Proposition

Let $I$ be an ideal in $k[x_1, \cdots, x_n]$.

1. In the special case when $\sqrt{I} = \ev{f_1, f_2}$, with $f_i^{m_i} \in I$, prove that $f^{m_1 + m_2 - 1} \in I$ for all $f \in \sqrt{I}$.
2. Now prove that for any $I$, there exists a single integer $m$ such that $f^m \in I$ for all $f \in \sqrt{I}$.

Solution

1

Let $f \in \sqrt{I}$. Then $f = a_1f_1 + a_2f_2$ for some $a_1, a_2 \in k[x_1, \cdots, x_n]$.

If $i \leq m_1 - 1$ and $m_1 + m_2 - 1 - i \leq m_2 - 1$, then $m_1 + m_2 - 1 \leq m_1 + m_2 - 2$, which is a contradiction. Therefore, either $i \geq m_1 - 1$ or $m_1 + m_2 - 1 - i \geq m_2 - 1$. This implies that $f_1^i \in I$ or $f_2^{m_1 + m_2 - 1 - i} \in I$, so $f^{m_1 + m_2 - 1} \in I$.

2

Let a finite set $\{ f_1, \cdots, f_n \} \subset \sqrt{I}$ with $f_i^{m_i} \in I$ be given. Let $M = \sum m_i$. Then we claim that the $(M - 1)$th power of any linear combination $a_1f_1 + \cdots + a_nf_n$ is in $I$.

When $n = 1$, this is obvious.

Suppose that we have proved this for some $n \in \mathbb{N}$. Let $\{ f_1, \cdots, f_{n + 1} \} \subset \sqrt{I}$ with $f_i^{m_i} \in I$ be given. Let $M = \sum m_i$.

Using the same argument as (1), $i \geq M - m_{n + 1} - 1$ or $M - i - 1 \geq m_{n + 1}$. If $i \geq M - m_{n + 1} - 1 = (\sum_{j=1}^{n} m_j) - 1$, $(a_1f_1 + \cdots + a_nf_n)^{i} \in I$ by the inductive hypothesis. If $M - i - 1 \geq m_{n + 1}$, then $(a_{n + 1}f_{n + 1})^{M - i - 1} \in I$. Therefore, the $(M - 1)$th power of any linear combination is in $I$.

By mathematical induction, this is true for any $n$.

By the Hilbert Basis Theorem on P.77 (Ideals, Varieties and Algorithms), every ideal is finitely generated. Therefore, for any $I$, $f^{(\sum m_i) - 1} \in I$ for all $f \in \sqrt{I}$ where $\sqrt{I} = \{ f_1, \cdots, f_n \}$ with $f_i^{m_i} \in I$.