Theorem primefld1cl

Description: The prime field contains the multiplicative neutral element of the division ring. (Contributed by Thierry Arnoux, 22-Aug-2023)

	Ref	Expression
Hypothesis	primefld1cl.1	⊢ 1 = ( 1r ‘ 𝑅 )
Assertion	primefld1cl	⊢ ( 𝑅 ∈ DivRing → 1 ∈ ∩ ( SubDRing ‘ 𝑅 ) )

Proof

Step	Hyp	Ref	Expression
1		primefld1cl.1	⊢ 1 = ( 1r ‘ 𝑅 )
2		issdrg	⊢ ( 𝑠 ∈ ( SubDRing ‘ 𝑅 ) ↔ ( 𝑅 ∈ DivRing ∧ 𝑠 ∈ ( SubRing ‘ 𝑅 ) ∧ ( 𝑅 ↾s 𝑠 ) ∈ DivRing ) )
3	2	simp2bi	⊢ ( 𝑠 ∈ ( SubDRing ‘ 𝑅 ) → 𝑠 ∈ ( SubRing ‘ 𝑅 ) )
4	3	a1i	⊢ ( 𝑅 ∈ DivRing → ( 𝑠 ∈ ( SubDRing ‘ 𝑅 ) → 𝑠 ∈ ( SubRing ‘ 𝑅 ) ) )
5	4	ssrdv	⊢ ( 𝑅 ∈ DivRing → ( SubDRing ‘ 𝑅 ) ⊆ ( SubRing ‘ 𝑅 ) )
6		eqid	⊢ ( Base ‘ 𝑅 ) = ( Base ‘ 𝑅 )
7	6	sdrgid	⊢ ( 𝑅 ∈ DivRing → ( Base ‘ 𝑅 ) ∈ ( SubDRing ‘ 𝑅 ) )
8	7	ne0d	⊢ ( 𝑅 ∈ DivRing → ( SubDRing ‘ 𝑅 ) ≠ ∅ )
9		subrgint	⊢ ( ( ( SubDRing ‘ 𝑅 ) ⊆ ( SubRing ‘ 𝑅 ) ∧ ( SubDRing ‘ 𝑅 ) ≠ ∅ ) → ∩ ( SubDRing ‘ 𝑅 ) ∈ ( SubRing ‘ 𝑅 ) )
10	5 8 9	syl2anc	⊢ ( 𝑅 ∈ DivRing → ∩ ( SubDRing ‘ 𝑅 ) ∈ ( SubRing ‘ 𝑅 ) )
11	1	subrg1cl	⊢ ( ∩ ( SubDRing ‘ 𝑅 ) ∈ ( SubRing ‘ 𝑅 ) → 1 ∈ ∩ ( SubDRing ‘ 𝑅 ) )
12	10 11	syl	⊢ ( 𝑅 ∈ DivRing → 1 ∈ ∩ ( SubDRing ‘ 𝑅 ) )