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 3 4 ply1bas B = Base 1 𝑜 mPoly H
9 6 2 7 8 subrgmpl 1 𝑜 On T SubRing R B SubRing 1 𝑜 mPoly R
10 5 9 mpan T SubRing R B SubRing 1 𝑜 mPoly R
11 eqidd T SubRing R Base S = Base S
12 eqid Base S = Base S
13 1 12 ply1bas Base S = Base 1 𝑜 mPoly R
14 13 a1i T SubRing R Base S = Base 1 𝑜 mPoly R
15 eqid + S = + S
16 1 6 15 ply1plusg + S = + 1 𝑜 mPoly R
17 16 a1i T SubRing R + S = + 1 𝑜 mPoly R
18 17 oveqdr T SubRing R x Base S y Base S x + S y = x + 1 𝑜 mPoly R y
19 eqid S = S
20 1 6 19 ply1mulr S = 1 𝑜 mPoly R
21 20 a1i T SubRing R S = 1 𝑜 mPoly R
22 21 oveqdr T SubRing R x Base S y Base S x S y = x 1 𝑜 mPoly R y
23 11 14 18 22 subrgpropd T SubRing R SubRing S = SubRing 1 𝑜 mPoly R
24 10 23 eleqtrrd T SubRing R B SubRing S