Metamath Proof Explorer


Theorem elirng

Description: Property for an element X of a field R to be integral over a subring S . (Contributed by Thierry Arnoux, 28-Jan-2025)

Ref Expression
Hypotheses irngval.o O = R evalSub 1 S
irngval.u U = R 𝑠 S
irngval.b B = Base R
irngval.0 0 ˙ = 0 R
elirng.r φ R CRing
elirng.s φ S SubRing R
Assertion elirng φ X R IntgRing S X B f Monic 1p U O f X = 0 ˙

Proof

Step Hyp Ref Expression
1 irngval.o O = R evalSub 1 S
2 irngval.u U = R 𝑠 S
3 irngval.b B = Base R
4 irngval.0 0 ˙ = 0 R
5 elirng.r φ R CRing
6 elirng.s φ S SubRing R
7 5 crngringd φ R Ring
8 3 subrgss S SubRing R S B
9 6 8 syl φ S B
10 1 2 3 4 7 9 irngval φ R IntgRing S = f Monic 1p U O f -1 0 ˙
11 10 eleq2d φ X R IntgRing S X f Monic 1p U O f -1 0 ˙
12 eliun X f Monic 1p U O f -1 0 ˙ f Monic 1p U X O f -1 0 ˙
13 11 12 bitrdi φ X R IntgRing S f Monic 1p U X O f -1 0 ˙
14 eqid R 𝑠 B = R 𝑠 B
15 eqid Base R 𝑠 B = Base R 𝑠 B
16 7 adantr φ f Monic 1p U R Ring
17 3 fvexi B V
18 17 a1i φ f Monic 1p U B V
19 eqid Poly 1 U = Poly 1 U
20 1 3 14 2 19 evls1rhm R CRing S SubRing R O Poly 1 U RingHom R 𝑠 B
21 5 6 20 syl2anc φ O Poly 1 U RingHom R 𝑠 B
22 eqid Base Poly 1 U = Base Poly 1 U
23 22 15 rhmf O Poly 1 U RingHom R 𝑠 B O : Base Poly 1 U Base R 𝑠 B
24 21 23 syl φ O : Base Poly 1 U Base R 𝑠 B
25 24 adantr φ f Monic 1p U O : Base Poly 1 U Base R 𝑠 B
26 eqid Monic 1p U = Monic 1p U
27 19 22 26 mon1pcl f Monic 1p U f Base Poly 1 U
28 27 adantl φ f Monic 1p U f Base Poly 1 U
29 25 28 ffvelcdmd φ f Monic 1p U O f Base R 𝑠 B
30 14 3 15 16 18 29 pwselbas φ f Monic 1p U O f : B B
31 ffn O f : B B O f Fn B
32 fniniseg O f Fn B X O f -1 0 ˙ X B O f X = 0 ˙
33 30 31 32 3syl φ f Monic 1p U X O f -1 0 ˙ X B O f X = 0 ˙
34 33 rexbidva φ f Monic 1p U X O f -1 0 ˙ f Monic 1p U X B O f X = 0 ˙
35 13 34 bitrd φ X R IntgRing S f Monic 1p U X B O f X = 0 ˙
36 r19.42v f Monic 1p U X B O f X = 0 ˙ X B f Monic 1p U O f X = 0 ˙
37 35 36 bitrdi φ X R IntgRing S X B f Monic 1p U O f X = 0 ˙