Let $\mathbb R[x]$ denote the ring of all polynomials with real coefficients. The mapping $f(x)\rightarrow f(1)$ is a ring homomorphism from $\mathbb R[x]$ onto $\mathbb R$.
Question:
Describe the kernel of this ring homomorphism.
Recall the definition of kernel of a $\left ( ring \right )$ homomorphism:
$\ker\left ( \varphi \right )=\left \{ f\left ( x \right ) \in\mathbb R\left [ x \right ] \mid \left [ f\left ( x \right ) \right ]\varphi=0 \right \}$
Now, we define:
Define
$\varphi:R\left [ x \right ]\rightarrow R$
$f\left ( x \right ) \mapsto f\left ( 1 \right )$
I was almost certain the kernel is the trivial kernel but my solution sheet dictates otherwise.
Hint is appreciated.
Thanks in advance
Best Answer
Hint: This is primarily a counterexample that shows the kernel to be non-trivial, but it also serves as a strong hint. Let $f(x) = x^2 - 3x + 2$. Then $\varphi(f(x)) = f(1) = 1 - 3 + 2 = 0 \implies f(x) \in \ker \varphi$.
Solution:
\begin{align*} \ker \varphi & = \{\, f(x) \in \mathbb R[x] \mid \varphi(f(x)) = 0 \in \mathbb R \,\}\\ & = \{\, f(x) \in \mathbb R[x] \mid f(1) = 0 \,\}\\ & = \{\, f(x) \in \mathbb R[x] \mid (x - 1)\ \text{is a factor of}\ f(x) \,\}\\ & = \{\, (x -1)g(x) \mid g(x) \in \mathbb R[x] \,\}\\ & = \langle x - 1 \rangle \end{align*}
That is, $\ker \varphi$ is the ideal generated by $x - 1$.