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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 !

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  • $\begingroup$ 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. $\endgroup$ – Mindlack Sep 11 at 11:09
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    $\begingroup$ 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. $\endgroup$ – ujsgeyrr1f0d0d0r0h1h0j0j_juj Sep 11 at 11:15
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$\mathbb{C}[X]\rightarrow\mathbb{Q}[X]$ 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.

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

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