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Metamath Proof Explorer

Theorem fldextsubrg

Description: Field extension implies a subring relation. (Contributed by Thierry Arnoux, 29-Jul-2023)

Ref Expression
Hypothesis fldextsubrg.1 ⊢ 𝑈 = ( Base ‘ 𝐹 )
Assertion fldextsubrg ( 𝐸 /FldExt 𝐹 → 𝑈 ∈ ( SubRing ‘ 𝐸 ) )

Proof

Step Hyp Ref Expression
1 fldextsubrg.1 ⊢ 𝑈 = ( Base ‘ 𝐹 )
2 fldextfld1 ⊢ ( 𝐸 /FldExt 𝐹 → 𝐸 ∈ Field )
3 fldextfld2 ⊢ ( 𝐸 /FldExt 𝐹 → 𝐹 ∈ Field )
4 brfldext ⊢ ( ( 𝐸 ∈ Field ∧ 𝐹 ∈ Field ) → ( 𝐸 /FldExt 𝐹 ↔ ( 𝐹 = ( 𝐸 ↾s ( Base ‘ 𝐹 ) ) ∧ ( Base ‘ 𝐹 ) ∈ ( SubRing ‘ 𝐸 ) ) ) )
5 2 3 4 syl2anc ⊢ ( 𝐸 /FldExt 𝐹 → ( 𝐸 /FldExt 𝐹 ↔ ( 𝐹 = ( 𝐸 ↾s ( Base ‘ 𝐹 ) ) ∧ ( Base ‘ 𝐹 ) ∈ ( SubRing ‘ 𝐸 ) ) ) )
6 5 ibi ⊢ ( 𝐸 /FldExt 𝐹 → ( 𝐹 = ( 𝐸 ↾s ( Base ‘ 𝐹 ) ) ∧ ( Base ‘ 𝐹 ) ∈ ( SubRing ‘ 𝐸 ) ) )
7 6 simprd ⊢ ( 𝐸 /FldExt 𝐹 → ( Base ‘ 𝐹 ) ∈ ( SubRing ‘ 𝐸 ) )
8 1 7 eqeltrid ⊢ ( 𝐸 /FldExt 𝐹 → 𝑈 ∈ ( SubRing ‘ 𝐸 ) )