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 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 ˙
Assertion ply1annidl φ Q LIdeal P

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 eqid p Base P O p A = p Base P O p A
10 1 2 3 4 5 6 7 8 9 ply1annidllem φ Q = p Base P O p A -1 0 ˙
11 eqid Base P = Base P
12 1 2 3 11 4 5 6 9 evls1maprhm φ p Base P O p A P RingHom R
13 4 crngringd φ R 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 O p A P RingHom R 0 ˙ LIdeal R p Base P O p A -1 0 ˙ LIdeal P
19 12 16 18 syl2anc φ p Base P O p A -1 0 ˙ LIdeal P
20 10 19 eqeltrd φ Q LIdeal P