Sufficient condition for an homogeneous ideal to be radical.

Question

Let $I$ be an homogeneous ideal. If for every $\textbf{homogeneous}$ polynomial $f\in k[x_0,x_1,\dots, x_n]$ such that $f^n\in I\Rightarrow f\in I$, then $I=\sqrt{I}$.

My attempt has been choosing an arbitrary polinomia $g\in k[x_0,x_1,\dots,x_n]$ such that $g^n\in I$ and try to prove that $g\in I$. For that purpuse, I decomposed $g$ in homogeneous components $g=\sum_i h_i$. Hence, $g^n=(\sum_i h_i)^n\in I$. But if I extend that, I cannot make sure that every $h_i^n\in I$, which would imply that $g\in I$.

How can I solve it?

Answer 1

Let $A$ be a graded commutative ring over the natural numbers (where $0\in\mathbb{N}$). An ideal $I\subset A$ is homogeneous if and only if $\forall f\in I$, the homogeneous components of $f$ are in $I$. We define the degree of any element $f\in A$ to be the maximum degree of its non-zero homogeneous components.

Suppose that $I\subset A$ is an homogeneous ideal such that for all homogeneous $f\in A$ with $f^r\in I$ for some $r\geq 1$, it holds $f\in I$. We show $I$ must be radical. Let $f\in A$ be arbitrary such that $f^r\in I$ for some $r\geq 1$. We prove that then $f\in I$ by induction on the degree $d=\deg f$. The case $d=0$ follows from the assumption. Now suppose $d>0$ and assume the result true for elements of $A$ of degree strictly less than $d$. The element $f\in A$ is of degree $d$, and we denote $f_{d}$ to its homogeneous component of maximum degree. The homogeneous component of $f^r$ in degree $rd$ is $f_{d}^r$. Since $f^r\in I$ and $I$ is homogeneous, $f_{d}^r\in I$, so $f_{d}\in I$ by the assumption. Denote $g=f-f_{d}$, which is of degree $< d$. By the binomial formula, $$ I\ni f^r=(g+f_{d})^r=g^r+\underbrace{\sum_{i=0}^{r-1}\binom{r}{i}g^if_{d}^{r-i}}_{\in I}. $$ So $g^r\in I$. By the induction hypothesis, $g\in I$ and therefore $f=g+f_{d}\in I$.

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If $A$ is graded instead over the integers, the result is still true: here I explain how to adapt the proof. In the accepted answer from the link, they show the result is false in general over arbitrary graduations.

Answer 2

$I$ is homogeneous if and only if for every $f\in I$, the homogeneous components of $f$ belong to $I$. Provided that the product of homogeneous polynomials is homogeneous, you can make sure that $h_i^n\in I\ \forall i$.