# How can I compute the radical $\sqrt{I}$ of an ideal $I$ of $R$?

Let $$R=\Bbb{C}[X,Y]$$ and $$I=(XY,Y^2)$$ an ideal of $$R$$. I want to compute $$\sqrt I$$.

I know that $$\sqrt I:=\{r\in R: \exists ~n\in \Bbb{N}, r^n\in I\}$$. I clearly see that in my case $$Y\in \sqrt I$$ since I can take $$n=2$$. In addition we also directly that $$X\notin \sqrt I$$. But now for the more complicated ones I don't see how to proceed. I mean if I take $$P\in R$$ then $$P^n\in I$$ for some $$n$$ if $$P^n=UXY+VY^2$$ for some $$U,V\in R$$. But then I need to find all such $$P$$ such that there exists an $$n$$ satisfying the equality before.

I thought about using the second binomial formula since $$UXY+VY^2$$ looks really familiar but also here I don't see how to proceed.

Is there like a general way how to approach such type of exercises?

Here's a useful piece of terminology: we say that an ideal $$I$$ is radical if $$\sqrt{I}=I$$, ie if $$x^n\in I$$ implies $$x\in I$$ for all $$x$$. Note that $$\sqrt{I}$$ is always radical, for any ideal $$I$$. (In other words, $$\sqrt{\sqrt{I}}=\sqrt{I}$$.)
Note that any prime ideal is radical. (Indeed, if $$P$$ is a prime ideal, you can prove by induction on $$n$$ that $$X^n\in P$$ implies $$X\in P$$ for all $$X$$, where the base case of $$n=2$$ follows from the definition of a prime ideal.) Now, you have shown that $$\langle Y\rangle\leqslant\sqrt{I}$$. On the other hand, $$\langle Y\rangle$$ is a prime ideal of $$R$$, since $$R/\langle Y\rangle\cong\mathbb{C}[X]$$. By the remark above, this means $$\langle Y\rangle$$ is a radical ideal. Moreover, $$I\leqslant\langle Y\rangle$$, since $$XY\in\langle Y\rangle$$ and $$Y^2\in\langle Y\rangle$$. Thus also $$\sqrt{I}\leqslant\sqrt{\langle Y\rangle}=\langle Y\rangle$$. (Here we use the general fact that, for any ideal $$I,J$$, if $$I\leqslant J$$ then $$\sqrt{I}\leqslant\sqrt{J}$$.) So $$\sqrt{I}=\langle Y\rangle$$, and we are done.
In terms of a general approach, here is a piece of advice I would recommend. If you're working with an ideal $$I$$, and you find an element $$a$$ such that $$a\in\sqrt{I}$$, then replace $$I$$ by $$I+\langle a\rangle$$ and work from there; often this will simplify the ideal you're working with. You can do this because we have $$\sqrt{I+\langle a\rangle}=\sqrt{I}$$ for any $$a\in\sqrt{I}$$ – can you see why this is the case? (Hint: use the two facts that $$\sqrt{\sqrt{I}}=\sqrt{I}$$ and that $$I\leqslant J$$ implies $$\sqrt{I}\leqslant\sqrt{J}$$.) In this case, when $$a=Y$$, we replace the ideal $$I$$ with the ideal $$\langle Y\rangle$$, and then conclude quickly since $$\langle Y\rangle$$ is prime. So, in a slogan, you don't have to stick with the ideal you start with! You can enlarge it as you find elements in its radical.
• Sorry what is $y$ in your first paragraph Commented Jul 3, 2022 at 19:46
• @Haveaniceday that would indeed be enough – but unfortunately in this case $I$ is not prime! however, what we exploit is that we can "easily" find a prime ideal $P$ such that $I\leqslant P\leqslant\sqrt{I}$. (in this case $P=\langle y\rangle$.) once you've done this, then you know $\sqrt{I}=P$! Commented Jul 3, 2022 at 20:02