Lemma 15.108.3. Let $A$ be a ring and let $I$ be a finitely generated ideal. Let $A \to C$ be a ring map such that for all $f \in I$ the ring map $A_ f \to C_ f$ is localization at an idempotent. Then there exists a surjection $A \to C'$ such that $A_ f \to (C \times C')_ f$ is an isomorphism for all $f \in I$.

Proof. Choose generators $f_1, \ldots , f_ r$ of $I$. Write

$C_{f_ i} = (A_{f_ i})_{e_ i}$

for some idempotent $e_ i \in A_{f_ i}$. Write $e_ i = a_ i/f_ i^ n$ for some $a_ i \in A$ and $n \geq 0$; we may use the same $n$ for all $i = 1, \ldots , r$. After replacing $a_ i$ by $f_ i^ ma_ i$ and $n$ by $n + m$ for a suitable $m \gg 0$, we may assume $a_ i^2 = f_ i^ n a_ i$ for all $i$. Since $e_ i$ maps to $1$ in $C_{f_ if_ j} = (A_{f_ if_ j})_{e_ j} = A_{f_ if_ ja_ j}$ we see that

$(f_ if_ ja_ j)^ N(f_ j^ n a_ i - f_ i^ na_ j) = 0$

for some $N$ (we can pick the same $N$ for all pairs $i, j$). Using $a_ j^2 = f_ j^ na_ j$ this gives

$f_ i^{N + n} f_ j^{N + nN} a_ j = f_ i^ N f_ j^{N + n} a_ ia_ j^ N$

After increasing $n$ to $n + N + nN$ and replacing $a_ i$ by $f_ i^{N + nN}a_ i$ we see that $f_ i^ n a_ j$ is in the ideal of $a_ i$ for all pairs $i, j$. Let $C' = A/(a_1, \ldots , a_ r)$. Then

$C'_{f_ i} = A_{f_ i}/(a_ i) = A_{f_ i}/(e_ i)$

because $a_ j$ is in the ideal generated by $a_ i$ after inverting $f_ i$. Since for an idempotent $e$ of a ring $B$ we have $B = B_ e \times B/(e)$ we see that the conclusion of the lemma holds for $f$ equal to one of $f_1, \ldots , f_ r$. Using glueing of functions, in the form of Algebra, Lemma 10.23.2, we conclude that the result holds for all $f \in I$. Namely, for $f \in I$ the elements $f_1, \ldots , f_ r$ generate the unit ideal in $A_ f$ so $A_ f \to (C \times C')_ f$ is an isomorphism if and only if this is the case after localizing at $f_1, \ldots , f_ r$. $\square$

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