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

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.

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.

• Thank you very much! Whenever you have time, I will be thankful if you can elaborate a little on 'if and only if' May 12, 2017 at 13:05
• I think that your answer is what I was looking for. Also, it shows that if $w$ is integral over $\mathbb{C}[x]$, then $x-a$ does not remain prime in $\mathbb{C}[x][w]$ (since $c_m(a)=1 \neq 0$), and in particular $x$ does not remain prime in $\mathbb{C}[x][x^{3/2}]$. May 14, 2017 at 12:49

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.

• The technical language I was omitting is that Spec$\mathbb{C}[x]$ rather than $\mathbb{C}[x]$ is the line, and the same for Spec$\mathbb{C}[x,w]$ being a two-dimensional complex space. I also realize that I'm basically lying about the dimensions of our space, but... I'm OK with that. Welcome to geometry :P May 12, 2017 at 3:49
• Part of the difference between your question and one of the questions you referenced is that they are proooobably considering a field which has extensions rather than the perfect, algebraically closed, etc. beauty which is $\mathbb{C}$... May 12, 2017 at 4:25
• Thank you very much for your answer and comments! I am not sure if I fully understood your answer (I will let you know later if I still have gaps in understanding it). May 12, 2017 at 13:06
• Perhaps it would help to read a version of Hilbert's Nullstellensatz at some point. Basically maximal ideals in $\mathbb{C}[x_1,...,x_n]$ are of the form $(x_1 - a_1,x_2 - a_2,...,x_n - a_n)$ for some point $(a_1,...,a_n)\in\mathbb{C}^n$ so that points $(a,b)\in\mathbb{C}^2$ correspond to maximal ideals of the form $(x - a, w - b)$ in $\mathbb{C}[x,w]$. Hence just giving one coordinate, namely just specifying that $x = a$ (equivalently mod'ing out by the ideal $(x-a)$) is like specifying the line above $x = a$ in $\mathbb{C}^2$. (This would be much easier with a picture.) May 12, 2017 at 15:41
• A line in the plane is irreducible (in that it cannot be written as a union of some finitely many irreducible components in the plane, e.g. a finite number of points or curves) so that the ideal corresponding to the line, $(x-a)$ in this case, is prime. (Prime ideals of $\mathbb{C}[x,w]$ correspond to irreducible subvarieties of $\mathbb{C}^2$. May 12, 2017 at 15:43