18
$\begingroup$

I want to find the irreducible components of the variety $V(X^2+Y^2-1, \ X^2-Z^2-1)\subset \mathbb{C}^3$ but I am completely stuck on how to do this. I have some useful results that can help me decompose $V(F)$ when $F$ is a single polynomial, but the problem seems much harder even with just two polynomials. Can someone please help me?

EDIT: In trying to answer this question, I knew it would be useful to know if the ideal $I=(X^2+Y^2-1, X^2-Z^2-1)$ was a prime ideal of $\mathbb{C}[X,Y,Z]$ but I'm finding it hard to describe the quotient ring. Is it a prime ideal?

$\endgroup$
  • $\begingroup$ Hint: $V(f, g) = V(f) \cap V(g)$. $\endgroup$ – xyzzyz Mar 31 '13 at 14:44
  • 7
    $\begingroup$ @xyzzyz I tried that, it gives that $V = V(X^2+Y^2-1) \bigcap V(X^2-Z^2-1) $ and those two algebraic sets are irreducible because those polynomials are irreducible. But that doesn't seem to answer my question about decomposing the set because its $\bigcap$ instead of $\bigcup.$ Is there something I'm missing? $\endgroup$ – Katie Dobbs Mar 31 '13 at 14:47
  • $\begingroup$ The same question was asked a a few days ago, I have answered it there. math.stackexchange.com/questions/341000 $\endgroup$ – Martin Brandenburg Apr 4 '13 at 15:50
19
$\begingroup$

Here is an ad hoc approach:

We know that a point $(x,y,z)\in\mathbb C^3$ is in the intersection iff $x^2+y^2-1=0$ and $x^2-z^2-1=0.$ In particular, we must have $x^2+y^2-1=x^2-z^2-1,$ which we simplify to $y^2+z^2=0.$ Thus the point must lie on one of the two hyperplanes $V(Y\pm iZ),$ i.e., $y=\pm iz.$

On the other hand, suppose that $(x,y,z)$ satisfies $x^2+y^2-1=0$ and $y=\pm iz.$ These imply that $x^2+(\pm iz)^2-1=x^2-z^2-1=0.$ Thus, we see that we can describe the intersection as $V(X^2+Y^2-1,Y-iZ)\cup V(X^2+Y^2-1,Y+iZ).$

Thus, we have reduced to finding the irreducible components of $V(X^2+Y^2-1,Y\pm iZ).$ The corresponding coordinate rings are isomorphic to $$\mathbb C[X,Y,Z]/(X^2+Y^2-1,Y\pm iZ)\cong\mathbb C[X,Y]/(X^2+Y^2-1),$$ which implies that $V(X^2+Y^2-1,Y\pm iZ)$ are irreducible.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ +1, also because your first sentence reminds me of Captain Haddock, Tintin's colourful friend :-) $\endgroup$ – Georges Elencwajg Mar 31 '13 at 16:34
  • 2
    $\begingroup$ "Blisterin' barnacles!" I'm amused to have reminded you of him, especially as Captain Haddock is my favourite character from the Tintin series! $\endgroup$ – Andrew Mar 31 '13 at 16:48
  • 2
    $\begingroup$ Dear Andrew: I'm pleasantly surprised, since I hadn't expected ( given your English sounding first name) that you even knew that hilarious character from a Belgian comic-book . $\endgroup$ – Georges Elencwajg Mar 31 '13 at 17:07
  • 2
    $\begingroup$ Dear @Georges, indeed I have a few of the comic-books, and watched the animated tv series as a child. All in English, however... I'm not sure whether Haddock said "blistering barnacles" in French :-) $\endgroup$ – Andrew Mar 31 '13 at 17:51
  • 1
    $\begingroup$ I've managed to show the injectivity now too! But I still lack intuition for this result. $\endgroup$ – Katie Dobbs Apr 1 '13 at 7:29
9
$\begingroup$

$\newcommand{\rad}{\text{rad}\hspace{1mm}} $

The ideal $(x^2 + y^2 - 1,x^2 - z^2 - 1)$ is equal to the ideal $(y^2 + z^2 ,x^2 - z^2 - 1)$. This is because \begin{eqnarray*} (y^2 + z^2) + (x^2 - z^2 - 1) &=& y^2 + x^2 - 1\\ (x^2 + y^2 - 1) - (x^2 - z^2 - 1) &=& y^2 + z^2. \end{eqnarray*}

Thus we get \begin{eqnarray*} V(x^2 + y^2 - 1,x^2 - z^2 - 1) &=& V( y^2 + z^2,x^2 - z^2 - 1) \\ &=& V\left( (y+zi)(y- zi),x^2 - z^2 - 1\right) \\ &=& V(y+zi,x^2 - z^2 - 1) \cup V(y-zi,x^2 - z^2 - 1).\end{eqnarray*} Now we claim that the ideals $(y+zi,x^2 - z^2 - 1)$ and $(y-zi,x^2 - z^2 - 1)$ are prime ideals. I will only show that the former is prime because the proof for the latter is similar. By the Third Isomorphism Theorem we have \begin{eqnarray*} \Bbb{C}[x,y,z]/(y+ zi,x^2 - z^2 - 1) &\cong& \Bbb{C}[x,y,z]/(y+zi) \bigg/ (y+ zi,x^2 - z^2 - 1)/ (y + zi) \\ &\cong& \Bbb{C}[x,z]/(x^2 - z^2 - 1)\end{eqnarray*}

because $\Bbb{C}[x,y,z]/(y + zi) \cong \Bbb{C}[x,z]$. At this point there are two ways to proceed, one of which is showing that $x^2 - z^2 - 1$ is irreducible. However there is a nicer approach which is the following. Writing \begin{eqnarray*} x &=& \frac{x+z}{2} + \frac{x-z}{2} \\ z &=& \frac{z + x}{2} + \frac{z-x}{2}\end{eqnarray*} this shows that $\Bbb{C}[x,z] = \Bbb{C}[x+z,x-z]$. Then upon factoring $x^2 - z^2 - 1$ as $(x+z)(x-z) - 1$ the quotient $\Bbb{C}[x,z]/(x^2 - z^2 - 1)$ is isomorphic to $\Bbb{C}[u][v]/(uv - 1)$ where $u = x+z, v = x-z$. Now recall that $$\left(\Bbb{C}[u] \right)[v]/(uv - 1) \cong \left(\Bbb{C}[u]\right)_{u} $$

where the subscript denotes localisation at the multiplicative set $\{1,u,u^2,u^3 \ldots \}$. Since the localisation of an integral domain is an integral domain, this completes the proof that $(y+zi,x^2 - z^2 - 1)$ is prime and hence a radical ideal.

Now use Hilbert's Nullstellensatz to complete the proof that your algebraic set decomposes into irreducibles as claimed in Andrew's answer.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I am sorry, maybe I am wrong, but it seems to me that you are using something like $I+(J\cap K)=(I+J)\cap (I+k)$, so a distributivity law for the sum, when $J,K$ are coprime. Which are the hypothesis to use this result? $\endgroup$ – LStefanello Mar 5 '19 at 13:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.