$\newcommand{\radical}{\operatorname{radical}}$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 $I$.

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

1 Answer 1


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 $x\notin (x^2)$.


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