I've been thinking about this exercise but I can't get the solution.
In $\mathbb{R}^3$ , I consider the usual axis:
$l_1=\{ x_1=x_2=0 \}$, $l_2=\{x_1=x_3=0\}$ and $l_3=\{ x_2=x_3=0 \}$.
Calculate generators for the ideal $I(l_1 \cup l_2 \cup l_3)=\{f\in \mathbb{R}[x_1,x_2,x_3]: f(a)=0, \forall a\in l_1 \cup l_2 \cup l_3 \}$.
My attemp:
Considering $H=\langle x_1x_2x_3\rangle$. If $h \in H \Rightarrow h=\sum _{finite} \lambda_j (x_1x_2x_3)$ and $h(a)=0, \forall a \in l_1 \cup l_2 \cup l_3.$ So $H \subseteq I(l_1 \cup l_2 \cup l_3)$. How could I continue?
Any help would be appreciate. Thank you in advance.
By the usual ideal-variety correspondence, you know that $I(l_1 \cup l_2 \cup l_3) = I(l_1) \cap I(l_2) \cap I(l_3)$. Thus what you want to compute is $$ \langle x, y \rangle \cap \langle x, z \rangle \cap \langle y,z \rangle .$$
Hint The intersection has three generators. It cannot be $H=\langle xyz\rangle$ because $H$ have dimension $2$, whereas we want dimension $1$.