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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.

user26857
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user137090
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2 Answers2

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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:

  1. 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 $P=(X,Z)$.

  2. 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.

user26857
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  • I don't understand your edit. Could you please explain? How can the nullstellensatz be used to find the radical if the field is closed? By the way +1 :) – Manos May 12 '14 at 20:59
  • The nullstellensatz says that when the field is algebraically closed, then the radical of $I$ is $I(V(I))$. But as far as i know the nullstellensatz does not say how to compute the radical. That's according to my understanding. – Manos May 12 '14 at 21:08
  • I see what you are saying. Could you please point me to an example which illustrates that computing $I(V(I))$ is simpler than computing the intersection of minimal primes? – Manos May 12 '14 at 21:17
  • @Manos If I understand well you are actually asking about the finding of the generators of the radical, while I'm talking about writing the radical as an intersection of primes. Is this right? – user26857 May 12 '14 at 22:14
  • Yes, that's right. – Manos May 13 '14 at 03:52
  • I am sorry, i still don't understand the edit. I just don't see why computing $I(V(I))$ is a simpler problem than finding the minimal primes of $I$. – Manos May 13 '14 at 14:52
  • How do you know that these equations are elementary and what does that mean? Also what are these equations? That's why i asked if you had an example which demonstrates that :) – Manos May 13 '14 at 18:52
  • @Manos You have an example on this page. But as I said, I'll look for more illustrative ones in the near future. – user26857 May 13 '14 at 19:49
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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,

Patrick Da Silva
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  • See http://math.stackexchange.com/questions/791789/computing-irreducible-components-of-algebraic-set/791954#791954 for generators of this intersection. – Manos May 12 '14 at 21:19