If (ideals) $J\subseteq I_1\cup\dots\cup I_n$ and $I_3,\dots,I_n$ are prime then $J\subseteq I_k$

Question

Let $R$ be a commutative ring with a unity. Let $I_1,\dots,I_n,J\subseteq R$ be ideals, $I_3,\dots,I_n$ are prine and $J\subseteq I_1\cup\dots\cup I_n$. Show that $\exists k$ s.t. $J\subseteq I_k$.

Attempt:

If $n=1,2$ then it is easily shown. Suppose $n=3$. Not sure how to porceed.

Answer 1

By induction we may assume that the union is minimal, so in particular $J$ is not included in any $\displaystyle\bigcup_{k\neq i} I_k$.

So let $x_i \in J\setminus\displaystyle\bigcup_{k\neq i}I_k$ (in particular $x_i \in I_i$)

Consider $x:=x_n+ x_1...x_{n-1}\in J$. There is $i$ with $x\in I_i$: if $i=n$, then $x_1...x_{n-1} \in I_n$, which implies ($I_n$ is prime because $n\geq 3$ because otherwise it's easy as you point out) that some $x_k \in I_n$ for some $k<n$: absurd.

Therefore $i<n$ but then $x_i\in I_i$ so $x_1...x_{n-1} \in I_i$ so $x_n\in I_i$, which is absurd.

Answer 2

Let's prove the contrapositive:

If $J$ is not contained in any $I_i$, then there is an element $x \in J$ not contained in any $I_i$.

This is exercise 1.1.6 in Matsumura's Commutative Ring Theory.

For $n = 2$, let $x \in J - I_1$, $y \in J - I_2$. Then all of $x$, $y$, $x + y$ are in $J$, but at least one of them is not in any of $I_1$ or $I_2$.

We now assume the result has already been proved for $n - 1$. We can also assume that there are no inclusions among $I_i$. Let $x$ be an element in $J$ but not in any of $I_1, \ldots, I_{n-1}$. Pick $y \in J(I_1 \ldots I_{n-1}) - I_n$. This is possible since $I_n$ is prime and none of $I_i$, $J$ is contained in $I_n$. Now both $x$ and $x + y$ are in $J$, but at least one of them is not in any of $I_i$.