Commutative Algebra/Intersection and prime chains or Krull theory

Definition 17.1:

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

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

Theorem 18.1 (prime avoidance):

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

Proof 1:

We prove the theorem directly. First consider the case ${\displaystyle n=2}$. Let ${\displaystyle a\in J\setminus I_{1}}$ and ${\displaystyle b\in J\setminus I_{2}}$. Then ${\displaystyle a\in I_{2}}$, ${\displaystyle b\in I_{1}}$ and ${\displaystyle a+b\in J}$. In case ${\displaystyle a+b\in I_{1}}$, we have ${\displaystyle a\in I_{1}}$ and in case ${\displaystyle a+b\in I_{2}}$ we have ${\displaystyle b\in I_{2}}$. Both are contradictions.

Now consider the case ${\displaystyle n>2}$. Without loss of generality, we may assume ${\displaystyle I_{1},I_{2}}$ are not prime and all the other ideals are prime. If ${\displaystyle J\subseteq I_{1}\cup I_{2}}$, the claim follows by what we already proved. Otherwise, there exists an element ${\displaystyle b\in J\cap (I_{3}\cup \cdots \cup I_{n}\setminus (I_{1}\cup I_{2}))}$. Without loss of generality, we may assume ${\displaystyle b\in J\cap (I_{3}\setminus (I_{1}\cup I_{2}))}$. We claim that ${\displaystyle J\subseteq I_{3}}$. First assume

Assume otherwise. If there exists ${\displaystyle a\in I_{1}}$ (or ${\displaystyle I_{2}}$), then .

INCOMPLETE

Proof 2:

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

${\displaystyle b:=a_{m}+a_{1}\cdots a_{m-1}a_{m+1}\cdots a_{n}}$.

For ${\displaystyle j\neq m}$, ${\displaystyle b}$ is not contained in ${\displaystyle I_{j}}$ because otherwise ${\displaystyle a_{m}}$ would be contained within ${\displaystyle I_{j}}$. For ${\displaystyle j=m}$, ${\displaystyle b}$ is also not contained within ${\displaystyle I_{j}}$, this time because otherwise ${\displaystyle a_{1}\cdots a_{m-1}a_{m+1}\cdots a_{n}\in I_{j}=I_{m}}$, contradicting ${\displaystyle I_{m}}$ being prime. Hence, we have a contradiction to the hypothesis.${\displaystyle \Box }$