Consider a ring $R$, an idela $I$. An element$z\in R$ is integral over $I$ if $z$ satisfies the equation $$z^n+a_1z^{n-1}+\ldots+a_{n-1}z+a_n=0,$$ where $a_i\in I^i$. We define the integral closure $\overline{I}$ of an ideal as the set of elements in $R$ which are integral over $I$ (and this is an ideal of $R$).
We say that $I$ is integrally closed if $I=\overline{I}$, and that $I$ is normal if all the powers $I^k$ are integrally closed (I'm using Villarreal's Monomial Algebras).
Question: I'm a bit confused, because from my point of view $I$ is integrally closed if and only if $I$ normal.
Proof: If $I$ is normal, then all the powers $I^k$ are integrally closed, hence imposing $k=1$ implies $I$ is integrally closed (maybe here I'm cheating a bit, but Villareal doesn't say anything like $k>1$.
On the other hand, if $I$ is integrally closed, then $I=\overline{I}$ by definition, hence $I^k=\overline{I}^k$ for any $k$.
I strongly suspect there's something wrong (strange to give $2$ different names for the same object), but I can't spot the mistake and I'd like to understand!
You say "On the other hand, if I is integrally closed, then $I=\bar{I}$ by definition, hence $I^k=\bar{I}^k$ for any k.", but $\bar{I^k} \ne \bar{I}^k$ in general so one cannot conclude from this that $I$ is normal. In fact, integrally closed ideals are frequently not normal. For a simple example, take $R=k[\![x,y]\!]/(x^2,xy)$ and set $\mathfrak{m}=(x,y)$. Then of course $\mathfrak{m}$ is integrally closed, but, when $k>1$, $\mathfrak{m}^k=(y^k)$. As the integral closure of any ideal always contains the nilradical of $R$, $x$ must be in the integral closure of any ideal, and so, in particular, $\mathfrak{m}^k$ is not integrally closed when $k>1$.