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 !

• No, every prime ideal of $\mathbb{Z}[X]$ is not principal. Example: $(2,X)$. For your maps, I would suggest considering the intersection of your prime ideals with the polynomial rings. – Mindlack Sep 11 at 11:09
• To determine $\mathbf{A}_{\mathbf{Z}}^1 = \textrm{Spec}(\mathbf{Z}[T])$ I would recommand to study the set-theoretic fibers of the morphism $\mathbf{A}_{\mathbf{Z}}^1 \to \textrm{Spec}(\mathbf{Z})$ : two cases : the fibers of non-zero primes ideals and the fibers of the zero ideal. Only the latter correspond to polynomial ideals generated by irreducible polynomials. – ujsgeyrr1f0d0d0r0h1h0j0j_juj Sep 11 at 11:15

$$\mathbb{C}[X]\rightarrow\mathbb{Q}[X]$$ 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.

• You're missing $\operatorname{Spec}$ in front of $\Bbb C[X]$ and $\Bbb Q[X]$. – KReiser Sep 11 at 18:43