Metamath Proof Explorer


Theorem fldgenfldext

Description: A subfield F extended with a set A forms a field extension. (Contributed by Thierry Arnoux, 22-Jun-2025)

Ref Expression
Hypotheses fldgenfldext.b B = Base E
fldgenfldext.k K = E 𝑠 F
fldgenfldext.l L = E 𝑠 E fldGen F A
fldgenfldext.e φ E Field
fldgenfldext.f φ F SubDRing E
fldgenfldext.1 φ A B
Assertion fldgenfldext φ L /FldExt K

Proof

Step Hyp Ref Expression
1 fldgenfldext.b B = Base E
2 fldgenfldext.k K = E 𝑠 F
3 fldgenfldext.l L = E 𝑠 E fldGen F A
4 fldgenfldext.e φ E Field
5 fldgenfldext.f φ F SubDRing E
6 fldgenfldext.1 φ A B
7 1 sdrgss F SubDRing E F B
8 5 7 syl φ F B
9 8 6 unssd φ F A B
10 1 4 9 fldgenfld φ E 𝑠 E fldGen F A Field
11 3 10 eqeltrid φ L Field
12 fldsdrgfld E Field F SubDRing E E 𝑠 F Field
13 4 5 12 syl2anc φ E 𝑠 F Field
14 2 13 eqeltrid φ K Field
15 3 oveq1i L 𝑠 F = E 𝑠 E fldGen F A 𝑠 F
16 ovexd φ E fldGen F A V
17 ressress E fldGen F A V F SubDRing E E 𝑠 E fldGen F A 𝑠 F = E 𝑠 E fldGen F A F
18 16 5 17 syl2anc φ E 𝑠 E fldGen F A 𝑠 F = E 𝑠 E fldGen F A F
19 15 18 eqtrid φ L 𝑠 F = E 𝑠 E fldGen F A F
20 4 flddrngd φ E DivRing
21 1 20 9 fldgenssid φ F A E fldGen F A
22 21 unssad φ F E fldGen F A
23 sseqin2 F E fldGen F A E fldGen F A F = F
24 22 23 sylib φ E fldGen F A F = F
25 24 oveq2d φ E 𝑠 E fldGen F A F = E 𝑠 F
26 19 25 eqtrd φ L 𝑠 F = E 𝑠 F
27 2 1 ressbas2 F B F = Base K
28 8 27 syl φ F = Base K
29 28 oveq2d φ L 𝑠 F = L 𝑠 Base K
30 26 29 eqtr3d φ E 𝑠 F = L 𝑠 Base K
31 2 30 eqtrid φ K = L 𝑠 Base K
32 11 fldcrngd φ L CRing
33 32 crngringd φ L Ring
34 14 fldcrngd φ K CRing
35 34 crngringd φ K Ring
36 2 35 eqeltrrid φ E 𝑠 F Ring
37 26 36 eqeltrd φ L 𝑠 F Ring
38 1 20 9 fldgenssv φ E fldGen F A B
39 3 1 ressbas2 E fldGen F A B E fldGen F A = Base L
40 38 39 syl φ E fldGen F A = Base L
41 22 40 sseqtrd φ F Base L
42 20 drngringd φ E Ring
43 sdrgsubrg F SubDRing E F SubRing E
44 eqid 1 E = 1 E
45 44 subrg1cl F SubRing E 1 E F
46 5 43 45 3syl φ 1 E F
47 22 46 sseldd φ 1 E E fldGen F A
48 3 1 44 ress1r E Ring 1 E E fldGen F A E fldGen F A B 1 E = 1 L
49 42 47 38 48 syl3anc φ 1 E = 1 L
50 49 46 eqeltrrd φ 1 L F
51 41 50 jca φ F Base L 1 L F
52 eqid Base L = Base L
53 eqid 1 L = 1 L
54 52 53 issubrg F SubRing L L Ring L 𝑠 F Ring F Base L 1 L F
55 33 37 51 54 syl21anbrc φ F SubRing L
56 28 55 eqeltrrd φ Base K SubRing L
57 brfldext L Field K Field L /FldExt K K = L 𝑠 Base K Base K SubRing L
58 57 biimpar L Field K Field K = L 𝑠 Base K Base K SubRing L L /FldExt K
59 11 14 31 56 58 syl22anc φ L /FldExt K