Computing the radical of an ideal 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. 
 A: 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.
A: You can just compute $\mathcal I [\mathcal Z [ (X^2-YZ,X(1-Z))]]$ ; assuming you're working over an algebraically closed field, notice that 
$$
x^2 - yz = 0 = x(1-z) \quad \Longleftrightarrow \quad x=y=0 \quad \text{ or } \quad x=z=0 \quad \text{ or } \quad y-x^2=0=z-1,
$$
so that 
$$
\mathcal Z[(X^2-YZ,X(1-Z))] = \mathcal Z[ (x,y) ] \cup \mathcal Z[ (x,z) ] \cup \mathcal Z[ (z-1,y-x^2)].
$$
Taking the ideal of zeros of this union gives you 
$$
\sqrt{(X^2-YZ,X(1-Z))} = (X,Y) \cap (X,Z) \cap (Z-1,Y-X^2)
$$
(I used the fact that all three of these ideals are prime, so that they are equal to their radicals ; to see it, compute the quotient ring and see that they are integral domains.)
If your question involves finding generators for this intersection, I'll think about a hint later if you ask for it.
Hope that helps,
