Prime elements of $\mathbb{C}[x]$ which remain prime in a simple algebraic ring extension

Question

Let $w$ be an algebraic element over $\mathbb{C}[x]$, with minimal polynomial $f(t)=c_mt^m+\cdots+c_1t+c_0$, $c_i \in \mathbb{C}[x]$.

Is it possible to find all prime elements of $\mathbb{C}[x]$ which remain prime in $R=\mathbb{C}[x][w]$? (in case $R$ has prime elements). Perhaps a partial answer should be in terms of the $c_j$'s.

Remarks: An irreducible element of $\mathbb{C}[x]$ is of the form $ax+b$, $0 \neq a,b \in \mathbb{C}$, because $\mathbb{C}$ is algebraically closed (we can replace $\mathbb{C}$ by any algebraically closed field). Also, $\mathbb{C}[x]$ is a UFD, so irreducibles= primes.

Non-example: $R= \mathbb{C}[x][x^{3/2}]$, $f(t)=t^2-x^3$ is the minimal polynomial of $w=x^{3/2}$. Notice that $x$ is prime in $\mathbb{C}[x]$, but it is not prime in $R$, because $xxx=x^{3/2}x^{3/2}$, namely, $x$ divides the product $x^{3/2}x^{3/2}$, but $x$ does not divide $x^{3/2}$. Actually, $R$ does not have prime elements, see this question.

This is a relevant question.

Answer 1

An element of the form $x-a\in\mathbb{C}[x]$ remains prime in $\mathbb{C}[x][w]$, where $w$ satisfies your equation, which I assume is irreducible, if and only if $c_m(a)=\cdots=c_2(a)=0$ and $c_1(a)\neq 0$.

I will elaborate, as you requested, on my answer. Let $f(w)\in\mathbb{C}[w]=A$ be any non-zero polynomial. Then $A/fA$ is a domain if and only if it is a field if and only if $f$ is linear.

In your case, $x-a$ is a prime in $R$ if and only $R/(x-a)R=\mathbb{C}[w]/f(a,w)$ is a domain, where $f(a,w)=c_m(a)w^m+\cdots +c_2(a)w^2+c_1(a)w+c_0(a)$. I hope the rest is clear.

Answer 2

OK let's do some algebraic geometry. I realize that wasn't the tag you gave, but that's what we're doing. Each $(x-a)\in\mathbb{C}[x]$ would correspond to a point $a$ on the complex line. Hence when you pass to $\mathbb{C}[x,w]$, $(x-a)$ now corresponds to not a point but in fact the (sort of, considering we're in four dimensions) vertical line over that point consisting of all points of the form $(a,y)$ where $y\in\mathbb{C}$. A vertical line is still irreducible in the plane, so $(x-a)$ must remain prime (as prime ideals correspond to irreducible varieties/geometric objects). I'm trying to dance around using the technical language of algebraic geometry (which you may not know) and the very intuitive picture which answers your question haha.