Metamath Proof Explorer


Theorem minplyann

Description: The minimal polynomial for A annihilates A (Contributed by Thierry Arnoux, 25-Apr-2025)

Ref Expression
Hypotheses ply1annig1p.o O = E evalSub 1 F
ply1annig1p.p P = Poly 1 E 𝑠 F
ply1annig1p.b B = Base E
ply1annig1p.e φ E Field
ply1annig1p.f φ F SubDRing E
ply1annig1p.a φ A B
minplyann.1 0 ˙ = 0 E
minplyann.m M = E minPoly F
Assertion minplyann φ O M A A = 0 ˙

Proof

Step Hyp Ref Expression
1 ply1annig1p.o O = E evalSub 1 F
2 ply1annig1p.p P = Poly 1 E 𝑠 F
3 ply1annig1p.b B = Base E
4 ply1annig1p.e φ E Field
5 ply1annig1p.f φ F SubDRing E
6 ply1annig1p.a φ A B
7 minplyann.1 0 ˙ = 0 E
8 minplyann.m M = E minPoly F
9 eqid q dom O | O q A = 0 ˙ = q dom O | O q A = 0 ˙
10 eqid RSpan P = RSpan P
11 eqid idlGen 1p E 𝑠 F = idlGen 1p E 𝑠 F
12 1 2 3 4 5 6 7 9 10 11 8 minplyval φ M A = idlGen 1p E 𝑠 F q dom O | O q A = 0 ˙
13 eqid E 𝑠 F = E 𝑠 F
14 13 sdrgdrng F SubDRing E E 𝑠 F DivRing
15 5 14 syl φ E 𝑠 F DivRing
16 4 fldcrngd φ E CRing
17 sdrgsubrg F SubDRing E F SubRing E
18 5 17 syl φ F SubRing E
19 1 2 3 16 18 6 7 9 ply1annidl φ q dom O | O q A = 0 ˙ LIdeal P
20 eqid LIdeal P = LIdeal P
21 2 11 20 ig1pcl E 𝑠 F DivRing q dom O | O q A = 0 ˙ LIdeal P idlGen 1p E 𝑠 F q dom O | O q A = 0 ˙ q dom O | O q A = 0 ˙
22 15 19 21 syl2anc φ idlGen 1p E 𝑠 F q dom O | O q A = 0 ˙ q dom O | O q A = 0 ˙
23 12 22 eqeltrd φ M A q dom O | O q A = 0 ˙
24 fveq2 q = M A O q = O M A
25 24 fveq1d q = M A O q A = O M A A
26 25 eqeq1d q = M A O q A = 0 ˙ O M A A = 0 ˙
27 26 elrab M A q dom O | O q A = 0 ˙ M A dom O O M A A = 0 ˙
28 23 27 sylib φ M A dom O O M A A = 0 ˙
29 28 simprd φ O M A A = 0 ˙