Metamath Proof Explorer


Theorem zrhchr

Description: The kernel of the homomorphism from the integers to a ring is injective if and only if the ring has characteristic 0 . (Contributed by Thierry Arnoux, 8-Nov-2017)

Ref Expression
Hypotheses zrhker.0 𝐵 = ( Base ‘ 𝑅 )
zrhker.1 𝐿 = ( ℤRHom ‘ 𝑅 )
zrhker.2 0 = ( 0g𝑅 )
Assertion zrhchr ( 𝑅 ∈ Ring → ( ( chr ‘ 𝑅 ) = 0 ↔ 𝐿 : ℤ –1-1𝐵 ) )

Proof

Step Hyp Ref Expression
1 zrhker.0 𝐵 = ( Base ‘ 𝑅 )
2 zrhker.1 𝐿 = ( ℤRHom ‘ 𝑅 )
3 zrhker.2 0 = ( 0g𝑅 )
4 eqid ( .g𝑅 ) = ( .g𝑅 )
5 eqid ( 1r𝑅 ) = ( 1r𝑅 )
6 2 4 5 zrhval2 ( 𝑅 ∈ Ring → 𝐿 = ( 𝑥 ∈ ℤ ↦ ( 𝑥 ( .g𝑅 ) ( 1r𝑅 ) ) ) )
7 f1eq1 ( 𝐿 = ( 𝑥 ∈ ℤ ↦ ( 𝑥 ( .g𝑅 ) ( 1r𝑅 ) ) ) → ( 𝐿 : ℤ –1-1𝐵 ↔ ( 𝑥 ∈ ℤ ↦ ( 𝑥 ( .g𝑅 ) ( 1r𝑅 ) ) ) : ℤ –1-1𝐵 ) )
8 6 7 syl ( 𝑅 ∈ Ring → ( 𝐿 : ℤ –1-1𝐵 ↔ ( 𝑥 ∈ ℤ ↦ ( 𝑥 ( .g𝑅 ) ( 1r𝑅 ) ) ) : ℤ –1-1𝐵 ) )
9 ringgrp ( 𝑅 ∈ Ring → 𝑅 ∈ Grp )
10 1 5 ringidcl ( 𝑅 ∈ Ring → ( 1r𝑅 ) ∈ 𝐵 )
11 eqid ( od ‘ 𝑅 ) = ( od ‘ 𝑅 )
12 eqid ( 𝑥 ∈ ℤ ↦ ( 𝑥 ( .g𝑅 ) ( 1r𝑅 ) ) ) = ( 𝑥 ∈ ℤ ↦ ( 𝑥 ( .g𝑅 ) ( 1r𝑅 ) ) )
13 1 11 4 12 odf1 ( ( 𝑅 ∈ Grp ∧ ( 1r𝑅 ) ∈ 𝐵 ) → ( ( ( od ‘ 𝑅 ) ‘ ( 1r𝑅 ) ) = 0 ↔ ( 𝑥 ∈ ℤ ↦ ( 𝑥 ( .g𝑅 ) ( 1r𝑅 ) ) ) : ℤ –1-1𝐵 ) )
14 9 10 13 syl2anc ( 𝑅 ∈ Ring → ( ( ( od ‘ 𝑅 ) ‘ ( 1r𝑅 ) ) = 0 ↔ ( 𝑥 ∈ ℤ ↦ ( 𝑥 ( .g𝑅 ) ( 1r𝑅 ) ) ) : ℤ –1-1𝐵 ) )
15 eqid ( chr ‘ 𝑅 ) = ( chr ‘ 𝑅 )
16 11 5 15 chrval ( ( od ‘ 𝑅 ) ‘ ( 1r𝑅 ) ) = ( chr ‘ 𝑅 )
17 16 eqeq1i ( ( ( od ‘ 𝑅 ) ‘ ( 1r𝑅 ) ) = 0 ↔ ( chr ‘ 𝑅 ) = 0 )
18 17 a1i ( 𝑅 ∈ Ring → ( ( ( od ‘ 𝑅 ) ‘ ( 1r𝑅 ) ) = 0 ↔ ( chr ‘ 𝑅 ) = 0 ) )
19 8 14 18 3bitr2rd ( 𝑅 ∈ Ring → ( ( chr ‘ 𝑅 ) = 0 ↔ 𝐿 : ℤ –1-1𝐵 ) )