Metamath Proof Explorer


Theorem coe1mon

Description: Coefficient vector of a monomial. (Contributed by Thierry Arnoux, 20-Feb-2025)

Ref Expression
Hypotheses ply1moneq.p P = Poly 1 R
ply1moneq.x X = var 1 R
ply1moneq.e × ˙ = mulGrp P
coe1mon.r φ R Ring
coe1mon.n φ N 0
coe1mon.0 0 ˙ = 0 R
coe1mon.1 1 ˙ = 1 R
Assertion coe1mon φ coe 1 N × ˙ X = k 0 if k = N 1 ˙ 0 ˙

Proof

Step Hyp Ref Expression
1 ply1moneq.p P = Poly 1 R
2 ply1moneq.x X = var 1 R
3 ply1moneq.e × ˙ = mulGrp P
4 coe1mon.r φ R Ring
5 coe1mon.n φ N 0
6 coe1mon.0 0 ˙ = 0 R
7 coe1mon.1 1 ˙ = 1 R
8 1 ply1sca R Ring R = Scalar P
9 4 8 syl φ R = Scalar P
10 9 fveq2d φ 1 R = 1 Scalar P
11 7 10 eqtrid φ 1 ˙ = 1 Scalar P
12 11 oveq1d φ 1 ˙ P N × ˙ X = 1 Scalar P P N × ˙ X
13 1 ply1lmod R Ring P LMod
14 4 13 syl φ P LMod
15 eqid mulGrp P = mulGrp P
16 eqid Base P = Base P
17 1 2 15 3 16 ply1moncl R Ring N 0 N × ˙ X Base P
18 4 5 17 syl2anc φ N × ˙ X Base P
19 eqid Scalar P = Scalar P
20 eqid P = P
21 eqid 1 Scalar P = 1 Scalar P
22 16 19 20 21 lmodvs1 P LMod N × ˙ X Base P 1 Scalar P P N × ˙ X = N × ˙ X
23 14 18 22 syl2anc φ 1 Scalar P P N × ˙ X = N × ˙ X
24 12 23 eqtrd φ 1 ˙ P N × ˙ X = N × ˙ X
25 24 fveq2d φ coe 1 1 ˙ P N × ˙ X = coe 1 N × ˙ X
26 eqid Base R = Base R
27 26 7 ringidcl R Ring 1 ˙ Base R
28 4 27 syl φ 1 ˙ Base R
29 6 26 1 2 20 15 3 coe1tm R Ring 1 ˙ Base R N 0 coe 1 1 ˙ P N × ˙ X = k 0 if k = N 1 ˙ 0 ˙
30 4 28 5 29 syl3anc φ coe 1 1 ˙ P N × ˙ X = k 0 if k = N 1 ˙ 0 ˙
31 25 30 eqtr3d φ coe 1 N × ˙ X = k 0 if k = N 1 ˙ 0 ˙