Metamath Proof Explorer


Theorem fldgenfld

Description: A generated subfield is a field. (Contributed by Thierry Arnoux, 11-Jan-2025)

Ref Expression
Hypotheses fldgenfld.1 𝐵 = ( Base ‘ 𝐹 )
fldgenfld.2 ( 𝜑𝐹 ∈ Field )
fldgenfld.3 ( 𝜑𝑆𝐵 )
Assertion fldgenfld ( 𝜑 → ( 𝐹s ( 𝐹 fldGen 𝑆 ) ) ∈ Field )

Proof

Step Hyp Ref Expression
1 fldgenfld.1 𝐵 = ( Base ‘ 𝐹 )
2 fldgenfld.2 ( 𝜑𝐹 ∈ Field )
3 fldgenfld.3 ( 𝜑𝑆𝐵 )
4 isfld ( 𝐹 ∈ Field ↔ ( 𝐹 ∈ DivRing ∧ 𝐹 ∈ CRing ) )
5 2 4 sylib ( 𝜑 → ( 𝐹 ∈ DivRing ∧ 𝐹 ∈ CRing ) )
6 5 simpld ( 𝜑𝐹 ∈ DivRing )
7 1 6 3 fldgensdrg ( 𝜑 → ( 𝐹 fldGen 𝑆 ) ∈ ( SubDRing ‘ 𝐹 ) )
8 fldsdrgfld ( ( 𝐹 ∈ Field ∧ ( 𝐹 fldGen 𝑆 ) ∈ ( SubDRing ‘ 𝐹 ) ) → ( 𝐹s ( 𝐹 fldGen 𝑆 ) ) ∈ Field )
9 2 7 8 syl2anc ( 𝜑 → ( 𝐹s ( 𝐹 fldGen 𝑆 ) ) ∈ Field )