Finding the irreducible components.

Question

Let $X$ be an algebraic set of the affine $3$-space $\mathbb{A}^{3}$ given by $x_{1}^{2}-x_{2}x_{3} = x_{1}x_{3}-x_{1}=0$. Find the irreducible components of $X$.

I can easily figure out that we can write $X$ as:

$X = \{ (x,x^{2},1) \vert x\in\mathbb{A}\}\cup\{(0,0,z)\vert z\in\mathbb{A}\} \cup \{ (0,z,0) \vert z\in\mathbb{A}\}$

These three sets are zero sets and thus closed. The solution was presented in a problem session, but no check for irreducibility was given.

Why is it obvious that $\{ (x,x^{2},1) \vert x\in\mathbb{A}\}$, $\{(0,0,z)\vert z\in\mathbb{A}\}$ and $\{ (0,z,0) \vert z\in\mathbb{A}\}$ are irreducible?

Each of the three sets is the (isomorphic) image of $\mathbf A^1$ under a regular map. The image of an irreducible set is irreducible. — , Sep 10 '14 at 11:41
Possible duplicate of Computing irreducible components of algebraic set — Lada Dudnikova, Aug 15 '19 at 18:51

score 2 · Accepted Answer · answered Sep 10 '14 at 12:00

An algebraic set $Z(\mathfrak a)$ is irreducible iff $\mathfrak a$ is prime. The components given correspond to $Z(x_3 - 1, x_1^2 - x_2)$, $Z(x_1, x_2)$ and $Z(x_1, x_3)$.

Since $k[x_1, x_2, x_3] / (x_1, x_2) \cong k[x_3]$ and $k[x_3]$ is an integral domain, it follows that $Z(x_1, x_2)$ is irreducible. Similarly, $Z(x_1, x_3)$ is irreducible.

Let $\mathfrak a = (x_3 - 1, x_1^2 - x_2)$. To see that $Z(\mathfrak a)$ is irreducible, consider the homomorphism $\varphi : k[x_1, x_2, x_3] \to k[t]$, $\varphi(x_1) = t, \varphi(x_2) = t^2, \varphi(x_3) = 1$. By direct computation, $\mathfrak a \subset \ker \varphi$. Using the polynomial division algorithm, one can see that $\ker \varphi \subset \mathfrak a$. The desired result now follows via the first isomorphism theorem.

Finding the irreducible components.

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