Metamath Proof Explorer


Theorem ply1annidllem

Description: Write the set Q of polynomials annihilating an element A as the kernel of the ring homomorphism F mapping polynomials p to their subring evaluation at a given point A . (Contributed by Thierry Arnoux, 9-Feb-2025)

Ref Expression
Hypotheses ply1annidl.o ⊢ O = R evalSub 1 S
ply1annidl.p ⊢ P = Poly 1 ⁡ R ↾ 𝑠 S
ply1annidl.b ⊢ B = Base R
ply1annidl.r ⊢ φ → R ∈ CRing
ply1annidl.s ⊢ φ → S ∈ SubRing ⁡ R
ply1annidl.a ⊢ φ → A ∈ B
ply1annidl.0 ⊢ 0 ˙ = 0 R
ply1annidl.q ⊢ Q = q ∈ dom ⁡ O | O ⁡ q ⁡ A = 0 ˙
ply1annidllem.f ⊢ F = p ∈ Base P ⟼ O ⁡ p ⁡ A
Assertion ply1annidllem ⊢ φ → Q = F -1 0 ˙

Proof

Step Hyp Ref Expression
1 ply1annidl.o ⊢ O = R evalSub 1 S
2 ply1annidl.p ⊢ P = Poly 1 ⁡ R ↾ 𝑠 S
3 ply1annidl.b ⊢ B = Base R
4 ply1annidl.r ⊢ φ → R ∈ CRing
5 ply1annidl.s ⊢ φ → S ∈ SubRing ⁡ R
6 ply1annidl.a ⊢ φ → A ∈ B
7 ply1annidl.0 ⊢ 0 ˙ = 0 R
8 ply1annidl.q ⊢ Q = q ∈ dom ⁡ O | O ⁡ q ⁡ A = 0 ˙
9 ply1annidllem.f ⊢ F = p ∈ Base P ⟼ O ⁡ p ⁡ A
10 nfv ⊢ Ⅎ p φ
11 fvexd ⊢ φ ∧ p ∈ Base P → O ⁡ p ⁡ A ∈ V
12 10 11 9 fnmptd ⊢ φ → F Fn Base P
13 eqid ⊢ Base P = Base P
14 1 2 13 4 5 evls1fn ⊢ φ → O Fn Base P
15 14 fndmd ⊢ φ → dom ⁡ O = Base P
16 15 fneq2d ⊢ φ → F Fn dom ⁡ O ↔ F Fn Base P
17 12 16 mpbird ⊢ φ → F Fn dom ⁡ O
18 fniniseg2 ⊢ F Fn dom ⁡ O → F -1 0 ˙ = q ∈ dom ⁡ O | F ⁡ q = 0 ˙
19 17 18 syl ⊢ φ → F -1 0 ˙ = q ∈ dom ⁡ O | F ⁡ q = 0 ˙
20 fveq2 ⊢ p = q → O ⁡ p = O ⁡ q
21 20 fveq1d ⊢ p = q → O ⁡ p ⁡ A = O ⁡ q ⁡ A
22 15 eleq2d ⊢ φ → q ∈ dom ⁡ O ↔ q ∈ Base P
23 22 biimpa ⊢ φ ∧ q ∈ dom ⁡ O → q ∈ Base P
24 fvexd ⊢ φ ∧ q ∈ dom ⁡ O → O ⁡ q ⁡ A ∈ V
25 9 21 23 24 fvmptd3 ⊢ φ ∧ q ∈ dom ⁡ O → F ⁡ q = O ⁡ q ⁡ A
26 25 eqeq1d ⊢ φ ∧ q ∈ dom ⁡ O → F ⁡ q = 0 ˙ ↔ O ⁡ q ⁡ A = 0 ˙
27 26 rabbidva ⊢ φ → q ∈ dom ⁡ O | F ⁡ q = 0 ˙ = q ∈ dom ⁡ O | O ⁡ q ⁡ A = 0 ˙
28 19 27 eqtr2d ⊢ φ → q ∈ dom ⁡ O | O ⁡ q ⁡ A = 0 ˙ = F -1 0 ˙
29 8 28 eqtrid ⊢ φ → Q = F -1 0 ˙