Metamath Proof Explorer


Theorem ply1idvr1

Description: The identity of a polynomial ring expressed as power of the polynomial variable. (Contributed by AV, 14-Aug-2019)

Ref Expression
Hypotheses ply1idvr1.p P = Poly 1 R
ply1idvr1.x X = var 1 R
ply1idvr1.n N = mulGrp P
ply1idvr1.e × ˙ = N
Assertion ply1idvr1 R Ring 0 × ˙ X = 1 P

Proof

Step Hyp Ref Expression
1 ply1idvr1.p P = Poly 1 R
2 ply1idvr1.x X = var 1 R
3 ply1idvr1.n N = mulGrp P
4 ply1idvr1.e × ˙ = N
5 eqid Base R = Base R
6 eqid 1 R = 1 R
7 5 6 ringidcl R Ring 1 R Base R
8 eqid P = P
9 eqid algSc P = algSc P
10 5 1 2 8 3 4 9 ply1scltm R Ring 1 R Base R algSc P 1 R = 1 R P 0 × ˙ X
11 7 10 mpdan R Ring algSc P 1 R = 1 R P 0 × ˙ X
12 1 ply1sca R Ring R = Scalar P
13 12 fveq2d R Ring 1 R = 1 Scalar P
14 13 oveq1d R Ring 1 R P 0 × ˙ X = 1 Scalar P P 0 × ˙ X
15 1 ply1lmod R Ring P LMod
16 0nn0 0 0
17 eqid Base P = Base P
18 1 2 3 4 17 ply1moncl R Ring 0 0 0 × ˙ X Base P
19 16 18 mpan2 R Ring 0 × ˙ X Base P
20 eqid Scalar P = Scalar P
21 eqid 1 Scalar P = 1 Scalar P
22 17 20 8 21 lmodvs1 P LMod 0 × ˙ X Base P 1 Scalar P P 0 × ˙ X = 0 × ˙ X
23 15 19 22 syl2anc R Ring 1 Scalar P P 0 × ˙ X = 0 × ˙ X
24 11 14 23 3eqtrrd R Ring 0 × ˙ X = algSc P 1 R
25 eqid 1 P = 1 P
26 1 9 6 25 ply1scl1 R Ring algSc P 1 R = 1 P
27 24 26 eqtrd R Ring 0 × ˙ X = 1 P