# Components of a variety in projective space

How do we find the irreducible components of the following projective variety in $\mathbb{P}^{3}$, $V(wy-x^{2},xz-y^{2})$?

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I'd start by looking for which coordinate subspaces are contained in the variety. After you account for those, consider the possibile solutions when or more coordinates are non-zero (so you can divide by them and solve for other coordinates). – Michael Joyce Apr 27 '12 at 19:24

The intersection $\bar C$ of your two quadrics is a priori a degree 4 curve. Let's analyze it.

In the affine space $\mathbb A^3 \subset \mathbb P^3$ corresponding to $w=1$ you are looking for the intersection of the quadrics $y-x^2=0$ and $xz-y^2$.
The relevant ideal is $I=(y-x^2,xz-y^2)=(y-x^2,x(z-x^3))$.
The affine intersection is thus $$C=V(I)=V(x,y-x^2)\cup V(y-x^2, z-x^3)=V(x,y)\cup V(y-x^2, z-x^3)=C_1\cup C_2$$ where $C_1=V(x,y)$ is a line while $C_2=V(y-x^2, z-x^3)$ is a twisted rational curve.
You must add the the intersection at infinity, that is the point(s) of $\bar C$ in the hyperplane $w=0$.
You obtain just the single point $x=y=w=0, z=1$ which is in the closure of both $C_1$ and $C_2$.

Summing up
The irreducible components of $\bar C \subset \mathbb P^3$ are the line $\bar C_1=V^{proj }(x,y)\subset \mathbb P^3$ and the twisted cubic curve $\bar C_2=C_2\cup \lbrace [0:0:1:0]\rbrace \subset \mathbb P^3$.
Notice that $\bar C_2$ is also the image of the morphism $$\mathbb P^1 \to \mathbb P^3:[u:v]\to [uv^2:u^2v:u^3:v^3]$$

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thanks, how do we argue that the line $\overline{C_{1}}$ is indeed irreducible? or how do we check this? – user10 Apr 30 '12 at 23:20
Every line in any projective space is isomorphic to $\mathbb P^1$, hence is irreducible. In the present case you can exhibit an isomorphism explicitly, namely $\mathbb P^1\stackrel {\cong}{\to} \bar C:[w:z]\mapsto [w:0:0:z]$ – Georges Elencwajg May 1 '12 at 7:32
thanks, but isn't $[0: 0 : 1 :0]$ a point of $V(x,y)$? why do we add it? wouldn't the irreducible components be $\{[1 : t : t^{2} : t^{3}] | t \in k\} \cup V(x,y)$? – user10 May 6 '12 at 21:48
Dear user 10, no $[0:0:1:0]$ is not on $V(x,y)$, but it is on $\overline {C_1}=V^{proj }(x,y)$: it is the point at infinity you have to add to $C_1$ in order to obtain $\overline {C_1}$. Coincidentally it is also the point at infinity you have to add to $C_2$ in order to obtain $\overline {C_2}$. (Of course "infinity" here is the plane $w=0$) – Georges Elencwajg May 11 '12 at 17:43

Using the fantastic MacAulay2 :

i1 : R = QQ[w..z]

o1 = R

o1 : PolynomialRing

i2 : primaryDecomposition ideal( w*y - x^2, x*z - y^2 )

2                    2
o2 = {ideal (y  - x*z, x*y - w*z, x  - w*y), ideal (y, x)}

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