# Irreducible components of the variety $V(X^2+Y^2-1,X^2-Z^2-1)\subset \mathbb{C}^3.$

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?

• Hint: $V(f, g) = V(f) \cap V(g)$. – xyzzyz Mar 31 '13 at 14:44
• @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? – Katie Dobbs Mar 31 '13 at 14:47
• The same question was asked a a few days ago, I have answered it there. math.stackexchange.com/questions/341000 – Martin Brandenburg Apr 4 '13 at 15:50

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.

• +1, also because your first sentence reminds me of Captain Haddock, Tintin's colourful friend :-) – Georges Elencwajg Mar 31 '13 at 16:34
• "Blisterin' barnacles!" I'm amused to have reminded you of him, especially as Captain Haddock is my favourite character from the Tintin series! – Andrew Mar 31 '13 at 16:48
• 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 . – Georges Elencwajg Mar 31 '13 at 17:07
• 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 :-) – Andrew Mar 31 '13 at 17:51
• I've managed to show the injectivity now too! But I still lack intuition for this result. – Katie Dobbs Apr 1 '13 at 7:29

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

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.

• 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? – LStefanello Mar 5 '19 at 13:50