# Irreducible components, dimension and degree of projective varieties

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.

• 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. May 9, 2023 at 7:07

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.)