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I have this problem given to me in my review session for my algebraic geometry final:

Describe the irreducible components and compute the degree and dimension of $V_p(x_0x_2-x_1^2, x_0x_3-x_1x_2)\subset \mathbb{P}^3$.

Unfortunately, my professor is not responding to my emails. What is meant by irreducible components? I tried eliminating certain terms and coming up with the basis monomials for $K[x_0, x_1, x_2, x_3]/(x_0x_2-x_1^2, x_0x_3-x_1x_2)$, but this approach seemed tedious. How do I find the first term of the Hilbert polynomial (and thus answer the degree and dimension question very quickly)?

Can anybody provide any hints/a solution/ideas to approach these kinds of problems? I am working with Gathmann 2014 edition: link.

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  • $\begingroup$ Every noetherian topological space can be writen as a finite union of irreducible subsets. These irreducible subsets are called irreucible components. For instance, the union of five lines have five irreducible components. To solve the problem, I would suggest intersecting your algebraic set with the three affine components of $\mathbb{P}^3$ and try and solve it there. That should hopefully help give you a global picture. $\endgroup$ May 9, 2023 at 7:07

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Let $X$ be the your projective variety, $V(x_0x_2-x_1^2, x_0x_3-x_1x_2)\subset \mathbb{P}^3$.

To get a feel for what $X$ looks like, I suggest we break up the $\mathbb P^3$ as $\mathbb A^3 \cup\mathbb P^2$, where

  • the $\mathbb A^3$ is $\{ [1: u: v: w] : (u, v, w) \in \mathbb A^3 \}$.
  • the $\mathbb P^2$ is $\{ [0: x: y: z] : [x: y: z] \in \mathbb P^2 \}$.

Hopefully it is easy to see that

  • $X \cap \mathbb A^3 = V(v-u^2, w-u^3) \subset \mathbb A^3$. Notice that this is isomorphic to $\mathbb A^1$. If $t$ is the coordinate on the $\mathbb A^1$, then the isomorphism from $\mathbb A^1$ to $V(v-u^2, w-u^3) \subset \mathbb A^3$ is given by $t \mapsto (t, t^2, t^3)$.
  • $X \cap \mathbb P^2 = V(x) \subset \mathbb P^2$. This is isomorphic to $\mathbb P^1$.

These observations should help you see that $X$ has two irreducible components $X_1$ and $X_2$, both of which are embeddings of $\mathbb P^1$ in $\mathbb P^3$.

  • $X_1$ (the first irreducible component of $X$) is the image of $\mathbb P^1$ in $\mathbb P^3$ under the third Veronese embedding, $[t_0 : t_1] \mapsto [t_0^3 : t_0^2 t_1 : t_0 t_1^2 : t_1^3]$. This component contains all the points in $X \cap \mathbb A^3$, plus the point $[0:0:0:1]$ in $X \cap \mathbb P^2$.
  • $X_2$ (the second irreducible component of $X$) is the image of $\mathbb P^1$ in $\mathbb P^3$ under the embedding $[t_0 : t_1] \mapsto [0:0:t_0: t_1]$. This component contains all the points in $X \cap \mathbb P^2$; it does not contain any points in $X \cap \mathbb A^3$.

Both irreducible components are isomorphic to $\mathbb P^1$. So $X$ has dimension $1$.

As for the degrees:

  • $X_1$, being the image of $\mathbb P^1$ in $\mathbb P^3$ under the third Veronese embedding, is a degree $3$ curve in $\mathbb P^3$. Indeed, if $\iota : \mathbb P^1 \to X_1 \subset \mathbb P^3$ is this Veronese embedding, and if $f = ax_0 + bx_1 + cx_2 + dx_3$ is the defining equation for any hyperplane in $\mathbb P^3$, then $\iota^\star(f) = at_0^3 + bt_0^2 t_1 + ct_0 t_1^2 + d t_1^3$ is a cubic on $\mathbb P^1$, which has three zeroes (counted with multiplicity).
  • Similarly, $X_2$ is a degree $1$ curve in $\mathbb P^3$. If $\iota : \mathbb P^1 \to X_2 \subset \mathbb P^3$ is the relevant embedding, and if $f = ax_0 + bx_1 + cx_2 + dx_3$ is the defining equation for any hyperplane in $\mathbb P^3$, then $\iota^\star(f) = ct_0 + d t_1$ is a linear polynomial on $\mathbb P^1$, which has one zero. (Unless $c = d = 0$, in which case $X_2$ is entirely contained in this hyperplane.)
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