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 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 ˙