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Let $A$ be a ring and let $\mathcal{F}$ be the inductive system of subrings of $A$ which are of finite type over $\mathbb{Z}$: $$ \mathcal{F} = \{ \mathbb{Z}[a_1,\dots,a_n] \subseteq A \mid n \geq 0, \ a_1, \dots, a_n \in A \}. $$ I'd like to know whether the following statement is true: For every $B \in \mathcal{F}$, there exists $C \in \mathcal{F}$ such that $C \supseteq B$ and the map $\mathrm{Spec}(A) \to \mathrm{Spec}(C)$, induced by the inclusion $C \hookrightarrow A$, is surjective.

Maybe EGA IV.8.3.8.(i) is useful, but I have no ideas to prove or disprove the statement. Any hint will be welcome. Thanks to all!

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up vote 1 down vote accepted

The statement is not true: if $A$ is a local ring, then $C$ must be local too. Now assume that $B$, thus $C$, contains at least one element that is transcendental over $\mathbb{Q}$. Then $C$ as a finitely generated $\mathbb{Z}$-algebra has infinitely many maximal ideals.

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Thanks! A simpler example is $A = \mathbb{Q}$, because every subring of $\mathbb{Q}$ of finite type over $\mathbb{Z}$ is of the form $\mathbb{Z}[1/d]$ and the map $\mathrm{Spec}(\mathbb{Q}) \to \mathrm{Spec}(\mathbb{Z}[1/d])$ cannot be surjective. – Andrea Oct 31 '12 at 12:31

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