1 17 18 19 20 asclval ZBaseScalarVAZ=ZV1V
22 16 21 syl φAZ=ZV1V
23 3 2 4 5 subrgply1 SSubRingRPSubRingV
24 eqid V𝑠P=V𝑠P
25 24 19 ressvsca PSubRingVV=V𝑠P
26 6 23 25 3syl φV=V𝑠P
27 26 oveqd φZV1V=ZV𝑠P1V
28 id φφ
29 20 subrg1cl PSubRingV1VP
30 6 23 29 3syl φ1VP
31 3 2 4 5 6 24 ressply1vsca φZS1VPZW1V=ZV𝑠P1V
32 28 7 30 31 syl12anc φZW1V=ZV𝑠P1V
33 27 32 eqtr4d φZV1V=ZW1V
34 2 subrgring SSubRingRURing
35 4 ply1lmod URingWLMod
36 6 34 35 3syl φWLMod
37 2 8 ressbas2 SBaseRS=BaseU
38 6 9 37 3syl φS=BaseU
39 7 38 eleqtrd φZBaseU
40 2 ovexi UV
41 4 ply1sca UVU=ScalarW
42 40 41 ax-mp U=ScalarW
43 eqid W=W
44 eqid BaseU=BaseU
45 5 42 43 44 lmodvscl WLModZBaseU1VPZW1VP
46 36 39 30 45 syl3anc φZW1VP
47 33 46 eqeltrd φZV1VP
48 22 47 eqeltrd φAZP