View source for Commutative Algebra/Intersection and prime chains or Krull theory

{{TextBox| M=0 | W=100% | BG=#FFFFFF |1='''Definition 17.1''':

Let <math>R</math> be a ring. The '''(Krull) dimension''' of <math>R</math> is defined to be
:<math>\dim R := \sup \{n \in \mathbb N | \exists p_0, \ldots, p_n \le R \text{ prime }: p_0\subsetneq p_1 \subsetneq \cdots \subsetneq p_n\}</math>.}}

{{TextBox| M=0 | W=100% | BG=#FFFFFF |1='''Theorem 18.1 (prime avoidance)''':

Let <math>J, I_1, \ldots, I_n</math> be ideals within a ring <math>R</math> such that at most two of the ideals <math>I_1, \ldots, I_n</math> are not prime ideals. If <math>J \subseteq \bigcup\limits_{k=1}^n I_k</math>, then there exists an <math>m \in \mathbb N</math> such that <math>J \subseteq I_m</math>.}}

'''Proof 1''':

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

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

Assume otherwise. If there exists <math>a \in I_1</math> (or <math>I_2</math>), then <math></math>.

INCOMPLETE

'''Proof 2''':

We prove the theorem by induction on <math>n</math>. The case <math>n=2</math> we take from the preceding proof. Let <math>n>2</math>. By induction, we have that <math>J</math> is not contained within any of <math>I_1 \cup \cdots \cup \hat{I_k} \cup \cdots \cup I_n</math>, where the hat symbol means that the <math>k</math>-th ideal is not counted in the union, for each <math>k \in \{1, \ldots, n\}</math>. Hence, we may choose for each <math>k \in \{1, \ldots, n\}</math> <math>a_k \in J \setminus I_1 \cup \cdots \cup \hat{I_k} \cup \cdots \cup I_n</math>. Since <math>n>2</math>, at least one of the ideals <math>I_1, \ldots, I_n</math> is prime; say <math>I_m</math> is this prime ideal. Consider the element of <math>J</math>
:<math>b := a_m + a_1 \cdots a_{m-1} a_{m+1} \cdots a_n</math>.
For <math>j \neq m</math>, <math>b</math> is not contained in <math>I_j</math> because otherwise <math>a_m</math> would be contained within <math>I_j</math>. For <math>j = m</math>, <math>b</math> is also not contained within <math>I_j</math>, this time because otherwise <math>a_1 \cdots a_{m-1} a_{m+1} \cdots a_n \in I_j = I_m</math>, contradicting <math>I_m</math> being prime. Hence, we have a contradiction to the hypothesis.{{QED}}

{{BookCat}}