Metamath Proof Explorer


Theorem fldextfld1

Description: A field extension is only defined if the extension is a field. (Contributed by Thierry Arnoux, 29-Jul-2023)

Ref Expression
Assertion fldextfld1 ( 𝐸 /FldExt 𝐹𝐸 ∈ Field )

Proof

Step Hyp Ref Expression
1 opabssxp { ⟨ 𝑒 , 𝑓 ⟩ ∣ ( ( 𝑒 ∈ Field ∧ 𝑓 ∈ Field ) ∧ ( 𝑓 = ( 𝑒s ( Base ‘ 𝑓 ) ) ∧ ( Base ‘ 𝑓 ) ∈ ( SubRing ‘ 𝑒 ) ) ) } ⊆ ( Field × Field )
2 df-br ( 𝐸 /FldExt 𝐹 ↔ ⟨ 𝐸 , 𝐹 ⟩ ∈ /FldExt )
3 2 biimpi ( 𝐸 /FldExt 𝐹 → ⟨ 𝐸 , 𝐹 ⟩ ∈ /FldExt )
4 df-fldext /FldExt = { ⟨ 𝑒 , 𝑓 ⟩ ∣ ( ( 𝑒 ∈ Field ∧ 𝑓 ∈ Field ) ∧ ( 𝑓 = ( 𝑒s ( Base ‘ 𝑓 ) ) ∧ ( Base ‘ 𝑓 ) ∈ ( SubRing ‘ 𝑒 ) ) ) }
5 3 4 eleqtrdi ( 𝐸 /FldExt 𝐹 → ⟨ 𝐸 , 𝐹 ⟩ ∈ { ⟨ 𝑒 , 𝑓 ⟩ ∣ ( ( 𝑒 ∈ Field ∧ 𝑓 ∈ Field ) ∧ ( 𝑓 = ( 𝑒s ( Base ‘ 𝑓 ) ) ∧ ( Base ‘ 𝑓 ) ∈ ( SubRing ‘ 𝑒 ) ) ) } )
6 1 5 sselid ( 𝐸 /FldExt 𝐹 → ⟨ 𝐸 , 𝐹 ⟩ ∈ ( Field × Field ) )
7 opelxp1 ( ⟨ 𝐸 , 𝐹 ⟩ ∈ ( Field × Field ) → 𝐸 ∈ Field )
8 6 7 syl ( 𝐸 /FldExt 𝐹𝐸 ∈ Field )