Integral closure of a monomial Recall that given an ideal $I$ in $R=k[x_1,\ldots,x_n]$, an element $r\in R$ is integral if $r$ satisfies an equation of the form
$$r^m+a_1r^{m-1}+\ldots+a_{m-1}r+a_m=0,$$
where $a_i\in I^i$ for any $i=1,\ldots,m$. The set (actually, an ideal) of integral closed elements is denoted by $\overline{I}$.
Goal: I want to prove the following result (taken from Villareal's "Monomial Algebra"): let $I$ be a monomial ideal of $R$. Then
$$\overline{I}=(x^a\mid x^{ma} \in I^m \text{ for some } m\geq 1).$$
Proof: I'm having trouble proving the $\subset$ inclusion. Consider $r=x^a\in\overline{I}$: by definition it satisfies the equation
$$r^n+a_1r^{n-1}+\ldots+a_{n-1}r+a_n=0,$$
where $a_i\in I^i$.
Now quoting "since $I$ is monomial ideal one obtains $r^m\in I^m$ for some $m\geq 1$. Observing that $\overline{I}$ is a monomial ideal the asserted equality follows."
I notice that for any $i=1,\ldots,n$ the element $a_ir^{m-i}$ belongs to $I$, therefore also $r^m\in I$. But apart from this I don't know how to continue: ok $r^m=(x^a)^m\in I$, but I don't have any costraint on which $I^t$ it belongs to.
Thanks in advance to anyone.
 A: Consider a monomial ideal $I$ of $R = k[x_1, \dots, x_n].$ Given any monomial $x^a$ in $R$ such that $x^{ma}$ is in $I^m$ for some $m \geq 1,$ it follows that $x^a$ satisfies the polynomial $p(t) = t^m - x^{ma},$ hence $x^a$ is integral over $I.$ Using the same argument for each monomial generator of $J,$ we have that $$J = (x^a \,|\, x^{ma} \in I^m \text{ for some } m \geq 1) \subseteq \overline I.$$
On the other hand, the integral closure of a monomial ideal is a monomial ideal. Consequently, we may consider a monomial generator $r = x^a = x_1^{f_1} x_2^{f_2} \cdots x_n^{f_n}$ of $\overline I.$ By definition of $\bar I,$ the monomial $x^a$ satisfies the equation $r^n + a_1 r^{n - 1} + \cdots + a_n = 0$ with $a_k$ in $I^k$ for each integer $1 \leq k \leq n.$ Observe that $R$ is a $\mathbb Z_{\geq 0}^n$-graded ring in the usual way (i.e., the degree of the monomial $x_i$ is the unit vector $e_i$ of $\mathbb Z_{\geq 0}^n$ with $1$ in the $i$th place and $0$s elsewhere). Considering that $I$ is a monomial ideal, it follows that $I$ is homogeneous with respect to the $\mathbb Z_{\geq 0}^n$-grading on $R,$ hence each homogeneous component of $a_k$ is an element of $I^k.$ Ultimately, we may group all of the homogeneous terms of degree $n \langle f_1, \dots, f_n \rangle$ to obtain an equation $$r^n + b_1 r^{n - 1} + \cdots + b_n = 0,$$ where $b_k$ is the homogeneous component of $a_k$ of degree $k \langle f_1, \dots, f_n \rangle.$ Consider some integer $1 \leq k \leq n$ such that $b_k r^{n - k}$ is nonzero. (One of them must be nonzero since $r$ is a nonzero monomial.) Observe that both $r^n$ and $b_k r^{n - k}$ lie in degree $n \langle f_1, \dots, f_n \rangle.$ Considering that the $n \langle f_1, \dots, f_n \rangle$-graded piece of $R$ is generated as a $k$-vector space by the monomial $r^n,$ it follows that $b_k r^{n - k} = cr^n$ for some nonzero scalar $c.$ Cancellation holds, as $R$ is an integral domain, hence we may eliminate $r^{n - k}$ from both sides to find that $r^k = c^{-1} b_k.$ But this says that $r^k - d_k = 0$ for some element $d_k$ in $I^k.$ Consequently, $r^k = x^{ka}$ is in $I^k$ for some $k \geq 1,$ as desired. QED.
