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Please help me to solve the following problem.

Let $L$ be a left ideal in a unital ring $R$, such that $L^2\neq L$. Prove that quotient map $\pi:R\to R/L$ have no right inverse homomorphism of $R$-modules.

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up vote 2 down vote accepted

Let $f\colon R/L \to R$ be such a right inverse, so that $\pi \circ f = \operatorname{id}_{R/L}$. I'll use a bar to denote the image of elements of $R$ in $R/L$. From $(\pi \circ f)(\bar1) = \bar1$ it follows that $f(\bar 1) = 1 + x$ for some $x \in L$. Select a $y \in L - L^2$. What can you say about $f(\bar y) = f(\bar 0) = 0$?

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$y=-yx\in L^2$. Contradiction – Norbert Mar 31 '12 at 20:52

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