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Metamath Proof Explorer

Theorem dvdsr1p

Description: Divisibility in a polynomial ring in terms of the remainder. (Contributed by Stefan O'Rear, 28-Mar-2015)

Ref Expression
Hypotheses dvdsq1p.p ⊢ 𝑃 = ( Poly1 ‘ 𝑅 )
dvdsq1p.d ⊢ ∥ = ( ∥r ‘ 𝑃 )
dvdsq1p.b ⊢ 𝐵 = ( Base ‘ 𝑃 )
dvdsq1p.c ⊢ 𝐶 = ( Unic1p ‘ 𝑅 )
dvdsr1p.z ⊢ 0 = ( 0g ‘ 𝑃 )
dvdsr1p.e ⊢ 𝐸 = ( rem1p ‘ 𝑅 )
Assertion dvdsr1p ( ( 𝑅 ∈ Ring ∧ 𝐹 ∈ 𝐵 ∧ 𝐺 ∈ 𝐶 ) → ( 𝐺 ∥ 𝐹 ↔ ( 𝐹 𝐸 𝐺 ) = 0 ) )

Proof

Step Hyp Ref Expression
1 dvdsq1p.p ⊢ 𝑃 = ( Poly1 ‘ 𝑅 )
2 dvdsq1p.d ⊢ ∥ = ( ∥r ‘ 𝑃 )
3 dvdsq1p.b ⊢ 𝐵 = ( Base ‘ 𝑃 )
4 dvdsq1p.c ⊢ 𝐶 = ( Unic1p ‘ 𝑅 )
5 dvdsr1p.z ⊢ 0 = ( 0g ‘ 𝑃 )
6 dvdsr1p.e ⊢ 𝐸 = ( rem1p ‘ 𝑅 )
7 1 ply1ring ⊢ ( 𝑅 ∈ Ring → 𝑃 ∈ Ring )
8 7 3ad2ant1 ⊢ ( ( 𝑅 ∈ Ring ∧ 𝐹 ∈ 𝐵 ∧ 𝐺 ∈ 𝐶 ) → 𝑃 ∈ Ring )
9 ringgrp ⊢ ( 𝑃 ∈ Ring → 𝑃 ∈ Grp )
10 8 9 syl ⊢ ( ( 𝑅 ∈ Ring ∧ 𝐹 ∈ 𝐵 ∧ 𝐺 ∈ 𝐶 ) → 𝑃 ∈ Grp )
11 simp2 ⊢ ( ( 𝑅 ∈ Ring ∧ 𝐹 ∈ 𝐵 ∧ 𝐺 ∈ 𝐶 ) → 𝐹 ∈ 𝐵 )
12 eqid ⊢ ( quot1p ‘ 𝑅 ) = ( quot1p ‘ 𝑅 )
13 12 1 3 4 q1pcl ⊢ ( ( 𝑅 ∈ Ring ∧ 𝐹 ∈ 𝐵 ∧ 𝐺 ∈ 𝐶 ) → ( 𝐹 ( quot1p ‘ 𝑅 ) 𝐺 ) ∈ 𝐵 )
14 1 3 4 uc1pcl ⊢ ( 𝐺 ∈ 𝐶 → 𝐺 ∈ 𝐵 )
15 14 3ad2ant3 ⊢ ( ( 𝑅 ∈ Ring ∧ 𝐹 ∈ 𝐵 ∧ 𝐺 ∈ 𝐶 ) → 𝐺 ∈ 𝐵 )
16 eqid ⊢ ( .r ‘ 𝑃 ) = ( .r ‘ 𝑃 )
17 3 16 ringcl ⊢ ( ( 𝑃 ∈ Ring ∧ ( 𝐹 ( quot1p ‘ 𝑅 ) 𝐺 ) ∈ 𝐵 ∧ 𝐺 ∈ 𝐵 ) → ( ( 𝐹 ( quot1p ‘ 𝑅 ) 𝐺 ) ( .r ‘ 𝑃 ) 𝐺 ) ∈ 𝐵 )
18 8 13 15 17 syl3anc ⊢ ( ( 𝑅 ∈ Ring ∧ 𝐹 ∈ 𝐵 ∧ 𝐺 ∈ 𝐶 ) → ( ( 𝐹 ( quot1p ‘ 𝑅 ) 𝐺 ) ( .r ‘ 𝑃 ) 𝐺 ) ∈ 𝐵 )
19 eqid ⊢ ( -g ‘ 𝑃 ) = ( -g ‘ 𝑃 )
20 3 5 19 grpsubeq0 ⊢ ( ( 𝑃 ∈ Grp ∧ 𝐹 ∈ 𝐵 ∧ ( ( 𝐹 ( quot1p ‘ 𝑅 ) 𝐺 ) ( .r ‘ 𝑃 ) 𝐺 ) ∈ 𝐵 ) → ( ( 𝐹 ( -g ‘ 𝑃 ) ( ( 𝐹 ( quot1p ‘ 𝑅 ) 𝐺 ) ( .r ‘ 𝑃 ) 𝐺 ) ) = 0 ↔ 𝐹 = ( ( 𝐹 ( quot1p ‘ 𝑅 ) 𝐺 ) ( .r ‘ 𝑃 ) 𝐺 ) ) )
21 10 11 18 20 syl3anc ⊢ ( ( 𝑅 ∈ Ring ∧ 𝐹 ∈ 𝐵 ∧ 𝐺 ∈ 𝐶 ) → ( ( 𝐹 ( -g ‘ 𝑃 ) ( ( 𝐹 ( quot1p ‘ 𝑅 ) 𝐺 ) ( .r ‘ 𝑃 ) 𝐺 ) ) = 0 ↔ 𝐹 = ( ( 𝐹 ( quot1p ‘ 𝑅 ) 𝐺 ) ( .r ‘ 𝑃 ) 𝐺 ) ) )
22 6 1 3 12 16 19 r1pval ⊢ ( ( 𝐹 ∈ 𝐵 ∧ 𝐺 ∈ 𝐵 ) → ( 𝐹 𝐸 𝐺 ) = ( 𝐹 ( -g ‘ 𝑃 ) ( ( 𝐹 ( quot1p ‘ 𝑅 ) 𝐺 ) ( .r ‘ 𝑃 ) 𝐺 ) ) )
23 11 15 22 syl2anc ⊢ ( ( 𝑅 ∈ Ring ∧ 𝐹 ∈ 𝐵 ∧ 𝐺 ∈ 𝐶 ) → ( 𝐹 𝐸 𝐺 ) = ( 𝐹 ( -g ‘ 𝑃 ) ( ( 𝐹 ( quot1p ‘ 𝑅 ) 𝐺 ) ( .r ‘ 𝑃 ) 𝐺 ) ) )
24 23 eqeq1d ⊢ ( ( 𝑅 ∈ Ring ∧ 𝐹 ∈ 𝐵 ∧ 𝐺 ∈ 𝐶 ) → ( ( 𝐹 𝐸 𝐺 ) = 0 ↔ ( 𝐹 ( -g ‘ 𝑃 ) ( ( 𝐹 ( quot1p ‘ 𝑅 ) 𝐺 ) ( .r ‘ 𝑃 ) 𝐺 ) ) = 0 ) )
25 1 2 3 4 16 12 dvdsq1p ⊢ ( ( 𝑅 ∈ Ring ∧ 𝐹 ∈ 𝐵 ∧ 𝐺 ∈ 𝐶 ) → ( 𝐺 ∥ 𝐹 ↔ 𝐹 = ( ( 𝐹 ( quot1p ‘ 𝑅 ) 𝐺 ) ( .r ‘ 𝑃 ) 𝐺 ) ) )
26 21 24 25 3bitr4rd ⊢ ( ( 𝑅 ∈ Ring ∧ 𝐹 ∈ 𝐵 ∧ 𝐺 ∈ 𝐶 ) → ( 𝐺 ∥ 𝐹 ↔ ( 𝐹 𝐸 𝐺 ) = 0 ) )