Metamath Proof Explorer


Theorem ressply10g

Description: A restricted polynomial algebra has the same group identity (zero polynomial). (Contributed by Thierry Arnoux, 20-Jan-2025)

Ref Expression
Hypotheses ressply.1 𝑆 = ( Poly1𝑅 )
ressply.2 𝐻 = ( 𝑅s 𝑇 )
ressply.3 𝑈 = ( Poly1𝐻 )
ressply.4 𝐵 = ( Base ‘ 𝑈 )
ressply.5 ( 𝜑𝑇 ∈ ( SubRing ‘ 𝑅 ) )
ressply10g.6 𝑍 = ( 0g𝑆 )
Assertion ressply10g ( 𝜑𝑍 = ( 0g𝑈 ) )

Proof

Step Hyp Ref Expression
1 ressply.1 𝑆 = ( Poly1𝑅 )
2 ressply.2 𝐻 = ( 𝑅s 𝑇 )
3 ressply.3 𝑈 = ( Poly1𝐻 )
4 ressply.4 𝐵 = ( Base ‘ 𝑈 )
5 ressply.5 ( 𝜑𝑇 ∈ ( SubRing ‘ 𝑅 ) )
6 ressply10g.6 𝑍 = ( 0g𝑆 )
7 eqid ( algSc ‘ 𝑆 ) = ( algSc ‘ 𝑆 )
8 eqid ( algSc ‘ 𝑈 ) = ( algSc ‘ 𝑈 )
9 1 7 2 3 5 8 subrg1ascl ( 𝜑 → ( algSc ‘ 𝑈 ) = ( ( algSc ‘ 𝑆 ) ↾ 𝑇 ) )
10 9 fveq1d ( 𝜑 → ( ( algSc ‘ 𝑈 ) ‘ ( 0g𝐻 ) ) = ( ( ( algSc ‘ 𝑆 ) ↾ 𝑇 ) ‘ ( 0g𝐻 ) ) )
11 eqid ( 0g𝐻 ) = ( 0g𝐻 )
12 eqid ( 0g𝑈 ) = ( 0g𝑈 )
13 2 subrgring ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) → 𝐻 ∈ Ring )
14 5 13 syl ( 𝜑𝐻 ∈ Ring )
15 3 8 11 12 14 ply1ascl0 ( 𝜑 → ( ( algSc ‘ 𝑈 ) ‘ ( 0g𝐻 ) ) = ( 0g𝑈 ) )
16 eqid ( 0g𝑅 ) = ( 0g𝑅 )
17 2 16 subrg0 ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) → ( 0g𝑅 ) = ( 0g𝐻 ) )
18 5 17 syl ( 𝜑 → ( 0g𝑅 ) = ( 0g𝐻 ) )
19 subrgsubg ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) → 𝑇 ∈ ( SubGrp ‘ 𝑅 ) )
20 16 subg0cl ( 𝑇 ∈ ( SubGrp ‘ 𝑅 ) → ( 0g𝑅 ) ∈ 𝑇 )
21 5 19 20 3syl ( 𝜑 → ( 0g𝑅 ) ∈ 𝑇 )
22 18 21 eqeltrrd ( 𝜑 → ( 0g𝐻 ) ∈ 𝑇 )
23 22 fvresd ( 𝜑 → ( ( ( algSc ‘ 𝑆 ) ↾ 𝑇 ) ‘ ( 0g𝐻 ) ) = ( ( algSc ‘ 𝑆 ) ‘ ( 0g𝐻 ) ) )
24 10 15 23 3eqtr3d ( 𝜑 → ( 0g𝑈 ) = ( ( algSc ‘ 𝑆 ) ‘ ( 0g𝐻 ) ) )
25 18 fveq2d ( 𝜑 → ( ( algSc ‘ 𝑆 ) ‘ ( 0g𝑅 ) ) = ( ( algSc ‘ 𝑆 ) ‘ ( 0g𝐻 ) ) )
26 subrgrcl ( 𝑇 ∈ ( SubRing ‘ 𝑅 ) → 𝑅 ∈ Ring )
27 5 26 syl ( 𝜑𝑅 ∈ Ring )
28 1 7 16 6 27 ply1ascl0 ( 𝜑 → ( ( algSc ‘ 𝑆 ) ‘ ( 0g𝑅 ) ) = 𝑍 )
29 24 25 28 3eqtr2rd ( 𝜑𝑍 = ( 0g𝑈 ) )