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=RevalSub1S
irngval.u U=R𝑠S
irngval.b B=BaseR
irngval.0 0˙=0R
irngval.r φRRing
irngval.s φSB
Assertion irngval φRIntgRingS=fMonic1pUOf-10˙

Proof

Step Hyp Ref Expression
1 irngval.o O=RevalSub1S
2 irngval.u U=R𝑠S
3 irngval.b B=BaseR
4 irngval.0 0˙=0R
5 irngval.r φRRing
6 irngval.s φSB
7 5 elexd φRV
8 3 fvexi BV
9 8 a1i φBV
10 9 6 ssexd φSV
11 fvexd φMonic1pUV
12 fvex OfV
13 12 cnvex Of-1V
14 13 imaex Of-10˙V
15 14 rgenw fMonic1pUOf-10˙V
16 iunexg Monic1pUVfMonic1pUOf-10˙VfMonic1pUOf-10˙V
17 11 15 16 sylancl φfMonic1pUOf-10˙V
18 oveq12 r=Rs=Sr𝑠s=R𝑠S
19 18 2 eqtr4di r=Rs=Sr𝑠s=U
20 19 fveq2d r=Rs=SMonic1pr𝑠s=Monic1pU
21 oveq12 r=Rs=SrevalSub1s=RevalSub1S
22 21 1 eqtr4di r=Rs=SrevalSub1s=O
23 22 fveq1d r=Rs=SrevalSub1sf=Of
24 23 cnveqd r=Rs=SrevalSub1sf-1=Of-1
25 simpl r=Rs=Sr=R
26 25 fveq2d r=Rs=S0r=0R
27 26 4 eqtr4di r=Rs=S0r=0˙
28 27 sneqd r=Rs=S0r=0˙
29 24 28 imaeq12d r=Rs=SrevalSub1sf-10r=Of-10˙
30 20 29 iuneq12d r=Rs=SfMonic1pr𝑠srevalSub1sf-10r=fMonic1pUOf-10˙
31 df-irng IntgRing=rV,sVfMonic1pr𝑠srevalSub1sf-10r
32 30 31 ovmpoga RVSVfMonic1pUOf-10˙VRIntgRingS=fMonic1pUOf-10˙
33 7 10 17 32 syl3anc φRIntgRingS=fMonic1pUOf-10˙