Function between $\operatorname{Spec}(\Bbb{C}[x])$ and $\operatorname{Spec}(\Bbb{R}[x])$ I was asked to describe $\operatorname{Spec}(\Bbb{C}[x])$ and $\operatorname{Spec}(\Bbb{R}[x])$. After describing them, I want to find the natural function $\phi:\operatorname{Spec}(\Bbb{C}[x])\rightarrow \operatorname{Spec}(\Bbb{R}[x])$ corresponding to the embedding $\psi:\Bbb{R}[x]\rightarrow\Bbb{C}[x]$. Now to my understanding, the embedding is simply $\psi(f(x))=f(x)$. So my attempt at describing this natural function is:
$\phi(\langle f\rangle)=
\begin{cases} 
      \langle 0\rangle & f=0\\
      \langle f\rangle & f\in\Bbb{R}[x] \\
      \langle f\bar f\rangle & f\in\Bbb{C}[x]\setminus\Bbb{R}[x] 
   \end{cases}
\
$
This feels the most natural to me, but am I missing something? There's no claim that this function has to be injective/surjective, but I can't see how it corresponds to the embedding. Any help would be appreciated.
 A: Given a field $K$, every element in $I\in \operatorname{Spec}(K[x])$ is either the ideal $(0)$ or corresponds to a unique monic irreducible polynomial $p$ such that $(p)=I$.
The map $\phi':\operatorname{Spec}(\mathbb C[x])\rightarrow\operatorname{Spec}(\mathbb R[x])$ corresponding to the embedding $j:\mathbb R[x]\hookrightarrow\mathbb C[x]$ sends the ideal $(0)$ to $(0)$, while an ideal $(p)$ (for $p$ a monic irreducible polynomial) is sent to the ideal $$\{f\in\mathbb R[x]|f=j(f)\in(p)\}=\{f\in\mathbb R[x]|f=qp,\mbox{ for some }q\in\mathbb C[x]\}
$$
As said in the comments, a monic irreducible polynomial $p\in\mathbb C[x]$ is a 1 degree polynomial $p=x-\theta$, for some $\theta\in\mathbb C$, so you just need to describe the ideal $\phi'(x-\theta)=\{f\in\mathbb R[x]|f\in (x-\theta)\}$
A: Hint:
$$\operatorname{Spec}(\mathbb{R}[X]) = \{(0)\} \cup \{(aX+b) \ \big| \ a,b\in \mathbb{R}, a\neq 0 \} \cup \{(aX^2+bX+c) \ \big| \ a,b,c \in \mathbb{R}, a\neq 0, \ b^2-4ac<0 \}$$
while
$$\operatorname{Spec}(\mathbb{C}[X]) = \{(0)\} \cup \{(aX+b)  \ \big| \ a,b \in \mathbb{C}, a\neq 0 \} $$
