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As for Peirce decompositions $R=Re\oplus Rf$, $R=eR\oplus fR$, and $R=eRe\oplus eRf\oplus fRe\oplus fRf$ about an arbitrary ring $R$ with idempotents $e$ and $f=1-e$, T. Y. Lam in his book "A First Course in Noncommutative Rings" asserts that the first (resp., second) is a decomposition of $R$ into left (resp., right) ideals, while the third is a decomposition of $R$ into additive subgroups, where $eRe$ and $fRf$ are, in fact, rings with identities $e$ and $f$, respectively.

Now, if a property $P$ which holds for the ring $R$ also holds for any quotient ring $R/I$ of $R$ , could we deduce that $P$ also holds for $eRe$, where $e^2=e\in R$?

Thanks for any (comprehensive) answer!

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    If $I$ is only a left *or* right ideal, then $R/I$ is not a ring, so I'm not sure what kind of property $P$ you're thinking of? – 2017-01-20
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    @Jeremy Rickard Sorry! You are right, and I edited the question. – 2017-01-20

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It's certainly not the case for all properties $P$.

For example, let $P$ be the property of not having a quotient ring isomorphic to $\mathbb{Q}$. If $R$ is the ring $M_2(\mathbb{Q})$ of $2\times 2$ matrices over $\mathbb{Q}$, then $R$ has property $P$, as does every quotient ring of $R$ (both of them!), but $eRe\cong\mathbb{Q}$ doesn't have property $P$ for $e=\begin{pmatrix}1&0\\0&0\end{pmatrix}$.