Finding Radical of an Ideal 
Given the ideal $J^\prime=\langle xy,xz-yz\rangle$, find it's radical.

I know that the ideal $\langle xy,yz,zx\rangle$ is radical ideal but that's not the case. How can I compute the radical here?
 A: The ideals $J^\prime=\langle xy,xz-yz\rangle\subset J= \langle xy,yz,zx \rangle$ have the same vanishing set in $\mathbb A^3 $, namely the union of the three coordinate axes, so that $\sqrt {J'}=\sqrt J$.
Since you know that $\sqrt J=J$, you can conclude that $$   \sqrt {J'}=J        $$
[The same result can be reached by an easy but uninteresting computation.]  
Edit: answering Danis's questions in his comment
1) Here  or there  are very general result implying  that $J$ is radical.
2) $J'$ is not radical because $xz\notin J'$ but $(xz)^2=xy\cdot z^2+(xz-yz)\cdot xz\in J'$.
A: I thought it might be helpful to spell out one of the ideas that is implicit in Georges's answer, namely:
Often in these examples one can just compute the irreducible components of the zero locus directly, and so determine the radical.
Your equations are $xy = 0$ and $(x-y) z = 0$.
From the first equation $x = 0$ or $y = 0$.  
If $x = 0$ then the second equation gives $yz = 0$, so $y = 0$ or $z = 0$.
So $(x,y)$ and $(x,z)$ cut out two of the components of the zero locus.
Going back to the other possibility for the first equation: we have $y = 0$,
and then the second equation gives $x z = 0$, i.e. $x = 0$ or $z = 0$.  So we get
$(x,y)$ (again) or $(y,z)$.
So the three components are cut out by $(x,y),$ $(x,z)$, or $(y,z).$
Thus the radical is equal to $(x,y) \cap (x,z) \cap (y,z)$, which you can check (e.g. by thinking directly about polynomials) is $(xy, xz, yz)$.  
