Showing the ideal $\left \langle yz,xz,yx+ay,x^2+ax \right \rangle$ is radical for all $a\neq 0$

Question

Let $I_a = \left \langle yz,xz,yx+ay,x^2+ax \right \rangle$ be an ideal of $k[x,y,z]$, where $a \neq 0$. Show that $I_a$ is radical. What is the geometric meaning of the elements in $\sqrt{I_0}\setminus I_0?$

The ideal $I_a$ is obtained by plugging $w=x+a$ in the ideal $I= \left \langle yz,xz,yw,xw\right \rangle$ of the closed set $X=\{x=y=0\} \cup \{z=w=0\}$. The ideal $I_a$ defines the closed set $\mathbb{V}(I_a)=\{x=y=0\}\cup\{x=-a,z=0\}$ which is the union of two non-intersecting lines. I know that generally for ideals that are not monomials it is hard to find the radical of an ideal without a computer, so the question probably calls for some trick or geometric insight.

By the method for monomial ideals, we know that $\sqrt{I_0}= \left \langle yz,xz,yx,x \right \rangle$, so its elements not in $I_0$ are exactly the elements $p_1yz+p_2xz+p_3yx+p_4x$ where $p_4 \in k$ is a constant.

Answer 1

$I_a = (x,y) \cdot (z,x+a)$ and $(x,y) + (z,x+a) =(x,y,z,x+a) = (1)$, since it contains $a \in k \setminus \{0\}$. By the Chinese remainder theorem $$k[x,y,z]/I_a \cong k[x,y,z]/(x,y) \times k[x,y,z]/(z,x+a) \cong k[z] \times k[y]$$ is the product of two domains, hence reduced.

Answer 2

I realized that the easiest way is to note that the ideal $\mathbb{I(V}(I_a))$ is radical (by the nullstellensatz), and $\mathbb{I(V}(I_a))=\mathbb{I}(\{x=y=0\}\cup\{x=-a,z=0\})=\mathbb{I}(\{x=y=0\})\cap\mathbb{I}(\{x=-a,z=0\})=(x,y)\cap(x+a,z)=I_a$ Hence this ideal is radical.

(The geometric meaning of the elements in $\sqrt {I_0}\setminus I_0$ is still unknown to me.)