Understanding the projective algebraic set $\mathcal Z(xz-y^2) \subset \mathbb P^2$. Is $\mathcal Z(xz-y^2) = \mathcal Z(y^2-z) \cup \mathcal Z(x,y)$?

Question

I would like to understand the projective algebraic set $\mathcal Z(xz-y^2) \subset \mathbb P^2$.

To try to do this, I am trying to understand what the points are by expanding the definition. I get $$ \begin{align*} \mathcal Z(xz-y^2) &= \{[a:b:c] \mid ac=b^2\}\\ &=\{[a:b:c] \mid a \ne 0, c=\frac{b^2}{a}\} \cup \{[0:0:c] \mid c \ne 0\}\\ &=\{[1:\frac{b}{a}: \frac{c}{a}] \mid a \ne 0, c=\frac{b^2}{a}\} \cup \mathcal Z(x,y)\\ &=\{[1:\frac{b}{a}: \frac{b^2}{a^2}] \mid a \ne 0, c=\frac{b^2}{a}\} \cup \mathcal Z(x,y)\\ &=\mathcal Z(y^2-z) \cup \mathcal Z(x,y). \end{align*} $$

This doesn't seem correct to me as now $$\mathcal Z(y^2-z) \cup \mathcal Z(x,y) = \mathcal Z(xy^2-xz) \cap \mathcal Z(y^3-yz).$$

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Is the above incorrect?

How can we see what $\mathcal Z(xz-y^2)$ is? The goal is to eventually look at the affine cone over $\mathcal Z(xz-y^2)$.

Answer 1

It is true that the variety equals $Z(y^2-z)\cup Z(x,y)$. Here, $Z(y^2-z)$ is the parabola inside $y,z$-plane (the open patch $\{[x:y:z]\mid x\neq 0\}$) and $Z(x,y)$ is the point $(0,0)$ inside $x,y$-plane. Together they give a circle inside the projective plane, which is $Z(xz-y^2)$.

However note that these are the patches of $Z(xz-y^2)$ inside different open sets. The last equality you wrote does not hold, as $Z(y^2-z)$ and $Z(x,y)$ are closed sets in different spaces, you cannot just write $Z(y^2-z)\cup Z(x,y)= Z(xy^2-xz)\cap Z(y^3-yz)$. Note that these last two polynomials are not projective and hence does not give a closed set in the projective plane.

Answer 2

There is the Veronese embedding:

$$v: \mathbb{P}^1 \rightarrow \mathbb{P}^2$$

defined by

$$v(u:v):=(u^2:uv:v^2).$$

The image $Im(v)=Z(xz-y^2)$ equals your variety. The image is irreducible hence you cannot write the set $Z(xz-y^2)$ as a union of two strict closed subsets.

Veronese embedding