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Is there a "general approach" to determine the minimal prime ideals over an ideal $J$?

I checked some books and didn't find a general approach. Maybe the theory of Gröbner bases is related to these type of questions, but I think this theory is beyond the scope of my introductory course.

The original exercise, which I had to solve, asked me to find the irreducible components, with respect to the Zariski topology, of the algebraic set $S=Z(x^2+y^2 -1, x^2 - z^2 -1)$ in $\mathbb{C}^3$.

I proposed that this question is equivalent to finding the minimal prime ideals of $J=(x^2+y^2 -1, x^2 - z^2 -1)$. This led me to the question whether there was a general approach for finding the minimal prime ideals over an ideal $I$?

It would be convenient if someone could illustrate me the general techinque, so I can imitate it for other cases also. Thanks in advance.

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    $\begingroup$ The minimal prime ideals of A containing J correspond to the minimal prime ideals of the quotient $A/J$. $\endgroup$ – user314 Mar 25 '13 at 20:09
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The minimal prime ideals of $A$ are precisely the generic points of $\mathrm{Spec}(A)$. Here, a point of a topological space $\eta \in X$ is called generic if $\eta \prec x \Rightarrow \eta = x$ with respect to the specialization preorder, i.e. $\eta$ has no proper generizations. In a soper space, generic points correspond 1:1 to the irreducible components via $\eta \mapsto \overline{\{\eta\}}$.

In your example, $S$ is the regular intersection of a cylinder with a hyperbolic surface, resulting in two circles. In fact, observe

$$\begin{eqnarray*} S&=&V(x^2+y^2-1,z^2+y^2)\\ &=&V(x^2+y^2-1,(z+iy)(z-iy))\\ &=&V(x^2+y^2-1,z+iy) \cup V(x^2+y^2-1,z-iy). \end{eqnarray*}$$ Both of these algebraic sets $\subseteq \mathbb{A}^3$ are isomorphic to the circle $\subseteq \mathbb{A}^2$. Hence these are the irreducible components.

You can also do this algebraically (but this is more complicated than the geometric approach, in my opinion) i.e. with your approach using minimal prime ideals. We have

$$\begin{eqnarray*} \Gamma(S)&=&k[x,y]/(x^2+y^2-1)[z]/((z+iy)(z-iy)) \\ &\stackrel{\mathrm{CRT}}{\cong}& k[x,y]/(x^2+y^2-1)[z]/(z+iy) \times k[x,y]/(x^2+y^2-1)[z]/(z-iy) \\ &\cong & k[x,y]/(x^2+y^2-1) \times k[x,y]/(x^2+y^2-1) \end{eqnarray*}$$

This is a product of two (isomorphic) integral domains. Hence there are two minimal prime ideals, namely $\langle (1,0) \rangle$ and $\langle (0,1) \rangle$.

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  • $\begingroup$ Thanks for the response. When you write $k[x,y]/(x^2+y^2-1)[z]/(z+iy)$ I have to interpret his as: $W[z]/\overline{(z+iy)}$ right? where $W=k[x,y]/(x^2+y^2-1)$ and we do $W[z]$ modulo $\overline{(z+iy)}$, the later attained by considering $(z+iy)$ over $W=k[x,y]/(x^2+y^2-1)$. The notation seems a bit odd. $\endgroup$ – yannickvda Mar 26 '13 at 8:57
  • $\begingroup$ Yes but I prefer not to overload the notation (as everyone else too). $\endgroup$ – Martin Brandenburg Mar 26 '13 at 10:02
  • $\begingroup$ I am reading the last part of your reply more closely now. You used $\Gamma(S)=k[x,y,z]/I(S)$ as a fundamental result in AG. However, with $S=Z(x^2 + y^2 -1, z^2 + y^2)$ and Hilbert-Nullstellensatz, we only have $I(S)=I(Z(J))=\sqrt{J}$. Why do you take the quotient by $J$ instead of $\sqrt{J}$? $\endgroup$ – yannickvda Mar 26 '13 at 11:44
  • $\begingroup$ When you suggest the isomorphism with the circle, we are speaking about all $(a_1,a_2,a_3)$ that are zeroes of $f(x,y,z)= \alpha (x^2 + y^2 -1) + \beta (z+iy)$. How is this solution-set isomorphic with a circle? Did you mean to say $\mathbb{P}^2$? I guess the equation in $\mathbb{P}^2$ becomes $\alpha (\bar{x}^2 + \bar{y}^2 - 1) + \beta ( 1 +i \bar{y}) = 0$, which after rewriting becomes $\alpha \bar{x}^2 + (\sqrt{\alpha} \bar{y} + \frac{\beta i}{2 \alpha})^2 -(\frac{\beta i}{2 \alpha})^2 + \beta - \alpha = 0$ which remotely resembles an ellipse after translation and rescaling. Not? $\endgroup$ – yannickvda Mar 26 '13 at 14:12
  • $\begingroup$ And as a last question, when you say there are two minimal prime ideals, namely $⟨(1,0)⟩$ and $⟨(0,1)⟩$. What do you mean by that? I guess that prime ideals in product rings $R \times R'$ are of the form $\mathfrak{p} \times R'$ and $R \times \mathfrak{p}'$ respectively. So you probably meant something else and were maybe implicitly referring to the generic points comment earlier. But that's not yet clear to me. $\endgroup$ – yannickvda Mar 26 '13 at 15:18

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