Metamath Proof Explorer


Theorem irngval

Description: The elements of a field R integral over a subset S . In the case of a subfield, those are the algebraic numbers over the field S within the field R . That is, the numbers X which are roots of monic polynomials P ( X ) with coefficients in 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
irngval.r φ R Ring
irngval.s φ S B
Assertion irngval φ R IntgRing S = f Monic 1p U O f -1 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 irngval.r φ R Ring
6 irngval.s φ S B
7 5 elexd φ R V
8 3 fvexi B V
9 8 a1i φ B V
10 9 6 ssexd φ S V
11 fvexd φ Monic 1p U V
12 fvex O f V
13 12 cnvex O f -1 V
14 13 imaex O f -1 0 ˙ V
15 14 rgenw f Monic 1p U O f -1 0 ˙ V
16 iunexg Monic 1p U V f Monic 1p U O f -1 0 ˙ V f Monic 1p U O f -1 0 ˙ V
17 11 15 16 sylancl φ f Monic 1p U O f -1 0 ˙ V
18 oveq12 r = R s = S r 𝑠 s = R 𝑠 S
19 18 2 eqtr4di r = R s = S r 𝑠 s = U
20 19 fveq2d r = R s = S Monic 1p r 𝑠 s = Monic 1p U
21 oveq12 r = R s = S r evalSub 1 s = R evalSub 1 S
22 21 1 eqtr4di r = R s = S r evalSub 1 s = O
23 22 fveq1d r = R s = S r evalSub 1 s f = O f
24 23 cnveqd r = R s = S r evalSub 1 s f -1 = O f -1
25 simpl r = R s = S r = R
26 25 fveq2d r = R s = S 0 r = 0 R
27 26 4 eqtr4di r = R s = S 0 r = 0 ˙
28 27 sneqd r = R s = S 0 r = 0 ˙
29 24 28 imaeq12d r = R s = S r evalSub 1 s f -1 0 r = O f -1 0 ˙
30 20 29 iuneq12d r = R s = S f Monic 1p r 𝑠 s r evalSub 1 s f -1 0 r = f Monic 1p U O f -1 0 ˙
31 df-irng IntgRing = r V , s V f Monic 1p r 𝑠 s r evalSub 1 s f -1 0 r
32 30 31 ovmpoga R V S V f Monic 1p U O f -1 0 ˙ V R IntgRing S = f Monic 1p U O f -1 0 ˙
33 7 10 17 32 syl3anc φ R IntgRing S = f Monic 1p U O f -1 0 ˙