Induced maps spec of $\mathbb{Z}[\boldsymbol{x}]$, $\mathbb{Q}[\boldsymbol{x}]$ and $\mathbb{C}[\boldsymbol{x}]$ I consider these three rings, $\mathbb{Z}[\boldsymbol{x}]$, $\mathbb{Q}[\boldsymbol{x}]$ and $\mathbb{C}[\boldsymbol{x}]$ with the natural inclusion :
$\mathbb{Z}[x] \hookrightarrow \mathbb{Q}[x] \hookrightarrow \mathbb{C}[x]$
Now, I know that $\phi : R \longrightarrow S$ a homomorphism of commutative rings,  then prime ideals in  S are mapped to prime ideals in R by $P \mapsto \phi^{-1}(P)$ and so I need to describe the induced maps : 
$ \operatorname{Spec}(\mathbb{C}[x]) \rightarrow \operatorname{Spec}(\mathbb{Q}[x]) \rightarrow \operatorname{Spec}(\mathbb{Z}[x]) $
And if i'm not mistaken, I know that : 
$\operatorname{Spec} \mathbb{Z}[\boldsymbol{x}]=\{(0),(f(x)) : f(x) \text { is an irreducible polynomial }\}$
$\operatorname{Spec} \mathbb{Q}[\boldsymbol{x}]=\{(0),(f(x)) : f(x) \text { is an irreducible polynomial }\} $
$ \operatorname{Spec} \mathbb{C}[x]=\{(0),(x-a) : a \in \mathbb{C}\} $
All help is appreciated, thanks !

$\mathbb{C}[X]\rightarrow\mathbb{Q}[X]$ is described as follows, if $a\in \mathbb{C}$ and is algebraic $f$ its minimal polynomial in $\mathbb{Q}[X]$, $\phi^{-1}(X-a)$  are the elements of $\mathbb{Q}[X]$ which are divided by $X-a$, it is is the ideal generated by the minimal polynomial of $a$.
If $a$ its trancendental,  show that only the image of the zero ideal is contained in $(X-a)$ since a root of a polynomial with rational coefficients is algebraic. 
