Metamath Proof Explorer


Theorem assafld

Description: If an algebra A of finite degree over a division ring K is an integral domain, then it is a field. Corollary of Proposition 2. in Chapter 5. of BourbakiAlg2 p. 113. (Contributed by Thierry Arnoux, 3-Aug-2025)

Ref Expression
Hypotheses assafld.k K = Scalar A
assafld.a φ A AssAlg
assafld.1 φ A IDomn
assafld.2 φ K DivRing
assafld.3 φ dim A 0
Assertion assafld φ A Field

Proof

Step Hyp Ref Expression
1 assafld.k K = Scalar A
2 assafld.a φ A AssAlg
3 assafld.1 φ A IDomn
4 assafld.2 φ K DivRing
5 assafld.3 φ dim A 0
6 3 idomringd φ A Ring
7 eqid Base A = Base A
8 eqid Unit A = Unit A
9 7 8 unitss Unit A Base A
10 9 a1i φ Unit A Base A
11 eqid 0 A = 0 A
12 3 idomdomd φ A Domn
13 domnnzr A Domn A NzRing
14 12 13 syl φ A NzRing
15 14 adantr φ 0 A Unit A A NzRing
16 simpr φ 0 A Unit A 0 A Unit A
17 8 11 15 16 unitnz φ 0 A Unit A 0 A 0 A
18 neirr ¬ 0 A 0 A
19 18 a1i φ 0 A Unit A ¬ 0 A 0 A
20 17 19 pm2.65da φ ¬ 0 A Unit A
21 ssdifsn Unit A Base A 0 A Unit A Base A ¬ 0 A Unit A
22 10 20 21 sylanbrc φ Unit A Base A 0 A
23 eqid RLReg A = RLReg A
24 2 adantr φ x Base A 0 A A AssAlg
25 4 adantr φ x Base A 0 A K DivRing
26 5 adantr φ x Base A 0 A dim A 0
27 12 adantr φ x Base A 0 A A Domn
28 simpr φ x Base A 0 A x Base A 0 A
29 28 eldifad φ x Base A 0 A x Base A
30 eldifsni x Base A 0 A x 0 A
31 28 30 syl φ x Base A 0 A x 0 A
32 7 23 11 domnrrg A Domn x Base A x 0 A x RLReg A
33 27 29 31 32 syl3anc φ x Base A 0 A x RLReg A
34 23 8 1 24 25 26 33 assarrginv φ x Base A 0 A x Unit A
35 22 34 eqelssd φ Unit A = Base A 0 A
36 7 8 11 isdrng A DivRing A Ring Unit A = Base A 0 A
37 6 35 36 sylanbrc φ A DivRing
38 3 idomcringd φ A CRing
39 isfld A Field A DivRing A CRing
40 37 38 39 sylanbrc φ A Field