I need help with this excercise



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> Let $k[X\_1,\ldots,X\_d]$ be the polynomial ring in $X\_1,\ldots,X\_d$ over a field $k$, and let $F\_1,\ldots,F\_m$ be forms of degree $n$. Assume that $(X\_1,\ldots,X\_d)=\sqrt{(F\_1,\ldots,F\_m)}$. Prove that $\overline{(F\_1,\ldots,F\_m)}=(X\_1,\ldots,X\_d)^n$.
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Clearly, $(X\_1,\ldots,X\_d)^n\subseteq \overline{(F\_1,\ldots,F\_m)}$. But the converse...?


**Definition(Integral closure)**: Let $R$ be a ring and $I$ an ideal of $R$. An element $x$ is said to be integral over $I$ if $x$ satisﬁes a monic equation
$x^n + i\_1x^{n−1} + ··· + i\_n = 0$ such that $i\_j ∈ I^j$ .The set of all elements that are integral over $I$ is called the integral closure of $I$,
and is denoted $\bar{I}$.



integral closure of m-primary ideals
The asked for inclusion $\overline{(F\_1,\ldots,F\_m)} \subseteq (X\_1,\ldots,X\_d)^n$ follows easily by considering the lowest non-zero homogeneous component. Formally: 


If $f = f\_0 + \cdots + f\_m$ is the decomposition of $f \in k[X\_1,...,X\_d], f\neq 0$ into homogeneous components ($\deg f\_k = k)$ then set $\ell(f) := \min\{ k \mid f\_k \neq 0\}$. Since we are working over a field, $\ell(fg)=\ell(f)+\ell(g)$ and $\ell(f+g)\ge \min\{\ell(f),\ell(g)\}$ if $f+g\neq 0$. 


Note that $(X\_1,\ldots,X\_d)^n$ is just the set of polynomials having lowest non-zero homogeneous component of degree greater or equal $n$. Hence, to show $f \in (X\_1,\ldots,X\_d)^n$ for $ f \in \overline{(F\_1,\ldots,F\_m)}, f\neq 0$, it suffices to show $\ell(f)\ge n$. We have an equation 
$$f^m + i\_1f^{m-1} + \cdots i\_{m-1}f + i\_m = 0\quad\text{ with }\quad i\_k \in (F\_1,\ldots,F\_m)^k.$$
In particular, 
$$\ell(f^m) = \ell(i\_1f^{m-1} + \cdots i\_{m-1}f + i\_m)\ge \ell(i\_kf^{m-k})\tag{$\ast$}$$
for some $i\_k \neq 0$ (exists, since $k[X\_1,...,X\_m]$ is a domain). $\ell(i\_k) \ge kn$ and $(\ast)$ yield 
$$m\, \ell(f) \ge \ell(i\_k) + (m-k)\ell(f) \ge nk + (m-k)\ell(f),$$ 
giving $\ell(f)\ge n$ since $k \ge 1$. **qed.** 


