The Stacks project

Lemma 15.101.3. 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 10.22.2, 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 $f_1, \ldots , f_ r$. $\square$


Comments (0)

There are also:

  • 2 comment(s) on Section 15.101: Branches of the completion

Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.




In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0C2A. Beware of the difference between the letter 'O' and the digit '0'.