Is there a way to characterize the prime ideals in $\mathbb{R}[x_1,x_2, \dots , x_n]$?

Question

I'm studying algebras which can be formed by the quotient of principal ideals in $\mathbb{R}[x_1, \dots , x_n]$, and thus would like to be able to determine which of said principal ideals are maximal, so I can determine which algebras will be integral domains, but I'm not sure of a good way of characterizing the prime ideals in $\mathbb{R}[x_1, \dots , x_n]$, or even just in $\mathbb{R}[x]$.

Is a good (i.e. better than just checking ideals case by case) characterization of such prime ideals known? Specifically for now I am looking only at principal ideals, but I'd also be interested to know a more general characterization of prime ideals in $\mathbb{R}[x_1, \dots , x_n]$, if possible.

Answer 1

Partial answer: Since $\mathbb{R}[x]$ is a PID, we can just focus on the prime/maximal elements. Define

$$A=\{ax+b : a \in \mathbb{R}_{\neq 0}\}$$

$$B=\{ax^2+bx+c : a \in \mathbb{R}_{\neq 0}, b \in \mathbb{R},c\in \mathbb{R} ,b^2-4ac < 0\}$$

Then $A \cup B$ is the set of prime elements of the ring $\mathbb{R}[x]$.

Claim 0. If $a \neq 0$, then $$\mathbb{R}[x]/(ax+b) \cong \mathbb{R}$$

Claim 1. If $a \neq 0$, then it seems possible that for $b^2-4ac<0$, we have: $$\mathbb{R}[x]/(ax^2+bx+c) \cong \mathbb{C}$$

I don't know if Claim 1 is actually true.

Proof of Claim 0. Write $$\left[x \mapsto \frac{-b}{a}\right] : \mathbb{R}[x] \rightarrow \mathbb{R}$$ for the unique such $\mathbb{R}$-algebra homomorphism that maps $x$ to $\frac{-b}{a}.$ Then the following are equivalent:

• $P \in \mathrm{ker}\left(\left[x \mapsto \frac{-b}{a}\right]\right)$
• $\left[x \mapsto \frac{-b}{a}\right]P = 0$
• $(x-\frac{-b}{a}) \mid P$
• $(x+\frac{b}{a}) \mid P$
• $ax+b \mid P$
• $P \in (ax+b)$

Hence we get an isomorphism of $\mathbb{R}$-algebras: $$\mathbb{R}[x]/(ax+b) \rightarrow \mathbb{R}$$