affine scheme that is finite type over $\mathbb{Z}$

Question

I have an affine scheme that is finite type over $\mathbb{Z}$, so by definition I can cover this Spec $A$ by Spec $B_i \ (1 \leq i \leq n)$ where each $B_i$ is a finitely generated $\mathbb{Z}$ algebra. I used the fact that Spec $A$ is quasi compact to get that we can cover Spec $A$ by finite number of these affine opens sets. Does it then follow that $A$ is also a finitely generated $\mathbb{Z}$ algebra? Thank you!

Answer 1

The statement is true. There is nothing specific to $\mathbb Z$ in the proof. It works for any ring $B$.

Suppose $\DeclareMathOperator{\Spec}{Spec} \Spec A$ has an affine open cover of the form $\{\Spec A_i\}$ where each $A_i$ is a finitely generated $B$-algebra. Then we can cover $\Spec A$ by basic open subsets of the form $\Spec A_{f_{ij}} = \Spec (A_i)_{g_{ij}}$, where $f_{ij} \in A$ and $g_{ij} \in A_i$. Since $\Spec A$ is quasi-compact, we can find a finite subcover $\{\Spec A_{h_1}, \ldots, \Spec A_{h_n}\}$.

We have reduced this proof to a problem of commutative algebra: If $A$ is a $B$-algebra such that $(h_1, \ldots, h_n) = A$ and each $A_{h_i}$ is a finitely generated $B$-algebra, then $A$ is a finitely generated $B$-algebra. To prove this, let $\{a_{i1}, \ldots, a_{im_i}, 1/h_i\}$ be generators of $A_{h_i}$ as a $B$-algebra, where $a_{i1}, \ldots, a_{im_i} \in A$. It is clear that we can always pick generators of this form. Let $a \in A$. For each $i$, the image of $a$ in $A_{h_i}$ has the form $p_i / h_i^{r_i}$ for some $p_i$ generated by $\{a_{i1}, \ldots, a_{im_i}\}$. It follows that $h_i^{s_i}(h_i^{r_i}a - p_i) = 0$ for some $s_i$. Let $M = \max_i (r_i + s_i)$. Since $(h_1, \ldots, h_n) = A$, we also have $(h_1^M, \ldots, h_n^M) = A$. Thus, $1 = \sum_{i=1}^n q_i h_i^M$ for some $q_i \in A$. We have $$ a = \sum_{i=1}^n a q_i h_i^M = \sum_{i=1}^n q_i h_i^{M - r_i} p_i. $$

Since each $p_i$ is generated by $\{a_{i1}, \ldots, a_{im_i}\}$ and $a$ arbitrary, it follows that $A$ is a finitely generated $B$-algebra.