# Sufficient condition for an homogeneous ideal to be radical.

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?

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 $$f=g+f_{d}\in I$$.
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.
$I$ 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 $h_i^n\in I\ \forall i$.