Recall that a ring $R$ is right Noetherian if for every chain $I_1 \subseteq I_2 \subseteq I_3 \subseteq \ldots$ of right ideal of $R$, there exists a positive integer $m$ such that $I_m = I_{m+1}=I_{m+2}=\ldots$.
Assume that $R$ is a right Noetherian ring and $e\in R$ is an idempotent. How can I prove that $eRe$ is a right Noetherian ring as well.
I appreciate any help. Thanks in advance.
Best Answer
Let $\mathfrak A$ be a right ideal of $eRe$. We will prove that $$ (\mathfrak AR) \cap eRe = \mathfrak A. $$ Indeed, if $\mathfrak A_0 := (\mathfrak AR) \cap eRe$, clearly $\mathfrak A_0 \supseteq \mathfrak A$; and \begin{align} \mathfrak A_0 &= \mathfrak A_0e \qquad (\textrm{since } \mathfrak A_0 \subseteq eRe) \\ &\subseteq \mathfrak A Re \qquad (\mathfrak A_0 \subseteq \mathfrak AR) \\ &= \mathfrak A eRe \qquad (\mathfrak A = \mathfrak Ae) \\ &= \mathfrak A \qquad (\textrm{$\mathfrak A$ is a right ideal of $eRe$}) \end{align} proving $\mathfrak A_0 = \mathfrak A$.
Conclude that $\mathfrak A \mapsto \mathfrak AR$ is an injective inclusion-preserving map from the right ideals of $eRe$ to those of $R$.
Thus, if $R$ is right Noetherian, then the same holds for $eRe$.
Reference:
T. Y. Lam. A First Course in Noncommutative Rings. Second Edition. Theorem 21.11 (p. 311).