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 $$$$.

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 $$I_1 \cup \cdots \cup \hat{I_k} \cup \cdots \cup I_n$$, where the hat symbol means that the $$k$$-th ideal is not counted in the union, for each $$k \in \{1, \ldots, n\}$$. Hence, we may choose for each $$k \in \{1, \ldots, n\}$$ $$a_k \in J \setminus I_1 \cup \cdots \cup \hat{I_k} \cup \cdots \cup I_n$$. Since $$n>2$$, at least one of the ideals $$I_1, \ldots, I_n$$ 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 $$j \neq m$$, $$b$$ 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.