Commutative Algebra/Intersection and prime chains or Krull theory

Proof 1:

We prove the theorem directly. First consider the case $n=2$. Let $a\in J\setminus I_1$ and $b\in J\setminus I_2$. Then $a\in I_2$, $b\in I_1$ and $a+b\in J$. In case $a+b\in I_1$, we have $a\in I_1$ and in case $a+b\in I_2$ we have $b\in I_2$. Both are contradictions.

Now consider the case $n>2$. Without loss of generality, we may assume $I_1,I_2$ are not prime and all the other ideals are prime. If $J\subseteq I_1\cup I_2$, the claim follows by what we already proved. Otherwise, there exists an element $b\in J\cap (I_3\cup \cdots \cup I_n\setminus (I_1\cup I_2))$. Without loss of generality, we may assume $b\in J\cap (I_3\setminus (I_1\cup I_2))$. We claim that $J\subseteq I_3$. First assume

Assume otherwise. If there exists $a\in I_1$ (or $I_2$), then ${\displaystyle }$.

INCOMPLETE

Proof 2:

We prove the theorem by induction on $n$. The case $n=2$ we take from the preceding proof. Let $n>2$. By induction, we have that $J$ is not contained within any of $I_1\cup \cdots \cup \widehat{I_k}\cup \cdots \cup I_n$, where the hat symbol means that the $k$-th ideal is not counted in the union, for each $k\in \{1,\dots ,n\}$. Hence, we may choose for each $k\in \{1,\dots ,n\}$ $a_k\in J\setminus I_1\cup \cdots \cup \widehat{I_k}\cup \cdots \cup I_n$. Since $n>2$, at least one of the ideals $I_1,\dots ,I_n$ is prime; say $I_m$ is this prime ideal. Consider the element of $J$

$b:=a_m+a_1\cdots a_{m-1}{a}_{m+1}\cdots a_n$.

For $j\ne m$, $b$ is not contained in $I_j$ because otherwise $a_m$ would be contained within $I_j$. For $j=m$, $b$ is also not contained within $I_j$, this time because otherwise $a_1\cdots a_{m-1}{a}_{m+1}\cdots a_n\in I_j=I_m$, contradicting $I_m$ being prime. Hence, we have a contradiction to the hypothesis.$◻$