1
$\begingroup$

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

$\endgroup$
2
  • 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

7
$\begingroup$

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 $x\notin (x^2)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.