Consider $K$ a field and a ring homomorphism $\phi:\mathbb{Z}[x]\to K$.
I guess this is wrong, because I was asked to give an example of precisely a field $K$ for which the kernel of the map above is not principal, but this is what I've thought so far: as long as $K$ is field (I had $K=\mathbb{Q},\mathbb{R}, \mathbb{C}$ in mind), $\ker{\phi}$ must be principal. The reason is that you consider taking $\mathbb{Z}[x]$ to its ring of fractions $\mathbb{Q}[x]$ and now consider the homorphism $\bar{\phi}:\mathbb{Q}[x]\to K$. Since $\mathbb{Q}[x]$ is an Euclidean domain, it's a PID, and so $\ker{\bar{\phi}}$ must be principal, generated by a polynomial of minimal degree (if not zero) having an element of $K$ as root. Now $\ker{\phi} = \ker{\bar{\phi}}\cap\mathbb{Z}[x]$ so it keeps being principal (if not zero).
Is my reasoning correct or is there a field $K$ for which the kernel of the map $\phi:\mathbb{Z}[x]\to K$ is not principal? I feel like I'm missing something important. Thank you so much!
Best Answer
What about $f:\mathbb{Z}[X] \rightarrow \mathbb{Z}/2\mathbb{Z}$ defined for all $P \in \mathbb{Z}[X]$ by $$f(P)=P(0) \quad \mod \text{ }2 \quad \quad ?$$