# Is a homogeneous ideal radical?

$$\newcommand{\radical}{\operatorname{radical}}$$I am self studying some introductory level algebraic geometry and I want to ask a naive question.

Let $$I\subset k[x_0,\ldots,x_n]$$ be a homogeneous ideal, then $$\radical(I)=I?$$

Recall $$\radical(I)=\{f\in k[x_0,\ldots,x_n] \mid \exists N \in \mathbb{Z}_{>0}, \text{ s.th. } f^N\in I\}$$

Certainly, $$\radical(I)\supset I$$

So enough to show another inclusion, if we pick any $$f=\sum_{i=0}^r a_i x^i \in \radical(I)$$

$$f^N = (\sum_{i=0}^r a_i x^i)^N=\sum_{j=0}^{rN} b_j x^j \in I$$ which is a homogeneous ideal.

Thus it follows that all the homogeneous component is in $$I$$, i.e. $$b_jx^j\in I$$ for all $$j=0,...,rN$$. And hence, all the homogenous component in $$f$$ are also in $$I$$ since they only vary up to multiplying a constant. Therefore, $$f$$ are also in $$I$$.

• How exactly are you concluding that the homogeneous components of $f$ are in $I$? Feb 19, 2021 at 3:09
• Yes, that is not quite true thank you
– Mike
Feb 19, 2021 at 3:23

No, it is not the case that every homogeneous ideal is radical. Here is an easy counterexample: $$(x^2)\subset k[x]$$. This is homogeneous, being generated by a homogeneous element, but not radical, since $$x\cdot x\in (x^2)$$ but $$x\notin (x^2)$$.