Lemma 10.168.2. Let $R \to A \to B$ be ring maps. Assume $A \to B$ faithfully flat of finite presentation. Then there exists a commutative diagram
with $R \to A_0$ of finite presentation, $A_0 \to B_0$ faithfully flat of finite presentation and $B = A \otimes _{A_0} B_0$.
Proof. We first prove the lemma with $R$ replaced $\mathbf{Z}$. By Lemma 10.168.1 there exists a diagram
where $A_0$ is of finite type over $\mathbf{Z}$, $B_0$ is flat of finite presentation over $A_0$ such that $B = A \otimes _{A_0} B_0$. As $A_0 \to B_0$ is flat of finite presentation we see that the image of $\mathop{\mathrm{Spec}}(B_0) \to \mathop{\mathrm{Spec}}(A_0)$ is open, see Proposition 10.41.8. Hence the complement of the image is $V(I_0)$ for some ideal $I_0 \subset A_0$. As $A \to B$ is faithfully flat the map $\mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ is surjective, see Lemma 10.39.16. Now we use that the base change of the image is the image of the base change. Hence $I_0A = A$. Pick a relation $\sum f_ i r_ i = 1$, with $r_ i \in A$, $f_ i \in I_0$. Then after enlarging $A_0$ to contain the elements $r_ i$ (and correspondingly enlarging $B_0$) we see that $A_0 \to B_0$ is surjective on spectra also, i.e., faithfully flat.
Thus the lemma holds in case $R = \mathbf{Z}$. In the general case, take the solution $A_0' \to B_0'$ just obtained and set $A_0 = A_0' \otimes _{\mathbf{Z}} R$, $B_0 = B_0' \otimes _{\mathbf{Z}} R$. $\square$
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