$IJ$ is the set of nilpotent elements

Question

Let $R$ be a commutative ring with identity which is Noetherian. Let $V(A)$ denote the set of all prime ideals of $R$ containing the ideal $A$. Suppose that $V(0) = V(I) \cup V(J)$ and $V(I) \cap V(J) = \emptyset$ for ideals $I$ and $J$ of $R$. Prove that,

1. $R = I + J$.
2. The ideal $IJ$ is the set of nilpotent elements.
3. There exists a positive integer $n$ such that $R \cong R/I^n \times R/J^n$.

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1. If $I + J \subseteq M$, for some maximal ideal $M$, then since $M$ is maximal it is a prime ideal containing both $I$ and $J$, contradicting $V(I) \cap V(J) = \emptyset$.

I'm having difficulties showing (2) and (3).

For (2), Let $\mathfrak{N}$ be the set of all nilpotent elements, I can show that in general, $\mathfrak{N} = \bigcap\mathfrak{p}_i$, i.e., intersection of all prime ideals. In our case, each $\mathfrak{p}_i$ contains $I$ or $J$, thus $IJ \subseteq I \cap J \subseteq \mathfrak{N}$. However I can't show that $\mathfrak{N} \subseteq IJ$.

Also, is $S = \{r \in R | r^m \neq 0\}$, then $S$ is multiplicatively closed. If $A = \{K \subset R |K\text{ is an ideal and } K \cap S = \emptyset \}$, then $A \neq \emptyset$, because $(0) \in A$. Thus $A$ has a maximal element $\mathcal{P}$ which is a prime ideal containing only/all nilpotent elements. But $V(0) = V(I) \cup V(J)$ implies either $I \subset \mathcal{P}, J \subset \mathcal{P}$ or $\mathcal{P} = (0)$. If $\mathcal{P} = (0)$, then $IJ = (0) = \mathfrak{N}$. If $I \subseteq \mathcal{P}$, then $I \subseteq J = R$, so $IJ = I$, but still $IJ \subset \mathcal{P}$ which doesn't till me much.

I haven't had much luck with (3) either. I think that I need to find $n$ such that $R = I^n + J^n$ but $I^nJ^n = (0)$.

Answer 1

Hints.

2. $V(IJ)=V(I)\cup V(J)=V(0)$, so $\sqrt{IJ}=\sqrt{0}$.

3. $R$ noetherian $\Rightarrow$ $IJ$ finitely generated, so $(IJ)^n=0$ for some $n$. Then use the CRT.