Metamath Proof Explorer


Theorem ply1annidl

Description: The set Q of polynomials annihilating an element A forms an ideal. (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 }
Assertion ply1annidl ( 𝜑𝑄 ∈ ( LIdeal ‘ 𝑃 ) )

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 eqid ( 𝑝 ∈ ( Base ‘ 𝑃 ) ↦ ( ( 𝑂𝑝 ) ‘ 𝐴 ) ) = ( 𝑝 ∈ ( Base ‘ 𝑃 ) ↦ ( ( 𝑂𝑝 ) ‘ 𝐴 ) )
10 1 2 3 4 5 6 7 8 9 ply1annidllem ( 𝜑𝑄 = ( ( 𝑝 ∈ ( Base ‘ 𝑃 ) ↦ ( ( 𝑂𝑝 ) ‘ 𝐴 ) ) “ { 0 } ) )
11 eqid ( Base ‘ 𝑃 ) = ( Base ‘ 𝑃 )
12 1 2 3 11 4 5 6 9 evls1maprhm ( 𝜑 → ( 𝑝 ∈ ( Base ‘ 𝑃 ) ↦ ( ( 𝑂𝑝 ) ‘ 𝐴 ) ) ∈ ( 𝑃 RingHom 𝑅 ) )
13 4 crngringd ( 𝜑𝑅 ∈ Ring )
14 eqid ( LIdeal ‘ 𝑅 ) = ( LIdeal ‘ 𝑅 )
15 14 7 lidl0 ( 𝑅 ∈ Ring → { 0 } ∈ ( LIdeal ‘ 𝑅 ) )
16 13 15 syl ( 𝜑 → { 0 } ∈ ( LIdeal ‘ 𝑅 ) )
17 eqid ( LIdeal ‘ 𝑃 ) = ( LIdeal ‘ 𝑃 )
18 17 rhmpreimaidl ( ( ( 𝑝 ∈ ( Base ‘ 𝑃 ) ↦ ( ( 𝑂𝑝 ) ‘ 𝐴 ) ) ∈ ( 𝑃 RingHom 𝑅 ) ∧ { 0 } ∈ ( LIdeal ‘ 𝑅 ) ) → ( ( 𝑝 ∈ ( Base ‘ 𝑃 ) ↦ ( ( 𝑂𝑝 ) ‘ 𝐴 ) ) “ { 0 } ) ∈ ( LIdeal ‘ 𝑃 ) )
19 12 16 18 syl2anc ( 𝜑 → ( ( 𝑝 ∈ ( Base ‘ 𝑃 ) ↦ ( ( 𝑂𝑝 ) ‘ 𝐴 ) ) “ { 0 } ) ∈ ( LIdeal ‘ 𝑃 ) )
20 10 19 eqeltrd ( 𝜑𝑄 ∈ ( LIdeal ‘ 𝑃 ) )