Metamath Proof Explorer


Theorem coe1mon

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

Ref Expression
Hypotheses ply1moneq.p 𝑃 = ( Poly1𝑅 )
ply1moneq.x 𝑋 = ( var1𝑅 )
ply1moneq.e = ( .g ‘ ( mulGrp ‘ 𝑃 ) )
coe1mon.r ( 𝜑𝑅 ∈ Ring )
coe1mon.n ( 𝜑𝑁 ∈ ℕ0 )
coe1mon.0 0 = ( 0g𝑅 )
coe1mon.1 1 = ( 1r𝑅 )
Assertion coe1mon ( 𝜑 → ( coe1 ‘ ( 𝑁 𝑋 ) ) = ( 𝑘 ∈ ℕ0 ↦ if ( 𝑘 = 𝑁 , 1 , 0 ) ) )

Proof

Step Hyp Ref Expression
1 ply1moneq.p 𝑃 = ( Poly1𝑅 )
2 ply1moneq.x 𝑋 = ( var1𝑅 )
3 ply1moneq.e = ( .g ‘ ( mulGrp ‘ 𝑃 ) )
4 coe1mon.r ( 𝜑𝑅 ∈ Ring )
5 coe1mon.n ( 𝜑𝑁 ∈ ℕ0 )
6 coe1mon.0 0 = ( 0g𝑅 )
7 coe1mon.1 1 = ( 1r𝑅 )
8 1 ply1sca ( 𝑅 ∈ Ring → 𝑅 = ( Scalar ‘ 𝑃 ) )
9 4 8 syl ( 𝜑𝑅 = ( Scalar ‘ 𝑃 ) )
10 9 fveq2d ( 𝜑 → ( 1r𝑅 ) = ( 1r ‘ ( Scalar ‘ 𝑃 ) ) )
11 7 10 eqtrid ( 𝜑1 = ( 1r ‘ ( Scalar ‘ 𝑃 ) ) )
12 11 oveq1d ( 𝜑 → ( 1 ( ·𝑠𝑃 ) ( 𝑁 𝑋 ) ) = ( ( 1r ‘ ( Scalar ‘ 𝑃 ) ) ( ·𝑠𝑃 ) ( 𝑁 𝑋 ) ) )
13 1 ply1lmod ( 𝑅 ∈ Ring → 𝑃 ∈ LMod )
14 4 13 syl ( 𝜑𝑃 ∈ LMod )
15 eqid ( mulGrp ‘ 𝑃 ) = ( mulGrp ‘ 𝑃 )
16 eqid ( Base ‘ 𝑃 ) = ( Base ‘ 𝑃 )
17 1 2 15 3 16 ply1moncl ( ( 𝑅 ∈ Ring ∧ 𝑁 ∈ ℕ0 ) → ( 𝑁 𝑋 ) ∈ ( Base ‘ 𝑃 ) )
18 4 5 17 syl2anc ( 𝜑 → ( 𝑁 𝑋 ) ∈ ( Base ‘ 𝑃 ) )
19 eqid ( Scalar ‘ 𝑃 ) = ( Scalar ‘ 𝑃 )
20 eqid ( ·𝑠𝑃 ) = ( ·𝑠𝑃 )
21 eqid ( 1r ‘ ( Scalar ‘ 𝑃 ) ) = ( 1r ‘ ( Scalar ‘ 𝑃 ) )
22 16 19 20 21 lmodvs1 ( ( 𝑃 ∈ LMod ∧ ( 𝑁 𝑋 ) ∈ ( Base ‘ 𝑃 ) ) → ( ( 1r ‘ ( Scalar ‘ 𝑃 ) ) ( ·𝑠𝑃 ) ( 𝑁 𝑋 ) ) = ( 𝑁 𝑋 ) )
23 14 18 22 syl2anc ( 𝜑 → ( ( 1r ‘ ( Scalar ‘ 𝑃 ) ) ( ·𝑠𝑃 ) ( 𝑁 𝑋 ) ) = ( 𝑁 𝑋 ) )
24 12 23 eqtrd ( 𝜑 → ( 1 ( ·𝑠𝑃 ) ( 𝑁 𝑋 ) ) = ( 𝑁 𝑋 ) )
25 24 fveq2d ( 𝜑 → ( coe1 ‘ ( 1 ( ·𝑠𝑃 ) ( 𝑁 𝑋 ) ) ) = ( coe1 ‘ ( 𝑁 𝑋 ) ) )
26 eqid ( Base ‘ 𝑅 ) = ( Base ‘ 𝑅 )
27 26 7 ringidcl ( 𝑅 ∈ Ring → 1 ∈ ( Base ‘ 𝑅 ) )
28 4 27 syl ( 𝜑1 ∈ ( Base ‘ 𝑅 ) )
29 6 26 1 2 20 15 3 coe1tm ( ( 𝑅 ∈ Ring ∧ 1 ∈ ( Base ‘ 𝑅 ) ∧ 𝑁 ∈ ℕ0 ) → ( coe1 ‘ ( 1 ( ·𝑠𝑃 ) ( 𝑁 𝑋 ) ) ) = ( 𝑘 ∈ ℕ0 ↦ if ( 𝑘 = 𝑁 , 1 , 0 ) ) )
30 4 28 5 29 syl3anc ( 𝜑 → ( coe1 ‘ ( 1 ( ·𝑠𝑃 ) ( 𝑁 𝑋 ) ) ) = ( 𝑘 ∈ ℕ0 ↦ if ( 𝑘 = 𝑁 , 1 , 0 ) ) )
31 25 30 eqtr3d ( 𝜑 → ( coe1 ‘ ( 𝑁 𝑋 ) ) = ( 𝑘 ∈ ℕ0 ↦ if ( 𝑘 = 𝑁 , 1 , 0 ) ) )