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=RevalSub1S
ply1annidl.p P=Poly1R𝑠S
ply1annidl.b B=BaseR
ply1annidl.r φRCRing
ply1annidl.s φSSubRingR
ply1annidl.a φAB
ply1annidl.0 0˙=0R
ply1annidl.q Q=qdomO|OqA=0˙
ply1annidllem.f F=pBasePOpA
Assertion ply1annidllem φQ=F-10˙

Proof

Step Hyp Ref Expression
1 ply1annidl.o O=RevalSub1S
2 ply1annidl.p P=Poly1R𝑠S
3 ply1annidl.b B=BaseR
4 ply1annidl.r φRCRing
5 ply1annidl.s φSSubRingR
6 ply1annidl.a φAB
7 ply1annidl.0 0˙=0R
8 ply1annidl.q Q=qdomO|OqA=0˙
9 ply1annidllem.f F=pBasePOpA
10 nfv pφ
11 fvexd φpBasePOpAV
12 10 11 9 fnmptd φFFnBaseP
13 eqid BaseP=BaseP
14 1 2 13 4 5 evls1fn φOFnBaseP
15 14 fndmd φdomO=BaseP
16 15 fneq2d φFFndomOFFnBaseP
17 12 16 mpbird φFFndomO
18 fniniseg2 FFndomOF-10˙=qdomO|Fq=0˙
19 17 18 syl φF-10˙=qdomO|Fq=0˙
20 fveq2 p=qOp=Oq
21 20 fveq1d p=qOpA=OqA
22 15 eleq2d φqdomOqBaseP
23 22 biimpa φqdomOqBaseP
24 fvexd φqdomOOqAV
25 9 21 23 24 fvmptd3 φqdomOFq=OqA
26 25 eqeq1d φqdomOFq=0˙OqA=0˙
27 26 rabbidva φqdomO|Fq=0˙=qdomO|OqA=0˙
28 19 27 eqtr2d φqdomO|OqA=0˙=F-10˙
29 8 28 eqtrid φQ=F-10˙