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Statement : If $R_{f_i} \rightarrow S_{f_i}$ is of finite type for $i=1,2, \dots,n$ such that $(f_1, \dots , f_n)=(1)$ ,then $R \rightarrow S$ is of finite type.

[QUESTION]

Is that true? I think It is not true (see below motive) but I can't find a counterexample.

[DEFINITION]

Let $R \rightarrow S$ be a ring homomorphism. If $S$ is a finitely generated $R$-algebra, then it is called "of finite type"

[MOTIVE]

I know three propositions :

(i) If $R_{f_i} \rightarrow S_{f_i}$ is integral ,then $R \rightarrow S$ is integral, where $(f_1, \dots , f_n)=(1)$

(ii) If $R_{f_i} \rightarrow S_{f_i}$ is finite ,then $R \rightarrow S$ is finite, where $(f_1, \dots , f_n)=(1)$

(iii) finite $\Leftrightarrow$ integral + of finite type

Many reference take the proposition (i),(ii) except "of finte type" version. So, I wonder "of finite type" version is not true.

hew
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    Does this help? https://stacks.math.columbia.edu/tag/00EO. Look at (7). You can adapt the proof explicitly or apply the result out of the box by noting that if $S_f$ is a finite type $R_f$-algebra, then also $S_f$ is a finite type $R$-algebra because $R_f \cong R[x]/(fx - 1)$ and compositions of finite type ring maps are of finite type. – Badam Baplan Jul 29 '20 at 06:18
  • @BadamBaplan Thanks you so much. You're right, this statement is true! – hew Jul 29 '20 at 08:14

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