Metamath Proof Explorer


Theorem qqh0

Description: The image of 0 by the QQHom homomorphism is the ring's zero. (Contributed by Thierry Arnoux, 22-Oct-2017)

Ref Expression
Hypotheses qqhval2.0 𝐵 = ( Base ‘ 𝑅 )
qqhval2.1 / = ( /r𝑅 )
qqhval2.2 𝐿 = ( ℤRHom ‘ 𝑅 )
Assertion qqh0 ( ( 𝑅 ∈ DivRing ∧ ( chr ‘ 𝑅 ) = 0 ) → ( ( ℚHom ‘ 𝑅 ) ‘ 0 ) = ( 0g𝑅 ) )

Proof

Step Hyp Ref Expression
1 qqhval2.0 𝐵 = ( Base ‘ 𝑅 )
2 qqhval2.1 / = ( /r𝑅 )
3 qqhval2.2 𝐿 = ( ℤRHom ‘ 𝑅 )
4 zssq ℤ ⊆ ℚ
5 0z 0 ∈ ℤ
6 4 5 sselii 0 ∈ ℚ
7 1 2 3 qqhvval ( ( ( 𝑅 ∈ DivRing ∧ ( chr ‘ 𝑅 ) = 0 ) ∧ 0 ∈ ℚ ) → ( ( ℚHom ‘ 𝑅 ) ‘ 0 ) = ( ( 𝐿 ‘ ( numer ‘ 0 ) ) / ( 𝐿 ‘ ( denom ‘ 0 ) ) ) )
8 6 7 mpan2 ( ( 𝑅 ∈ DivRing ∧ ( chr ‘ 𝑅 ) = 0 ) → ( ( ℚHom ‘ 𝑅 ) ‘ 0 ) = ( ( 𝐿 ‘ ( numer ‘ 0 ) ) / ( 𝐿 ‘ ( denom ‘ 0 ) ) ) )
9 1z 1 ∈ ℤ
10 gcd0id ( 1 ∈ ℤ → ( 0 gcd 1 ) = ( abs ‘ 1 ) )
11 9 10 ax-mp ( 0 gcd 1 ) = ( abs ‘ 1 )
12 abs1 ( abs ‘ 1 ) = 1
13 11 12 eqtri ( 0 gcd 1 ) = 1
14 0cn 0 ∈ ℂ
15 14 div1i ( 0 / 1 ) = 0
16 15 eqcomi 0 = ( 0 / 1 )
17 13 16 pm3.2i ( ( 0 gcd 1 ) = 1 ∧ 0 = ( 0 / 1 ) )
18 1nn 1 ∈ ℕ
19 qnumdenbi ( ( 0 ∈ ℚ ∧ 0 ∈ ℤ ∧ 1 ∈ ℕ ) → ( ( ( 0 gcd 1 ) = 1 ∧ 0 = ( 0 / 1 ) ) ↔ ( ( numer ‘ 0 ) = 0 ∧ ( denom ‘ 0 ) = 1 ) ) )
20 6 5 18 19 mp3an ( ( ( 0 gcd 1 ) = 1 ∧ 0 = ( 0 / 1 ) ) ↔ ( ( numer ‘ 0 ) = 0 ∧ ( denom ‘ 0 ) = 1 ) )
21 17 20 mpbi ( ( numer ‘ 0 ) = 0 ∧ ( denom ‘ 0 ) = 1 )
22 21 simpli ( numer ‘ 0 ) = 0
23 22 fveq2i ( 𝐿 ‘ ( numer ‘ 0 ) ) = ( 𝐿 ‘ 0 )
24 21 simpri ( denom ‘ 0 ) = 1
25 24 fveq2i ( 𝐿 ‘ ( denom ‘ 0 ) ) = ( 𝐿 ‘ 1 )
26 23 25 oveq12i ( ( 𝐿 ‘ ( numer ‘ 0 ) ) / ( 𝐿 ‘ ( denom ‘ 0 ) ) ) = ( ( 𝐿 ‘ 0 ) / ( 𝐿 ‘ 1 ) )
27 drngring ( 𝑅 ∈ DivRing → 𝑅 ∈ Ring )
28 eqid ( 0g𝑅 ) = ( 0g𝑅 )
29 3 28 zrh0 ( 𝑅 ∈ Ring → ( 𝐿 ‘ 0 ) = ( 0g𝑅 ) )
30 eqid ( 1r𝑅 ) = ( 1r𝑅 )
31 3 30 zrh1 ( 𝑅 ∈ Ring → ( 𝐿 ‘ 1 ) = ( 1r𝑅 ) )
32 29 31 oveq12d ( 𝑅 ∈ Ring → ( ( 𝐿 ‘ 0 ) / ( 𝐿 ‘ 1 ) ) = ( ( 0g𝑅 ) / ( 1r𝑅 ) ) )
33 27 32 syl ( 𝑅 ∈ DivRing → ( ( 𝐿 ‘ 0 ) / ( 𝐿 ‘ 1 ) ) = ( ( 0g𝑅 ) / ( 1r𝑅 ) ) )
34 drnggrp ( 𝑅 ∈ DivRing → 𝑅 ∈ Grp )
35 1 28 grpidcl ( 𝑅 ∈ Grp → ( 0g𝑅 ) ∈ 𝐵 )
36 34 35 syl ( 𝑅 ∈ DivRing → ( 0g𝑅 ) ∈ 𝐵 )
37 1 2 30 dvr1 ( ( 𝑅 ∈ Ring ∧ ( 0g𝑅 ) ∈ 𝐵 ) → ( ( 0g𝑅 ) / ( 1r𝑅 ) ) = ( 0g𝑅 ) )
38 27 36 37 syl2anc ( 𝑅 ∈ DivRing → ( ( 0g𝑅 ) / ( 1r𝑅 ) ) = ( 0g𝑅 ) )
39 33 38 eqtrd ( 𝑅 ∈ DivRing → ( ( 𝐿 ‘ 0 ) / ( 𝐿 ‘ 1 ) ) = ( 0g𝑅 ) )
40 26 39 syl5eq ( 𝑅 ∈ DivRing → ( ( 𝐿 ‘ ( numer ‘ 0 ) ) / ( 𝐿 ‘ ( denom ‘ 0 ) ) ) = ( 0g𝑅 ) )
41 40 adantr ( ( 𝑅 ∈ DivRing ∧ ( chr ‘ 𝑅 ) = 0 ) → ( ( 𝐿 ‘ ( numer ‘ 0 ) ) / ( 𝐿 ‘ ( denom ‘ 0 ) ) ) = ( 0g𝑅 ) )
42 8 41 eqtrd ( ( 𝑅 ∈ DivRing ∧ ( chr ‘ 𝑅 ) = 0 ) → ( ( ℚHom ‘ 𝑅 ) ‘ 0 ) = ( 0g𝑅 ) )