Let $I$ be an ideal in a Noetherian ring $R$ such that $\mathrm{gr}_I(R)$ is a reduced ring. Prove that $I$ is normal.
What is the error in this proof?
We proceed by induction over $n$.
If $n=1$ we show that $\overline{I} \subseteq I$. Let $r$ be an element in $\overline{I}$. We have that $\overline{I} \subseteq \sqrt{I}$, thus there exists $l$ such that $r^l \in I$. Then, $([r]_0)^l=[r^l]_0=0$. As $\mathrm{gr}_I(R)$ is a reduced ring, then $[r]_0=0$. Therefore, $r \in I$.
For $n>1$, we show that $\overline{I^n} \subseteq I^n$. We know that $I^n \subseteq \overline{I^n}$ is a reduction. Hence, there exists $l$ such that $\overline{I^n}^{l+1}=I^n\overline{I^n}^l$. Then, $$\overline{I^n}^{l+1}\subseteq I^n\overline{I^{n-1}}^l=I^n(I^{n-1})^l=I^{l(n-1)+n}$$ by inductive hypothesis. Hence, for every $r\in \overline{I^n}$, $r^{l+1} \in I^{l(n-1)+n}$. Thus, $([r]_{n-1})^{l+1}=[r^{l+1}]_{(l+1)(n-1)}=[r^{l+1}]_{l(n-1)+n-1}=0$. As $\mathrm{gr}_I(R)$ is a reduced ring, then $[r]_{n-1}=0$. Therefore, $r \in I^n$.