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Metamath Proof Explorer

Theorem fldextsdrg

Description: Deduce sub-division-ring from field extension. (Contributed by Thierry Arnoux, 26-Oct-2025)

Ref Expression
Hypotheses fldextsdrg.1 ⊢ 𝐵 = ( Base ‘ 𝐹 )
fldextsdrg.2 ⊢ ( 𝜑 → 𝐸 /FldExt 𝐹 )
Assertion fldextsdrg ( 𝜑 → 𝐵 ∈ ( SubDRing ‘ 𝐸 ) )

Proof

Step Hyp Ref Expression
1 fldextsdrg.1 ⊢ 𝐵 = ( Base ‘ 𝐹 )
2 fldextsdrg.2 ⊢ ( 𝜑 → 𝐸 /FldExt 𝐹 )
3 fldextfld1 ⊢ ( 𝐸 /FldExt 𝐹 → 𝐸 ∈ Field )
4 2 3 syl ⊢ ( 𝜑 → 𝐸 ∈ Field )
5 4 flddrngd ⊢ ( 𝜑 → 𝐸 ∈ DivRing )
6 1 fldextsubrg ⊢ ( 𝐸 /FldExt 𝐹 → 𝐵 ∈ ( SubRing ‘ 𝐸 ) )
7 2 6 syl ⊢ ( 𝜑 → 𝐵 ∈ ( SubRing ‘ 𝐸 ) )
8 fldextress ⊢ ( 𝐸 /FldExt 𝐹 → 𝐹 = ( 𝐸 ↾s ( Base ‘ 𝐹 ) ) )
9 2 8 syl ⊢ ( 𝜑 → 𝐹 = ( 𝐸 ↾s ( Base ‘ 𝐹 ) ) )
10 1 oveq2i ⊢ ( 𝐸 ↾s 𝐵 ) = ( 𝐸 ↾s ( Base ‘ 𝐹 ) )
11 9 10 eqtr4di ⊢ ( 𝜑 → 𝐹 = ( 𝐸 ↾s 𝐵 ) )
12 fldextfld2 ⊢ ( 𝐸 /FldExt 𝐹 → 𝐹 ∈ Field )
13 2 12 syl ⊢ ( 𝜑 → 𝐹 ∈ Field )
14 11 13 eqeltrrd ⊢ ( 𝜑 → ( 𝐸 ↾s 𝐵 ) ∈ Field )
15 14 flddrngd ⊢ ( 𝜑 → ( 𝐸 ↾s 𝐵 ) ∈ DivRing )
16 issdrg ⊢ ( 𝐵 ∈ ( SubDRing ‘ 𝐸 ) ↔ ( 𝐸 ∈ DivRing ∧ 𝐵 ∈ ( SubRing ‘ 𝐸 ) ∧ ( 𝐸 ↾s 𝐵 ) ∈ DivRing ) )
17 5 7 15 16 syl3anbrc ⊢ ( 𝜑 → 𝐵 ∈ ( SubDRing ‘ 𝐸 ) )