integral closure of m-primary ideals

Question

I need help with this excercise

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 $\overline{(F_1,\ldots,F_m)}=(X_1,\ldots,X_d)^n$.

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 $\bar{I}$.

Answer 1

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.