Metamath Proof Explorer


Theorem mplcrng

Description: The polynomial ring is a commutative ring. (Contributed by Mario Carneiro, 9-Jan-2015)

Ref Expression
Hypothesis mplgrp.p ⊢ 𝑃 = ( 𝐼 mPoly 𝑅 )
Assertion mplcrng ( ( 𝐼 ∈ 𝑉 ∧ 𝑅 ∈ CRing ) → 𝑃 ∈ CRing )

Proof

Step Hyp Ref Expression
1 mplgrp.p ⊢ 𝑃 = ( 𝐼 mPoly 𝑅 )
2 eqid ⊢ ( 𝐼 mPwSer 𝑅 ) = ( 𝐼 mPwSer 𝑅 )
3 simpl ⊢ ( ( 𝐼 ∈ 𝑉 ∧ 𝑅 ∈ CRing ) → 𝐼 ∈ 𝑉 )
4 simpr ⊢ ( ( 𝐼 ∈ 𝑉 ∧ 𝑅 ∈ CRing ) → 𝑅 ∈ CRing )
5 2 3 4 psrcrng ⊢ ( ( 𝐼 ∈ 𝑉 ∧ 𝑅 ∈ CRing ) → ( 𝐼 mPwSer 𝑅 ) ∈ CRing )
6 eqid ⊢ ( Base ‘ 𝑃 ) = ( Base ‘ 𝑃 )
7 crngring ⊢ ( 𝑅 ∈ CRing → 𝑅 ∈ Ring )
8 7 adantl ⊢ ( ( 𝐼 ∈ 𝑉 ∧ 𝑅 ∈ CRing ) → 𝑅 ∈ Ring )
9 2 1 6 3 8 mplsubrg ⊢ ( ( 𝐼 ∈ 𝑉 ∧ 𝑅 ∈ CRing ) → ( Base ‘ 𝑃 ) ∈ ( SubRing ‘ ( 𝐼 mPwSer 𝑅 ) ) )
10 1 2 6 mplval2 ⊢ 𝑃 = ( ( 𝐼 mPwSer 𝑅 ) ↾s ( Base ‘ 𝑃 ) )
11 10 subrgcrng ⊢ ( ( ( 𝐼 mPwSer 𝑅 ) ∈ CRing ∧ ( Base ‘ 𝑃 ) ∈ ( SubRing ‘ ( 𝐼 mPwSer 𝑅 ) ) ) → 𝑃 ∈ CRing )
12 5 9 11 syl2anc ⊢ ( ( 𝐼 ∈ 𝑉 ∧ 𝑅 ∈ CRing ) → 𝑃 ∈ CRing )