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Metamath Proof Explorer

Theorem ply1scl0

Description: The zero scalar is zero. (Contributed by Stefan O'Rear, 29-Mar-2015)

Ref Expression
Hypotheses ply1scl.p ⊢ 𝑃 = ( Poly1 ‘ 𝑅 )
ply1scl.a ⊢ 𝐴 = ( algSc ‘ 𝑃 )
ply1scl0.z ⊢ 0 = ( 0g ‘ 𝑅 )
ply1scl0.y ⊢ 𝑌 = ( 0g ‘ 𝑃 )
Assertion ply1scl0 ( 𝑅 ∈ Ring → ( 𝐴 ‘ 0 ) = 𝑌 )

Proof

Step Hyp Ref Expression
1 ply1scl.p ⊢ 𝑃 = ( Poly1 ‘ 𝑅 )
2 ply1scl.a ⊢ 𝐴 = ( algSc ‘ 𝑃 )
3 ply1scl0.z ⊢ 0 = ( 0g ‘ 𝑅 )
4 ply1scl0.y ⊢ 𝑌 = ( 0g ‘ 𝑃 )
5 1 ply1sca ⊢ ( 𝑅 ∈ Ring → 𝑅 = ( Scalar ‘ 𝑃 ) )
6 5 fveq2d ⊢ ( 𝑅 ∈ Ring → ( 0g ‘ 𝑅 ) = ( 0g ‘ ( Scalar ‘ 𝑃 ) ) )
7 3 6 eqtrid ⊢ ( 𝑅 ∈ Ring → 0 = ( 0g ‘ ( Scalar ‘ 𝑃 ) ) )
8 7 fveq2d ⊢ ( 𝑅 ∈ Ring → ( 𝐴 ‘ 0 ) = ( 𝐴 ‘ ( 0g ‘ ( Scalar ‘ 𝑃 ) ) ) )
9 eqid ⊢ ( Scalar ‘ 𝑃 ) = ( Scalar ‘ 𝑃 )
10 1 ply1lmod ⊢ ( 𝑅 ∈ Ring → 𝑃 ∈ LMod )
11 1 ply1ring ⊢ ( 𝑅 ∈ Ring → 𝑃 ∈ Ring )
12 2 9 10 11 ascl0 ⊢ ( 𝑅 ∈ Ring → ( 𝐴 ‘ ( 0g ‘ ( Scalar ‘ 𝑃 ) ) ) = ( 0g ‘ 𝑃 ) )
13 8 12 eqtrd ⊢ ( 𝑅 ∈ Ring → ( 𝐴 ‘ 0 ) = ( 0g ‘ 𝑃 ) )
14 13 4 eqtr4di ⊢ ( 𝑅 ∈ Ring → ( 𝐴 ‘ 0 ) = 𝑌 )