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 evalSub1 𝑆 )
ply1annidl.p 𝑃 = ( Poly1 ‘ ( 𝑅s 𝑆 ) )
ply1annidl.b 𝐵 = ( Base ‘ 𝑅 )
ply1annidl.r ( 𝜑𝑅 ∈ CRing )
ply1annidl.s ( 𝜑𝑆 ∈ ( SubRing ‘ 𝑅 ) )
ply1annidl.a ( 𝜑𝐴𝐵 )
ply1annidl.0 0 = ( 0g𝑅 )
ply1annidl.q 𝑄 = { 𝑞 ∈ dom 𝑂 ∣ ( ( 𝑂𝑞 ) ‘ 𝐴 ) = 0 }
ply1annidllem.f 𝐹 = ( 𝑝 ∈ ( Base ‘ 𝑃 ) ↦ ( ( 𝑂𝑝 ) ‘ 𝐴 ) )
Assertion ply1annidllem ( 𝜑𝑄 = ( 𝐹 “ { 0 } ) )

Proof

Step Hyp Ref Expression
1 ply1annidl.o 𝑂 = ( 𝑅 evalSub1 𝑆 )
2 ply1annidl.p 𝑃 = ( Poly1 ‘ ( 𝑅s 𝑆 ) )
3 ply1annidl.b 𝐵 = ( Base ‘ 𝑅 )
4 ply1annidl.r ( 𝜑𝑅 ∈ CRing )
5 ply1annidl.s ( 𝜑𝑆 ∈ ( SubRing ‘ 𝑅 ) )
6 ply1annidl.a ( 𝜑𝐴𝐵 )
7 ply1annidl.0 0 = ( 0g𝑅 )
8 ply1annidl.q 𝑄 = { 𝑞 ∈ dom 𝑂 ∣ ( ( 𝑂𝑞 ) ‘ 𝐴 ) = 0 }
9 ply1annidllem.f 𝐹 = ( 𝑝 ∈ ( Base ‘ 𝑃 ) ↦ ( ( 𝑂𝑝 ) ‘ 𝐴 ) )
10 nfv 𝑝 𝜑
11 fvexd ( ( 𝜑𝑝 ∈ ( Base ‘ 𝑃 ) ) → ( ( 𝑂𝑝 ) ‘ 𝐴 ) ∈ V )
12 10 11 9 fnmptd ( 𝜑𝐹 Fn ( Base ‘ 𝑃 ) )
13 eqid ( Base ‘ 𝑃 ) = ( Base ‘ 𝑃 )
14 1 2 13 4 5 evls1fn ( 𝜑𝑂 Fn ( Base ‘ 𝑃 ) )
15 14 fndmd ( 𝜑 → dom 𝑂 = ( Base ‘ 𝑃 ) )
16 15 fneq2d ( 𝜑 → ( 𝐹 Fn dom 𝑂𝐹 Fn ( Base ‘ 𝑃 ) ) )
17 12 16 mpbird ( 𝜑𝐹 Fn dom 𝑂 )
18 fniniseg2 ( 𝐹 Fn dom 𝑂 → ( 𝐹 “ { 0 } ) = { 𝑞 ∈ dom 𝑂 ∣ ( 𝐹𝑞 ) = 0 } )
19 17 18 syl ( 𝜑 → ( 𝐹 “ { 0 } ) = { 𝑞 ∈ dom 𝑂 ∣ ( 𝐹𝑞 ) = 0 } )
20 fveq2 ( 𝑝 = 𝑞 → ( 𝑂𝑝 ) = ( 𝑂𝑞 ) )
21 20 fveq1d ( 𝑝 = 𝑞 → ( ( 𝑂𝑝 ) ‘ 𝐴 ) = ( ( 𝑂𝑞 ) ‘ 𝐴 ) )
22 15 eleq2d ( 𝜑 → ( 𝑞 ∈ dom 𝑂𝑞 ∈ ( Base ‘ 𝑃 ) ) )
23 22 biimpa ( ( 𝜑𝑞 ∈ dom 𝑂 ) → 𝑞 ∈ ( Base ‘ 𝑃 ) )
24 fvexd ( ( 𝜑𝑞 ∈ dom 𝑂 ) → ( ( 𝑂𝑞 ) ‘ 𝐴 ) ∈ V )
25 9 21 23 24 fvmptd3 ( ( 𝜑𝑞 ∈ dom 𝑂 ) → ( 𝐹𝑞 ) = ( ( 𝑂𝑞 ) ‘ 𝐴 ) )
26 25 eqeq1d ( ( 𝜑𝑞 ∈ dom 𝑂 ) → ( ( 𝐹𝑞 ) = 0 ↔ ( ( 𝑂𝑞 ) ‘ 𝐴 ) = 0 ) )
27 26 rabbidva ( 𝜑 → { 𝑞 ∈ dom 𝑂 ∣ ( 𝐹𝑞 ) = 0 } = { 𝑞 ∈ dom 𝑂 ∣ ( ( 𝑂𝑞 ) ‘ 𝐴 ) = 0 } )
28 19 27 eqtr2d ( 𝜑 → { 𝑞 ∈ dom 𝑂 ∣ ( ( 𝑂𝑞 ) ‘ 𝐴 ) = 0 } = ( 𝐹 “ { 0 } ) )
29 8 28 eqtrid ( 𝜑𝑄 = ( 𝐹 “ { 0 } ) )