Metamath Proof Explorer


Theorem minplyelirng

Description: If the minimial polynomial F of an element X of a field R has nonnegative degree, then X is integral. (Contributed by Thierry Arnoux, 26-Oct-2025)

Ref Expression
Hypotheses minplyelirng.b B = Base R
minplyelirng.m M = R minPoly S
minplyelirng.d D = deg 1 R 𝑠 S
minplyelirng.r φ R Field
minplyelirng.s φ S SubDRing R
minplyelirng.a φ A B
minplyelirng.1 φ D M A 0
Assertion minplyelirng φ A R IntgRing S

Proof

Step Hyp Ref Expression
1 minplyelirng.b B = Base R
2 minplyelirng.m M = R minPoly S
3 minplyelirng.d D = deg 1 R 𝑠 S
4 minplyelirng.r φ R Field
5 minplyelirng.s φ S SubDRing R
6 minplyelirng.a φ A B
7 minplyelirng.1 φ D M A 0
8 fveq2 m = M A R evalSub 1 S m = R evalSub 1 S M A
9 8 fveq1d m = M A R evalSub 1 S m A = R evalSub 1 S M A A
10 9 eqeq1d m = M A R evalSub 1 S m A = 0 R R evalSub 1 S M A A = 0 R
11 eqid 0 Poly 1 R = 0 Poly 1 R
12 sdrgsubrg S SubDRing R S SubRing R
13 5 12 syl φ S SubRing R
14 eqid R 𝑠 S = R 𝑠 S
15 14 subrgring S SubRing R R 𝑠 S Ring
16 13 15 syl φ R 𝑠 S Ring
17 eqid R evalSub 1 S = R evalSub 1 S
18 eqid Poly 1 R 𝑠 S = Poly 1 R 𝑠 S
19 eqid 0 R = 0 R
20 eqid q dom R evalSub 1 S | R evalSub 1 S q A = 0 R = q dom R evalSub 1 S | R evalSub 1 S q A = 0 R
21 eqid RSpan Poly 1 R 𝑠 S = RSpan Poly 1 R 𝑠 S
22 eqid idlGen 1p R 𝑠 S = idlGen 1p R 𝑠 S
23 17 18 1 4 5 6 19 20 21 22 2 minplycl φ M A Base Poly 1 R 𝑠 S
24 eqid 0 Poly 1 R 𝑠 S = 0 Poly 1 R 𝑠 S
25 eqid Base Poly 1 R 𝑠 S = Base Poly 1 R 𝑠 S
26 3 18 24 25 deg1nn0clb R 𝑠 S Ring M A Base Poly 1 R 𝑠 S M A 0 Poly 1 R 𝑠 S D M A 0
27 26 biimpar R 𝑠 S Ring M A Base Poly 1 R 𝑠 S D M A 0 M A 0 Poly 1 R 𝑠 S
28 16 23 7 27 syl21anc φ M A 0 Poly 1 R 𝑠 S
29 eqid Poly 1 R = Poly 1 R
30 29 14 18 25 13 11 ressply10g φ 0 Poly 1 R = 0 Poly 1 R 𝑠 S
31 28 30 neeqtrrd φ M A 0 Poly 1 R
32 eqid Monic 1p R 𝑠 S = Monic 1p R 𝑠 S
33 1 11 4 5 2 6 31 32 minplynzm1p φ M A Monic 1p R 𝑠 S
34 17 18 1 4 5 6 19 2 minplyann φ R evalSub 1 S M A A = 0 R
35 10 33 34 rspcedvdw φ m Monic 1p R 𝑠 S R evalSub 1 S m A = 0 R
36 4 fldcrngd φ R CRing
37 17 14 1 19 36 13 elirng φ A R IntgRing S A B m Monic 1p R 𝑠 S R evalSub 1 S m A = 0 R
38 6 35 37 mpbir2and φ A R IntgRing S