(January 2014) For the polynomial ring \(R=\mathbb{R}[x]\), define \[ I=\{f\in R\mid f(2)=f(5)=0\}. \] Prove that \(I\) is an ideal of \(R\).
Solution
Proof. (Nonempty) Note that 0∈ R gives the constant polynomial 0 ∈R. Since 0(2) = 0(5) = 0, we have 0∈I. (Closed under−) Given f,g ∈I, we have f(2) =f(5) =g(2) =g(5) = 0. So (f−g)(2) =f(2)−g(2) = 0− 0 = 0, and (f−g)(5) =f(5)−g(5) = 0− 0 = 0, and hence f−g∈I. (Sticky) Given f∈I and g∈R, we have f(2) =f(5) = 0. So (gf )(2) =g(2)f(2) =g(2)· 0 = 0, and (gf )(5) =g(5)f(5) =g(5)· 0 = 0. Thus, gf∈I. Since R is commutative, we also have fg =gf∈I. QED
The Division Algorithm. Know the statement of the division algorithm in k[x]. Here is a problem that uses the division algorithm.