Metamath Proof Explorer


Theorem rng2idl1cntr

Description: The unity of a two-sided ideal of a non-unital ring is central, i.e., an element of the center of the multiplicative semigroup of the non-unital ring. This is part of the proof given in MathOverflow, which seems to be sufficient to show that F given below (see rngqiprngimf ) is an isomorphism. In our proof, however we show that F is linear regarding the multiplication ( rngqiprnglin ) via rngqiprnglinlem1 instead. (Contributed by AV, 13-Feb-2025)

Ref Expression
Hypotheses rng2idl1cntr.r ⊢ φ → R ∈ Rng
rng2idl1cntr.i ⊢ φ → I ∈ 2Ideal ⁡ R
rng2idl1cntr.j ⊢ J = R ↾ 𝑠 I
rng2idl1cntr.u ⊢ φ → J ∈ Ring
rng2idl1cntr.1 ⊢ 1 ˙ = 1 J
rng2idl1cntr.m ⊢ M = mulGrp R
Assertion rng2idl1cntr ⊢ φ → 1 ˙ ∈ Cntr ⁡ M

Proof

Step Hyp Ref Expression
1 rng2idl1cntr.r ⊢ φ → R ∈ Rng
2 rng2idl1cntr.i ⊢ φ → I ∈ 2Ideal ⁡ R
3 rng2idl1cntr.j ⊢ J = R ↾ 𝑠 I
4 rng2idl1cntr.u ⊢ φ → J ∈ Ring
5 rng2idl1cntr.1 ⊢ 1 ˙ = 1 J
6 rng2idl1cntr.m ⊢ M = mulGrp R
7 eqid ⊢ Base R = Base R
8 3 7 ressbasss ⊢ Base J ⊆ Base R
9 eqid ⊢ Base J = Base J
10 9 5 ringidcl ⊢ J ∈ Ring → 1 ˙ ∈ Base J
11 4 10 syl ⊢ φ → 1 ˙ ∈ Base J
12 8 11 sselid ⊢ φ → 1 ˙ ∈ Base R
13 1 adantr ⊢ φ ∧ x ∈ Base R → R ∈ Rng
14 12 adantr ⊢ φ ∧ x ∈ Base R → 1 ˙ ∈ Base R
15 simpr ⊢ φ ∧ x ∈ Base R → x ∈ Base R
16 eqid ⊢ ⋅ R = ⋅ R
17 7 16 rngass ⊢ R ∈ Rng ∧ 1 ˙ ∈ Base R ∧ x ∈ Base R ∧ 1 ˙ ∈ Base R → 1 ˙ ⋅ R x ⋅ R 1 ˙ = 1 ˙ ⋅ R x ⋅ R 1 ˙
18 13 14 15 14 17 syl13anc ⊢ φ ∧ x ∈ Base R → 1 ˙ ⋅ R x ⋅ R 1 ˙ = 1 ˙ ⋅ R x ⋅ R 1 ˙
19 eqid ⊢ ⋅ J = ⋅ J
20 4 adantr ⊢ φ ∧ x ∈ Base R → J ∈ Ring
21 1 2 3 4 7 16 5 rngqiprngghmlem1 ⊢ φ ∧ x ∈ Base R → 1 ˙ ⋅ R x ∈ Base J
22 9 19 5 20 21 ringridmd ⊢ φ ∧ x ∈ Base R → 1 ˙ ⋅ R x ⋅ J 1 ˙ = 1 ˙ ⋅ R x
23 3 16 ressmulr ⊢ I ∈ 2Ideal ⁡ R → ⋅ R = ⋅ J
24 2 23 syl ⊢ φ → ⋅ R = ⋅ J
25 24 oveqd ⊢ φ → 1 ˙ ⋅ R x ⋅ R 1 ˙ = 1 ˙ ⋅ R x ⋅ J 1 ˙
26 25 eqeq1d ⊢ φ → 1 ˙ ⋅ R x ⋅ R 1 ˙ = 1 ˙ ⋅ R x ↔ 1 ˙ ⋅ R x ⋅ J 1 ˙ = 1 ˙ ⋅ R x
27 26 adantr ⊢ φ ∧ x ∈ Base R → 1 ˙ ⋅ R x ⋅ R 1 ˙ = 1 ˙ ⋅ R x ↔ 1 ˙ ⋅ R x ⋅ J 1 ˙ = 1 ˙ ⋅ R x
28 22 27 mpbird ⊢ φ ∧ x ∈ Base R → 1 ˙ ⋅ R x ⋅ R 1 ˙ = 1 ˙ ⋅ R x
29 2 2idllidld ⊢ φ → I ∈ LIdeal ⁡ R
30 eqid ⊢ LIdeal ⁡ R = LIdeal ⁡ R
31 7 30 lidlss ⊢ I ∈ LIdeal ⁡ R → I ⊆ Base R
32 3 7 ressbas2 ⊢ I ⊆ Base R → I = Base J
33 32 eqcomd ⊢ I ⊆ Base R → Base J = I
34 29 31 33 3syl ⊢ φ → Base J = I
35 34 29 eqeltrd ⊢ φ → Base J ∈ LIdeal ⁡ R
36 2 3 9 2idlbas ⊢ φ → Base J = I
37 ringrng ⊢ J ∈ Ring → J ∈ Rng
38 4 37 syl ⊢ φ → J ∈ Rng
39 3 38 eqeltrrid ⊢ φ → R ↾ 𝑠 I ∈ Rng
40 1 2 39 rng2idlsubrng ⊢ φ → I ∈ SubRng ⁡ R
41 36 40 eqeltrd ⊢ φ → Base J ∈ SubRng ⁡ R
42 subrngsubg ⊢ Base J ∈ SubRng ⁡ R → Base J ∈ SubGrp ⁡ R
43 eqid ⊢ 0 R = 0 R
44 43 subg0cl ⊢ Base J ∈ SubGrp ⁡ R → 0 R ∈ Base J
45 41 42 44 3syl ⊢ φ → 0 R ∈ Base J
46 1 35 45 3jca ⊢ φ → R ∈ Rng ∧ Base J ∈ LIdeal ⁡ R ∧ 0 R ∈ Base J
47 11 anim1ci ⊢ φ ∧ x ∈ Base R → x ∈ Base R ∧ 1 ˙ ∈ Base J
48 43 7 16 30 rnglidlmcl ⊢ R ∈ Rng ∧ Base J ∈ LIdeal ⁡ R ∧ 0 R ∈ Base J ∧ x ∈ Base R ∧ 1 ˙ ∈ Base J → x ⋅ R 1 ˙ ∈ Base J
49 46 47 48 syl2an2r ⊢ φ ∧ x ∈ Base R → x ⋅ R 1 ˙ ∈ Base J
50 9 19 5 20 49 ringlidmd ⊢ φ ∧ x ∈ Base R → 1 ˙ ⋅ J x ⋅ R 1 ˙ = x ⋅ R 1 ˙
51 24 oveqd ⊢ φ → 1 ˙ ⋅ R x ⋅ R 1 ˙ = 1 ˙ ⋅ J x ⋅ R 1 ˙
52 51 eqeq1d ⊢ φ → 1 ˙ ⋅ R x ⋅ R 1 ˙ = x ⋅ R 1 ˙ ↔ 1 ˙ ⋅ J x ⋅ R 1 ˙ = x ⋅ R 1 ˙
53 52 adantr ⊢ φ ∧ x ∈ Base R → 1 ˙ ⋅ R x ⋅ R 1 ˙ = x ⋅ R 1 ˙ ↔ 1 ˙ ⋅ J x ⋅ R 1 ˙ = x ⋅ R 1 ˙
54 50 53 mpbird ⊢ φ ∧ x ∈ Base R → 1 ˙ ⋅ R x ⋅ R 1 ˙ = x ⋅ R 1 ˙
55 18 28 54 3eqtr3d ⊢ φ ∧ x ∈ Base R → 1 ˙ ⋅ R x = x ⋅ R 1 ˙
56 55 ralrimiva ⊢ φ → ∀ x ∈ Base R 1 ˙ ⋅ R x = x ⋅ R 1 ˙
57 ssidd ⊢ φ → Base R ⊆ Base R
58 6 7 mgpbas ⊢ Base R = Base M
59 6 16 mgpplusg ⊢ ⋅ R = + M
60 eqid ⊢ Cntz ⁡ M = Cntz ⁡ M
61 58 59 60 elcntz ⊢ Base R ⊆ Base R → 1 ˙ ∈ Cntz ⁡ M ⁡ Base R ↔ 1 ˙ ∈ Base R ∧ ∀ x ∈ Base R 1 ˙ ⋅ R x = x ⋅ R 1 ˙
62 57 61 syl ⊢ φ → 1 ˙ ∈ Cntz ⁡ M ⁡ Base R ↔ 1 ˙ ∈ Base R ∧ ∀ x ∈ Base R 1 ˙ ⋅ R x = x ⋅ R 1 ˙
63 12 56 62 mpbir2and ⊢ φ → 1 ˙ ∈ Cntz ⁡ M ⁡ Base R
64 58 60 cntrval ⊢ Cntz ⁡ M ⁡ Base R = Cntr ⁡ M
65 63 64 eleqtrdi ⊢ φ → 1 ˙ ∈ Cntr ⁡ M