# Ring Theory (Lift)

Let $R$ be a ring, $A$ is an ideal of $R$, and assume that $A\subseteq J(R)$. Let $r\longmapsto \overline{r}$ denote the coset map $R\rightarrow \overline{R}=R/A$. The following are equivalent:

(1) $A$ is lifting.

(2) If $\overline{R}=\overline{K_1}\oplus...\oplus \overline{K_n}$ where the $\overline{K_i}$ are left ideals of $\overline{R}$, there exists a decomposable $R=L_1\oplus...\oplus L_n$ where the $L_i$ are left ideals of $R$ such that $\overline{L_i}=\overline{K_i}$ for each $i=1,2,...,n$.

(3) Same as $(2)$ except $n=2$.

Do you have any idea or any source for finding a prove?

Edit by Arturo Magidin: Above is the original question. The new phrasing is below:

Let $R$ be a ring, $A$ is an ideal of $R$, and assume that $A\subseteq J(R)$. Let $r\longmapsto \overline{r}$ denote the coset map $R\rightarrow \overline{R}=R/A$ If $A$ is lifting, for $\overline{R}=\overline{K_1}\oplus \overline{K_2}$ where the $\overline{K_i}$ are left ideals of $\overline{R}$, there exists a decomposable $R=L_1\oplus L_2$ where the $L_i$ are left ideals of $R$ such that $\overline{L_i}=\overline{K_i}$ for each $i=1,2$.

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What is a lifting in this context? – Arturo Magidin Dec 14 '10 at 17:31
@ Arturo Magidin: " If $A$ is an ideal of $R$, and if $r+A$ is an idempotent in $R/A$, we say that $r+A$ can be lifted to $R$ if there exists an idempotent $e^2=e \in R$ such that $e+A=r+A$, that is if $e-r \in A$. We say that idempotents can be lifted moduleo $A$, or that $A$ is lifting, if every idempotent in $R/A$ can be lifted. – Arash Dec 14 '10 at 17:48
please stop editing questions in a way that renders existing answers disconnected with the new text. – Mariano Suárez-Alvarez Dec 15 '10 at 4:49
This question now contains no question. – Alex B. Dec 15 '10 at 5:41
I've started a meta discussion on these perplexing edits. meta.math.stackexchange.com/questions/1343/… – Arturo Magidin Dec 15 '10 at 6:29

Suppose you can lift idempotents modulo $A$. If $\bar R=I_1\oplus I_2$ as a left $\bar R$-module, there is an idempotent $e\in\operatorname{End}_{\bar R}(\bar R)=\bar R$, given by projecting onto $I_1$, and we have $I_1=\bar R\cdot e$ and $I_2=\bar R\cdot(1-e)$. Lift $e$ to an idempotent $\tilde e\in R$, and let $J_1=R\cdot \tilde e$ and $J_2=R\cdot(1-\tilde e)$. Etc.
This will give $(1)\implies(3)$. The rest of the implications are of the same spirit.