TITLE: Computing the radical of an ideal
QUESTION [7 upvotes]: What is the best way to compute $\sqrt{(X^2-YZ,X(1-Z))}$ ? 
This is after using Nullstellensatz by the way as I thought it would be easier to compute a radical than finding the vanishing ideal.

REPLY [7 votes]: In this case there is no need to work over an algebraically closed field to get $$\sqrt{(X^2-YZ,X(1-Z))} = (X,Y) \cap (X,Z) \cap (Z-1,Y-X^2).$$

The radical of an ideal is the intersection of the minimal primes over it.  

Let $P$ be a minimal prime over $(X^2-YZ,X(1-Z))$. Since $X(1-Z)\in P$ there are two cases to consider: 

If $X\in P$, then from $X^2-YZ\in P$ we get $Y\in P$ or $Z\in P$, so $P=(X,Y)$ or $P=(X,Z)$. 
If $1-Z\in P$, since $X^2-YZ=X^2+Y(1-Z)-Y\in P$ it follows that $X^2-Y\in P$, so $P=(X^2-Y,1-Z)$. (Well, now we have to know that $(X^2-Y,1-Z)$ is a prime ideal, but this is easy: $K[X,Y,Z]/(X^2-Y,1-Z)\simeq K[X,Y]/(Y-X^2)\simeq K[X]$.)

Edit. If the field is algebraically closed I think is easier to find the radical of an ideal (as an intersection of primes) by using the Nullstellensatz.