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Metamath Proof Explorer

Theorem subrgply1

Description: A subring of the base ring induces a subring of polynomials. (Contributed by Mario Carneiro, 3-Jul-2015)

Ref Expression
Hypotheses subrgply1.s ⊢ 𝑆 = ( Poly1 ‘ 𝑅 )
subrgply1.h ⊢ 𝐻 = ( 𝑅 ↾s 𝑇 )
subrgply1.u ⊢ 𝑈 = ( Poly1 ‘ 𝐻 )
subrgply1.b ⊢ 𝐵 = ( Base ‘ 𝑈 )
Assertion subrgply1 ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) → 𝐵 ∈ ( SubRing ‘ 𝑆 ) )

Proof

Step Hyp Ref Expression
1 subrgply1.s ⊢ 𝑆 = ( Poly1 ‘ 𝑅 )
2 subrgply1.h ⊢ 𝐻 = ( 𝑅 ↾s 𝑇 )
3 subrgply1.u ⊢ 𝑈 = ( Poly1 ‘ 𝐻 )
4 subrgply1.b ⊢ 𝐵 = ( Base ‘ 𝑈 )
5 1on ⊢ 1o ∈ On
6 eqid ⊢ ( 1o mPoly 𝑅 ) = ( 1o mPoly 𝑅 )
7 eqid ⊢ ( 1o mPoly 𝐻 ) = ( 1o mPoly 𝐻 )
8 3 4 ply1bas ⊢ 𝐵 = ( Base ‘ ( 1o mPoly 𝐻 ) )
9 6 2 7 8 subrgmpl ⊢ ( ( 1o ∈ On ∧ 𝑇 ∈ ( SubRing ‘ 𝑅 ) ) → 𝐵 ∈ ( SubRing ‘ ( 1o mPoly 𝑅 ) ) )
10 5 9 mpan ⊢ ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) → 𝐵 ∈ ( SubRing ‘ ( 1o mPoly 𝑅 ) ) )
11 eqidd ⊢ ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) → ( Base ‘ 𝑆 ) = ( Base ‘ 𝑆 ) )
12 eqid ⊢ ( Base ‘ 𝑆 ) = ( Base ‘ 𝑆 )
13 1 12 ply1bas ⊢ ( Base ‘ 𝑆 ) = ( Base ‘ ( 1o mPoly 𝑅 ) )
14 13 a1i ⊢ ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) → ( Base ‘ 𝑆 ) = ( Base ‘ ( 1o mPoly 𝑅 ) ) )
15 eqid ⊢ ( +g ‘ 𝑆 ) = ( +g ‘ 𝑆 )
16 1 6 15 ply1plusg ⊢ ( +g ‘ 𝑆 ) = ( +g ‘ ( 1o mPoly 𝑅 ) )
17 16 a1i ⊢ ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) → ( +g ‘ 𝑆 ) = ( +g ‘ ( 1o mPoly 𝑅 ) ) )
18 17 oveqdr ⊢ ( ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) ∧ ( 𝑥 ∈ ( Base ‘ 𝑆 ) ∧ 𝑦 ∈ ( Base ‘ 𝑆 ) ) ) → ( 𝑥 ( +g ‘ 𝑆 ) 𝑦 ) = ( 𝑥 ( +g ‘ ( 1o mPoly 𝑅 ) ) 𝑦 ) )
19 eqid ⊢ ( .r ‘ 𝑆 ) = ( .r ‘ 𝑆 )
20 1 6 19 ply1mulr ⊢ ( .r ‘ 𝑆 ) = ( .r ‘ ( 1o mPoly 𝑅 ) )
21 20 a1i ⊢ ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) → ( .r ‘ 𝑆 ) = ( .r ‘ ( 1o mPoly 𝑅 ) ) )
22 21 oveqdr ⊢ ( ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) ∧ ( 𝑥 ∈ ( Base ‘ 𝑆 ) ∧ 𝑦 ∈ ( Base ‘ 𝑆 ) ) ) → ( 𝑥 ( .r ‘ 𝑆 ) 𝑦 ) = ( 𝑥 ( .r ‘ ( 1o mPoly 𝑅 ) ) 𝑦 ) )
23 11 14 18 22 subrgpropd ⊢ ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) → ( SubRing ‘ 𝑆 ) = ( SubRing ‘ ( 1o mPoly 𝑅 ) ) )
24 10 23 eleqtrrd ⊢ ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) → 𝐵 ∈ ( SubRing ‘ 𝑆 ) )