Find the irreducible components of $V(yw - x^2, zw - xy)$.

Question

Find the irreducible components of $V(yw - x^2, zw - xy)$, where $yw - x^2 $and $zw - xy $ denote the homogenization of $f_1 = y-x^2$ and $f_2 = z-xy$.

attempt: $V(yw - x^2, zw - xy) = V(yw - x^2)\cap V( wz - xy)$.

I don't know how to simplify anymore. Since I can't factor anything . Can someone please help? or verify this is correct.Any feedback would really help. Thank you!

Answer 1

In the following, I shall denote projective coordinates by square brackets. $V$ will be the given set.

1. Study first what are the points with $w=0$: you will get that their equations become $x^2 = 0$ and $xy=0$.

   1.1 If $y=0$ then you get $[0,0,z,0]$. Since $z \ne 0$ (because $[0,0,0,0]$ is not an element of the projective space), and since projective coordinates are defined only up to a non-zero multiplicative factor, the above equations only give you the point $[0,0,1,0]$.

   1.2 If $y \ne 0$ then you may divide by it, obtaining the points $[0, 1, \frac z y, 0]$. Letting $t = \frac y z$, this gives the subset $\{[0,1,t,0] \mid t \in K\}$ - an affine line embedded in the projective space.

   Notice that the point obtained at 1.1 is just the point at infinity of the straight line obtained at 1.2. This means that the points obtained so far, corresponding to $w=0$, form a projective line, which is clearly irreducible. Call it $L$.

2. Study now the points corresponding to $w \ne 0$ (this amounts to deprojectivizing your equations and therefore working in the affine coordinate patch of the first three coordinates) : letting $s = \frac x w$, $t = \frac y w$ and $u = \frac z w$, the equations become $\begin{cases}s^2 - t = 0 \\ u - st = 0 \end{cases}$, which are equivalent to $\begin{cases}s^2 - t = 0 \\ u - s^3 = 0 \end{cases}$, which are exactly the affine curve $\{(s,s^2,s^3) \mid s \in K\}$ - which is clearly irreducible. Call it $C$.

It follows that $V = L \cup C$ (and we even have $L \cap C = \emptyset$). The problem is that $C$ is closed in $\Bbb A^3$, but not in $\Bbb P^3$ and not in $V$, and the irreducible components must be closed in $V$. Nevertheless, since $V$ and $L$ are closed in $\Bbb P^3$, it follows that $V = \overline V = \overline {L \cup C} = \overline L \cup \overline C = L \cup \overline C$. Since $L$ is irreducible, it remains to show that $\overline C$ is irreducible too. But the projective closure of an irreducible affine set is itself irreducible, hence $\overline C$ is irreducible. It is also obvious (by inspecting their equations) that $C \ne V$ and $C \ne L$, so the decomposition of $V$ is indeed $L \cup \overline C$.

Answer 2

Some thoughts before I catch my train.

You have

$$\frac{x}{w}=\frac{y}{x}=\frac{z}{y}=t$$

Thus the points are

$$[tw,t^2w,t^3w,w]$$

Set $w\neq 0$ this gives a component. If $w=0$ the from the original equations we see $x=0$ so you have

$$[0,y,z,0]$$ and this is a line.

Now all you have to do is find if there are other components by looking at when various coordinates are zero. Gotta go, hope this helps.

Answer 3

Let $V_0=V\cap \mathbb A^3=V(f_1,f_2)\subset \mathbb A^3$ be the affine part of your variety $V$.
The irreducible components of $V$ consist of the closures $\overline {V_i}$ of the irreducible components $V_i$ of $V_0$ and the irreducible components of $V_\infty =V\cap \{W=0\}$ not included in any $\overline {V_i}$.
In your case $V_0$ is already irreducible since it is the image of the regular map $\mathbb A^1 \to \mathbb A^3:x\mapsto (x,x^2,x^3)$ whose source $\mathbb A^1 $ is irreducible.
Thus the first irreducible component of $V$ is $\overline {V_0}=V_0\cup \{(0:0:1:0)\}$.
Since $V_\infty=V(w,x^2, xy)=V(x,w)$ (topologically), an irreducible set not included in $\overline {V_0}$, we get the second irreducible component $V_\infty$.
Summing up: The ireducible components of $V$ are $\overline {V_0}$ and $V_\infty$.

Geometric explanation
The variety $V$ is the intersection of the two quadrics $V(yw - x^2)$ and $V(zw - xy)$ which happen to be tangent along their common line $V_\infty$.