Is there a field $K$ for which the kernel of the map $\mathbb{Z}[x]\to K$ is NOT principal? 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!
 A: What about $f:\mathbb{Z}[X] \rightarrow \mathbb{Z}/2\mathbb{Z}$ defined for all $P \in \mathbb{Z}[X]$ by
$$f(P)=P(0) \quad \mod \text{ }2 \quad \quad ?$$
A: If the map $\mathbb{Z}[x] \to K$ happens to be surjective, then its kernel will be a maximal ideal. So we could first ask if there are any maximal ideals of $\mathbb{Z}[x]$ which are not principal. How do we make a maximal ideal of $\mathbb{Z}[x]$? We can just take a maximal ideal of $\mathbb{Z}$, e.g. $(2)$, and "lift" it to $\mathbb{Z}[x]$ by adding in the generator $x$. The quotient $\mathbb{Z}[x]/(2,x)$ is isomorphic to $\mathbb{Z}/(2)$, which is a field, so $(2,x)$ is a maximal ideal of $\mathbb{Z}[x]$. Why can't this ideal be principal?
Note: this is "the same" as @TheSilverDoe's answer.
Note 2: Any map $f : \mathbb{Z}[x] \to K$ gives a surjection $\mathbb{Z}[x] \to \operatorname{img}(f)$ with the same kernel, and $\operatorname{img}(f)$ is a domain. Conversely, every surjection $\mathbb{Z}[x] \to R$ with $R$ a domain gives a map $\mathbb{Z}[x] \to \operatorname{Frac}(R)$ with the same kernel, and $\operatorname{Frac}(R)$ is a field. Thus, it's equivalent to find a surjection from $\mathbb{Z}[x]$ to a domain with non-principal kernel, which is equivalent to finding a non-principal prime ideal of $\mathbb{Z}$. In particular, I focused on maximal ideals, but if this didn't work you could think about more general primes!
