Commutative Algebra/Intersection and prime chains or Krull theory

Definition 17.1:

Let $R$ be a ring. The (Krull) dimension of $R$ is defined to be

$\dim R:=\sup\{n\in \mathbb {N} |\exists p_{0},\ldots ,p_{n}\leq R{\text{ prime }}:p_{0}\subsetneq p_{1}\subsetneq \cdots \subsetneq p_{n}\}$.

Theorem 18.1 (prime avoidance):

Let $J,I_{1},\ldots ,I_{n}$ be ideals within a ring $R$ such that at most two of the ideals $I_{1},\ldots ,I_{n}$ are not prime ideals. If $J\subseteq \bigcup \limits _{k=1}^{n}I_{k}$, then there exists an $m\in \mathbb {N}$ such that $J\subseteq I_{m}$.

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 $I_{1}\cup \cdots \cup {\hat {I_{k}}}\cup \cdots \cup I_{n}$, 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.$\Box$