Metamath Proof Explorer


Theorem fldsdrgfldext2

Description: A sub-sub-division-ring of a field forms a field extension. (Contributed by Thierry Arnoux, 19-Oct-2025)

Ref Expression
Hypotheses fldsdrgfldext.1 G = F 𝑠 A
fldsdrgfldext.2 φ F Field
fldsdrgfldext.3 φ A SubDRing F
fldsdrgfldext2.b φ B SubDRing G
fldsdrgfldext2.h H = F 𝑠 B
Assertion fldsdrgfldext2 φ G /FldExt H

Proof

Step Hyp Ref Expression
1 fldsdrgfldext.1 G = F 𝑠 A
2 fldsdrgfldext.2 φ F Field
3 fldsdrgfldext.3 φ A SubDRing F
4 fldsdrgfldext2.b φ B SubDRing G
5 fldsdrgfldext2.h H = F 𝑠 B
6 eqid G 𝑠 B = G 𝑠 B
7 fldsdrgfld F Field A SubDRing F F 𝑠 A Field
8 2 3 7 syl2anc φ F 𝑠 A Field
9 1 8 eqeltrid φ G Field
10 6 9 4 fldsdrgfldext φ G /FldExt G 𝑠 B
11 eqid Base G = Base G
12 11 sdrgss B SubDRing G B Base G
13 4 12 syl φ B Base G
14 eqid Base F = Base F
15 14 sdrgss A SubDRing F A Base F
16 1 14 ressbas2 A Base F A = Base G
17 3 15 16 3syl φ A = Base G
18 13 17 sseqtrrd φ B A
19 ressabs A SubDRing F B A F 𝑠 A 𝑠 B = F 𝑠 B
20 3 18 19 syl2anc φ F 𝑠 A 𝑠 B = F 𝑠 B
21 1 oveq1i G 𝑠 B = F 𝑠 A 𝑠 B
22 20 21 5 3eqtr4g φ G 𝑠 B = H
23 10 22 breqtrd φ G /FldExt H