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 ) )