I am trying to prove the following:
Let $X = \{(a_1, a_2, a_3) \in \mathbb{A}^3\; | \; a_1 = 0 \text{ or } a_2 = 0 \text{ or } a_3 = 0\}$ determine $I(X)$ and the irreducible components of $X$.
My attempt: The set $X$ can be decomposed as the union of three zero sets, namely:
$$X = Z(x) \cup Z(y) \cup Z(z) $$
where
$$Z(x) = \{ (a_1, a_2, a_3) \in \mathbb{A}^3\; | \; a_1 = 0 \},$$
$$Z(y) = \{ (a_1, a_2, a_3) \in \mathbb{A}^3\; | \; a_2 = 0 \},$$
$$Z(z) = \{ (a_1, a_2, a_3) \in \mathbb{A}^3\; | \; a_3 = 0 \}.$$
Now, the union of zero sets can be written as the product of the each zero set, hence we have:
$$Z(x) \cup Z(y) \cup Z(z) = Z(xyz) = Z(\langle xyz\rangle).$$ $$=\{ f\in k[x,y,z]: f^k\in \langle xyz\rangle,\, k\in \mathbb{N}\} $$
The ideal $I$ can be found by noticing that $$I(x) = \sqrt{\langle xyz\rangle}$$ where the $\sqrt{}$ denotes the radical of the ideal.
If $f^k\in \langle xyz\rangle \implies f^k = \sum a_i (xyz)$
$\implies f^k\in \langle x\rangle \cap \langle y\rangle \cap \langle z\rangle $
$\implies f\in \sqrt{\langle x\rangle} \cap \sqrt{\langle y\rangle} \cap \sqrt{\langle z\rangle} $
However, $\langle x\rangle$, $\langle y \rangle$ and $\langle z\rangle$ are prime, so:
$\implies f\in \langle x\rangle \cap \langle y\rangle \cap \langle z\rangle $
$\textbf{irreducible components}:$ $Z(x), Z(y), Z(z)$
$\textbf{I(x)}: ? $