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 = ( Poly1𝑅 )
subrgply1.h 𝐻 = ( 𝑅s 𝑇 )
subrgply1.u 𝑈 = ( Poly1𝐻 )
subrgply1.b 𝐵 = ( Base ‘ 𝑈 )
Assertion subrgply1 ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) → 𝐵 ∈ ( SubRing ‘ 𝑆 ) )

Proof

Step Hyp Ref Expression
1 subrgply1.s 𝑆 = ( Poly1𝑅 )
2 subrgply1.h 𝐻 = ( 𝑅s 𝑇 )
3 subrgply1.u 𝑈 = ( Poly1𝐻 )
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 eqid ( PwSer1𝐻 ) = ( PwSer1𝐻 )
9 3 8 4 ply1bas 𝐵 = ( Base ‘ ( 1o mPoly 𝐻 ) )
10 6 2 7 9 subrgmpl ( ( 1o ∈ On ∧ 𝑇 ∈ ( SubRing ‘ 𝑅 ) ) → 𝐵 ∈ ( SubRing ‘ ( 1o mPoly 𝑅 ) ) )
11 5 10 mpan ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) → 𝐵 ∈ ( SubRing ‘ ( 1o mPoly 𝑅 ) ) )
12 eqidd ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) → ( Base ‘ 𝑆 ) = ( Base ‘ 𝑆 ) )
13 eqid ( PwSer1𝑅 ) = ( PwSer1𝑅 )
14 eqid ( Base ‘ 𝑆 ) = ( Base ‘ 𝑆 )
15 1 13 14 ply1bas ( Base ‘ 𝑆 ) = ( Base ‘ ( 1o mPoly 𝑅 ) )
16 15 a1i ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) → ( Base ‘ 𝑆 ) = ( Base ‘ ( 1o mPoly 𝑅 ) ) )
17 eqid ( +g𝑆 ) = ( +g𝑆 )
18 1 6 17 ply1plusg ( +g𝑆 ) = ( +g ‘ ( 1o mPoly 𝑅 ) )
19 18 a1i ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) → ( +g𝑆 ) = ( +g ‘ ( 1o mPoly 𝑅 ) ) )
20 19 oveqdr ( ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) ∧ ( 𝑥 ∈ ( Base ‘ 𝑆 ) ∧ 𝑦 ∈ ( Base ‘ 𝑆 ) ) ) → ( 𝑥 ( +g𝑆 ) 𝑦 ) = ( 𝑥 ( +g ‘ ( 1o mPoly 𝑅 ) ) 𝑦 ) )
21 eqid ( .r𝑆 ) = ( .r𝑆 )
22 1 6 21 ply1mulr ( .r𝑆 ) = ( .r ‘ ( 1o mPoly 𝑅 ) )
23 22 a1i ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) → ( .r𝑆 ) = ( .r ‘ ( 1o mPoly 𝑅 ) ) )
24 23 oveqdr ( ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) ∧ ( 𝑥 ∈ ( Base ‘ 𝑆 ) ∧ 𝑦 ∈ ( Base ‘ 𝑆 ) ) ) → ( 𝑥 ( .r𝑆 ) 𝑦 ) = ( 𝑥 ( .r ‘ ( 1o mPoly 𝑅 ) ) 𝑦 ) )
25 12 16 20 24 subrgpropd ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) → ( SubRing ‘ 𝑆 ) = ( SubRing ‘ ( 1o mPoly 𝑅 ) ) )
26 11 25 eleqtrrd ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) → 𝐵 ∈ ( SubRing ‘ 𝑆 ) )