Theorem qqhvq

Description: The image of a quotient by the QQHom homomorphism. (Contributed by Thierry Arnoux, 28-Oct-2017)

	Ref	Expression
Hypotheses	qqhval2.0	⊢ 𝐵 = ( Base ‘ 𝑅 )
	qqhval2.1	⊢ / = ( /r ‘ 𝑅 )
	qqhval2.2	⊢ 𝐿 = ( ℤRHom ‘ 𝑅 )
Assertion	qqhvq	⊢ ( ( ( 𝑅 ∈ DivRing ∧ ( chr ‘ 𝑅 ) = 0 ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ∧ 𝑌 ≠ 0 ) ) → ( ( ℚHom ‘ 𝑅 ) ‘ ( 𝑋 / 𝑌 ) ) = ( ( 𝐿 ‘ 𝑋 ) / ( 𝐿 ‘ 𝑌 ) ) )

Proof

Step	Hyp	Ref	Expression
1		qqhval2.0	⊢ 𝐵 = ( Base ‘ 𝑅 )
2		qqhval2.1	⊢ / = ( /r ‘ 𝑅 )
3		qqhval2.2	⊢ 𝐿 = ( ℤRHom ‘ 𝑅 )
4		zssq	⊢ ℤ ⊆ ℚ
5		simpr1	⊢ ( ( ( 𝑅 ∈ DivRing ∧ ( chr ‘ 𝑅 ) = 0 ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ∧ 𝑌 ≠ 0 ) ) → 𝑋 ∈ ℤ )
6	4 5	sselid	⊢ ( ( ( 𝑅 ∈ DivRing ∧ ( chr ‘ 𝑅 ) = 0 ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ∧ 𝑌 ≠ 0 ) ) → 𝑋 ∈ ℚ )
7		simpr2	⊢ ( ( ( 𝑅 ∈ DivRing ∧ ( chr ‘ 𝑅 ) = 0 ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ∧ 𝑌 ≠ 0 ) ) → 𝑌 ∈ ℤ )
8	4 7	sselid	⊢ ( ( ( 𝑅 ∈ DivRing ∧ ( chr ‘ 𝑅 ) = 0 ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ∧ 𝑌 ≠ 0 ) ) → 𝑌 ∈ ℚ )
9		simpr3	⊢ ( ( ( 𝑅 ∈ DivRing ∧ ( chr ‘ 𝑅 ) = 0 ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ∧ 𝑌 ≠ 0 ) ) → 𝑌 ≠ 0 )
10		qdivcl	⊢ ( ( 𝑋 ∈ ℚ ∧ 𝑌 ∈ ℚ ∧ 𝑌 ≠ 0 ) → ( 𝑋 / 𝑌 ) ∈ ℚ )
11	6 8 9 10	syl3anc	⊢ ( ( ( 𝑅 ∈ DivRing ∧ ( chr ‘ 𝑅 ) = 0 ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ∧ 𝑌 ≠ 0 ) ) → ( 𝑋 / 𝑌 ) ∈ ℚ )
12	1 2 3	qqhvval	⊢ ( ( ( 𝑅 ∈ DivRing ∧ ( chr ‘ 𝑅 ) = 0 ) ∧ ( 𝑋 / 𝑌 ) ∈ ℚ ) → ( ( ℚHom ‘ 𝑅 ) ‘ ( 𝑋 / 𝑌 ) ) = ( ( 𝐿 ‘ ( numer ‘ ( 𝑋 / 𝑌 ) ) ) / ( 𝐿 ‘ ( denom ‘ ( 𝑋 / 𝑌 ) ) ) ) )
13	11 12	syldan	⊢ ( ( ( 𝑅 ∈ DivRing ∧ ( chr ‘ 𝑅 ) = 0 ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ∧ 𝑌 ≠ 0 ) ) → ( ( ℚHom ‘ 𝑅 ) ‘ ( 𝑋 / 𝑌 ) ) = ( ( 𝐿 ‘ ( numer ‘ ( 𝑋 / 𝑌 ) ) ) / ( 𝐿 ‘ ( denom ‘ ( 𝑋 / 𝑌 ) ) ) ) )
14	1 2 3	qqhval2lem	⊢ ( ( ( 𝑅 ∈ DivRing ∧ ( chr ‘ 𝑅 ) = 0 ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ∧ 𝑌 ≠ 0 ) ) → ( ( 𝐿 ‘ ( numer ‘ ( 𝑋 / 𝑌 ) ) ) / ( 𝐿 ‘ ( denom ‘ ( 𝑋 / 𝑌 ) ) ) ) = ( ( 𝐿 ‘ 𝑋 ) / ( 𝐿 ‘ 𝑌 ) ) )
15	13 14	eqtrd	⊢ ( ( ( 𝑅 ∈ DivRing ∧ ( chr ‘ 𝑅 ) = 0 ) ∧ ( 𝑋 ∈ ℤ ∧ 𝑌 ∈ ℤ ∧ 𝑌 ≠ 0 ) ) → ( ( ℚHom ‘ 𝑅 ) ‘ ( 𝑋 / 𝑌 ) ) = ( ( 𝐿 ‘ 𝑋 ) / ( 𝐿 ‘ 𝑌 ) ) )