Let $R$ be a noetherian local ring and $I$ an ideal with $\operatorname{ht}I=\mu(I)$. Prove that $I$ is basic. (Recall that an ideal $I$ is basic when it has no proper reduction.)
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Let $J$ be a minimal reduction of $I$. By our assumption, we have $ht(J)\leq\mu(J)\leq\ell(I)\leq\mu(I)=ht(I)=ht(J)$. Thus $\mu(J)=\mu(I)$. From the exact sequence $$0\longrightarrow\frac{J}{mI\cap J}\longrightarrow\frac{I}{mI}\longrightarrow\frac{I}{J+mI}\longrightarrow 0$$ and this fact that $mI\cap J=mJ$, we have $I=J+mI$. Therefore $I=J$.
