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Let $R$ be a ring, let $I$ be an ideal of $R$, and let $u\in R$ be idempotent modulo $I$ (that is, $u^{2}-u\in I$). Then $u$ can be lifted to an idempotent in $R$ in case there is an idempotent $e$ in $R$ with $e-u \in I$. My question is that if $R$ is the ring of $n\times n$ upper triangular matrice over a field $\mathbb{Q}$ and if $J$ is the ideal of matrices having zero on the diagonal, then every idempotent modulo J can be lifted to an idempotent in $R$

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    Hint: Show that $R/J$ is isomorphic to $\mathbf{Q}^n$. One component per diagonal element. Finding the idempotents in the latter ring is easy.2011-12-15

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Yes. Every idempotent in $R/I$ can be lifted to $R$.

Suppose $A+I$ is an idempotent of $R/I$. Then $(A+I)^2=A+I$ so that $A^2+I=A+I$ and thus $A^2-A \in I$. Since $A^2-A$ is strictly upper-triangular, it is nilpotent. Thus $(A^2-A)^k=0$ for some $k>0$. This implies that the minimal polynomial for $A$ divides $(x^2-x)^k$. Therefore, the eigenvalues of $A$ are 0 and 1.

Now $A$ is upper-triangular and its eigenvalues are 0's and 1's. Thus the diagonal of $A$ is a bunch of 0's and 1's. Let $B$ be the matrix $A$ minus it's diagonal. So $B$ is strictly upper-triangular ($B \in I$). Notice that $A-B$ is just the diagonal of $A$. Since this is a bunch of 0's and 1's (idempotents), $(A-B)^2=(A-B)$ (i.e. $A-B$ is an idempotent of $R$). Finally, $A+I = (A-B)+I$ (since $A$ and $A-B$ differ by an element of $I$).