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 = Poly 1 R
subrgply1.h H = R 𝑠 T
subrgply1.u U = Poly 1 H
subrgply1.b B = Base U
Assertion subrgply1 T SubRing R B SubRing S

Proof

Step Hyp Ref Expression
1 subrgply1.s S = Poly 1 R
2 subrgply1.h H = R 𝑠 T
3 subrgply1.u U = Poly 1 H
4 subrgply1.b B = Base U
5 1on 1 𝑜 On
6 eqid 1 𝑜 mPoly R = 1 𝑜 mPoly R
7 eqid 1 𝑜 mPoly H = 1 𝑜 mPoly H
8 eqid PwSer 1 H = PwSer 1 H
9 3 8 4 ply1bas B = Base 1 𝑜 mPoly H
10 6 2 7 9 subrgmpl 1 𝑜 On T SubRing R B SubRing 1 𝑜 mPoly R
11 5 10 mpan T SubRing R B SubRing 1 𝑜 mPoly R
12 eqidd T SubRing R Base S = Base S
13 eqid PwSer 1 R = PwSer 1 R
14 eqid Base S = Base S
15 1 13 14 ply1bas Base S = Base 1 𝑜 mPoly R
16 15 a1i T SubRing R Base S = Base 1 𝑜 mPoly R
17 eqid + S = + S
18 1 6 17 ply1plusg + S = + 1 𝑜 mPoly R
19 18 a1i T SubRing R + S = + 1 𝑜 mPoly R
20 19 oveqdr T SubRing R x Base S y Base S x + S y = x + 1 𝑜 mPoly R y
21 eqid S = S
22 1 6 21 ply1mulr S = 1 𝑜 mPoly R
23 22 a1i T SubRing R S = 1 𝑜 mPoly R
24 23 oveqdr T SubRing R x Base S y Base S x S y = x 1 𝑜 mPoly R y
25 12 16 20 24 subrgpropd T SubRing R SubRing S = SubRing 1 𝑜 mPoly R
26 11 25 eleqtrrd T SubRing R B SubRing S