Metamath Proof Explorer


Theorem pf1subrg

Description: Polynomial functions are a subring. (Contributed by Mario Carneiro, 19-Mar-2015) (Revised by Mario Carneiro, 6-May-2015)

Ref Expression
Hypotheses pf1const.b ⊢ B = Base R
pf1const.q ⊢ Q = ran ⁡ eval 1 ⁡ R
Assertion pf1subrg ⊢ R ∈ CRing → Q ∈ SubRing ⁡ R ↑ 𝑠 B

Proof

Step Hyp Ref Expression
1 pf1const.b ⊢ B = Base R
2 pf1const.q ⊢ Q = ran ⁡ eval 1 ⁡ R
3 eqid ⊢ eval 1 ⁡ R = eval 1 ⁡ R
4 eqid ⊢ Poly 1 ⁡ R = Poly 1 ⁡ R
5 eqid ⊢ R ↑ 𝑠 B = R ↑ 𝑠 B
6 3 4 5 1 evl1rhm ⊢ R ∈ CRing → eval 1 ⁡ R ∈ Poly 1 ⁡ R RingHom R ↑ 𝑠 B
7 eqid ⊢ Base Poly 1 ⁡ R = Base Poly 1 ⁡ R
8 eqid ⊢ Base R ↑ 𝑠 B = Base R ↑ 𝑠 B
9 7 8 rhmf ⊢ eval 1 ⁡ R ∈ Poly 1 ⁡ R RingHom R ↑ 𝑠 B → eval 1 ⁡ R : Base Poly 1 ⁡ R ⟶ Base R ↑ 𝑠 B
10 ffn ⊢ eval 1 ⁡ R : Base Poly 1 ⁡ R ⟶ Base R ↑ 𝑠 B → eval 1 ⁡ R Fn Base Poly 1 ⁡ R
11 fnima ⊢ eval 1 ⁡ R Fn Base Poly 1 ⁡ R → eval 1 ⁡ R Base Poly 1 ⁡ R = ran ⁡ eval 1 ⁡ R
12 6 9 10 11 4syl ⊢ R ∈ CRing → eval 1 ⁡ R Base Poly 1 ⁡ R = ran ⁡ eval 1 ⁡ R
13 12 2 eqtr4di ⊢ R ∈ CRing → eval 1 ⁡ R Base Poly 1 ⁡ R = Q
14 4 ply1assa ⊢ R ∈ CRing → Poly 1 ⁡ R ∈ AssAlg
15 assaring ⊢ Poly 1 ⁡ R ∈ AssAlg → Poly 1 ⁡ R ∈ Ring
16 7 subrgid ⊢ Poly 1 ⁡ R ∈ Ring → Base Poly 1 ⁡ R ∈ SubRing ⁡ Poly 1 ⁡ R
17 14 15 16 3syl ⊢ R ∈ CRing → Base Poly 1 ⁡ R ∈ SubRing ⁡ Poly 1 ⁡ R
18 rhmima ⊢ eval 1 ⁡ R ∈ Poly 1 ⁡ R RingHom R ↑ 𝑠 B ∧ Base Poly 1 ⁡ R ∈ SubRing ⁡ Poly 1 ⁡ R → eval 1 ⁡ R Base Poly 1 ⁡ R ∈ SubRing ⁡ R ↑ 𝑠 B
19 6 17 18 syl2anc ⊢ R ∈ CRing → eval 1 ⁡ R Base Poly 1 ⁡ R ∈ SubRing ⁡ R ↑ 𝑠 B
20 13 19 eqeltrrd ⊢ R ∈ CRing → Q ∈ SubRing ⁡ R ↑ 𝑠 B