Let $R$ be a commutative ring with $1$ such that its uniform dimension is infinity, equivalently, $$\sup\{k \mid R \text{ contains a direct sum of $k$ nonzero ideals }\}=\infty.$$ How can we construct an ideal $I$ of $R$ and an element $r\in R\setminus I$ such that the ring $(R/I)_{r+I}$ has infinitely many idempotents, where $(R/I)_{r+I}$ is the localization of the ring $R/I$ at the subset $\{r^n+I\mid n\in \mathbb{N}\}$ of $R/I$?
Infinite uniform dimension $\Rightarrow$ infinitely many idempotents in a localization of a quotient
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1 Answer
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In general, it is not possible to produce such an ideal $I$ and element $r \in R \setminus I$ from your assumption.
An apparent obstacle is the class of rings of infinite uniform dimension having exactly one prime ideal. If a ring $R$ has a unique prime ideal, then the same is true of every nonzero quotient and localization. In particular, $R$ doesn't have any idempotents, nor does any ring resulting from $R$ by quotients and localizations.
On the other hand, it's easy to produce rings with a unique prime ideal which also have infinite uniform dimension. You can just adjoin an infinite number of mutually orthogonal nilpotents to a field, e.g. $R = k[x_1, x_2, \ldots]/ (x_{i}x_j)_{0 \leq i \leq j}$ contains $\bigoplus_i x_iR$.
