Metamath Proof Explorer


Theorem ply1pid

Description: The polynomials over a field are a PID. (Contributed by Stefan O'Rear, 29-Mar-2015)

Ref Expression
Hypothesis ply1lpir.p 𝑃 = ( Poly1𝑅 )
Assertion ply1pid ( 𝑅 ∈ Field → 𝑃 ∈ PID )

Proof

Step Hyp Ref Expression
1 ply1lpir.p 𝑃 = ( Poly1𝑅 )
2 fldidom ( 𝑅 ∈ Field → 𝑅 ∈ IDomn )
3 1 ply1idom ( 𝑅 ∈ IDomn → 𝑃 ∈ IDomn )
4 2 3 syl ( 𝑅 ∈ Field → 𝑃 ∈ IDomn )
5 isfld ( 𝑅 ∈ Field ↔ ( 𝑅 ∈ DivRing ∧ 𝑅 ∈ CRing ) )
6 5 simplbi ( 𝑅 ∈ Field → 𝑅 ∈ DivRing )
7 1 ply1lpir ( 𝑅 ∈ DivRing → 𝑃 ∈ LPIR )
8 6 7 syl ( 𝑅 ∈ Field → 𝑃 ∈ LPIR )
9 df-pid PID = ( IDomn ∩ LPIR )
10 9 elin2 ( 𝑃 ∈ PID ↔ ( 𝑃 ∈ IDomn ∧ 𝑃 ∈ LPIR ) )
11 4 8 10 sylanbrc ( 𝑅 ∈ Field → 𝑃 ∈ PID )