The Krull dimension of a commutative ring; another view

Posted: November 3, 2011 in Elementary Algebra; Problems & Solutions, Rings and Modules
Tags: ,

We will assume that R is a commutative ring with 1. We will allow the case R=(0), i.e. where 1_R=0_R. The goal is to describe  the Krull dimension of R in terms of elements of R and not prime ideals of R. The reference is this paper: A Short Proof for the Krull Dimension of a Polynomial Ring. Recall that the Krull dimension of R is the largest integer n \geq 0 for which there exist prime ideals P_i, \ 0 \leq i \leq n, of R such that P_0 \subset P_1 \subset \ldots \subset P_n. Then we write \text{K.dim}(R)=n. If there is no such integer, then we define \text{K.dim}(R)=\infty and if R=(0), we define \text{K.dim}(R)=-1.

Notation. Let x \in R and set S_x = \{x^n(1+ax): \ n \geq 0, \ a \in R \}. Clearly S_x is multiplicatively closed. Let R_x be the localization of R at S_x. If 0 \in S_x, then we define R_x =(0).

Problem 1. Let x \in R and suppose that \mathfrak{m} is a maximal ideal of R. Then \mathfrak{m} \cap S_x \neq \emptyset. Moreover, if P \subset \mathfrak{m} is a prime ideal and x \in \mathfrak{m} \setminus P, then P \cap S_x = \emptyset.

Solution. if x \in \mathfrak{m}, then x \in \mathfrak{m} \cap S_x and we are done. Otherwise, \mathfrak{m} + Rx = R and thus 1 + ax \in \mathfrak{m} for some a \in R. Then 1+ax \in \mathfrak{m} \cap S_x. For the second part, suppose to the contrary that x^n(1+ax) \in P for some integer n \geq 0 and a \in R. Then, since x \notin P, we have 1+ax \in P \subset \mathfrak{m} and thus 1 \in \mathfrak{m} because x \in \mathfrak{m}. \Box

Problem 2. \text{K.dim}(R)=0 if and only if R_x=\{0\} for all x \in R.

Solution. As we already mentioned, R_x = \{0\} means 0 \in S_x. Suppose that 0 \notin S_x for some x \in R. Then there exists a prime ideal P of R such that P \cap S_x = \emptyset, because S_x is multiplicatively closed. By Problem 1, P is not a maximal ideal and thus \text{K.dim}(R) > 0. Conversely, suppose that 0 \in S_x for all x \in R and P is a prime ideal of R. Let \mathfrak{m} be  a maximal ideal of R which contains P. Choose x \in \mathfrak{m} \setminus P. By Problem 1, P \cap S_x =\emptyset, contradicting 0 \in P \cap S_x. \ \Box

Problem 3. Let n \geq 0 be an integer. Then \text{K.dim}(R) \leq n if and only if \text{K.dim}(R_x) \leq n-1 for all x \in R.

Solution. The case n=0 was done in Problem 2. So we’ll assume that n \geq 1. Suppose that \text{K.dim}(R) \leq n and let x \in R. A prime ideal of R_x is in the form PR_x, where P is a prime ideal of R and P \cap S_x = \emptyset. Also, if PR_x and QR_x are two prime ideals of R_x with PR_x \subset QR_x, then P \subset Q. Now, suppose to the contrary that \text{K.dim}(R_x) \geq n and consider the chain P_0R_x \subset P_1R_x \subset \ldots \subset P_n R_x of prime ideals of R_x. By Problem 1, P_n is not a maximal ideal and so there exists a prime ideal P_{n+1} such that P_n \subset P_{n+1} and that will give us the chain P_0 \subset P_1 \subset \ldots \subset P_{n+1} of prime ideals of R, contradicting \text{K.dim}(R) \leq n. Conversely, suppose that \text{K.dim}(R_x) \leq n-1 for all x \in R. Suppose also, to the contrary, that there exists a chain P_0 \subset P_1 \subset \ldots \subset P_{n+1} of prime ideals of R. Let x \in P_{n+1} \setminus P_n. By the second part of Problem 1, P_n \cap S_x = \emptyset and thus P_0R_x \subset P_1R_x \subset \ldots \subset P_nR_x is a chain of prime ideals of R_x, contradicting \text{K.dim}(R_x) \leq n-1. \ \Box

Problem 4. Let n \geq 0 be an integer. Then \text{K.dim}(R) \leq n if and only if for every x_0, x_1, \ldots , x_n \in R there exist integers k_0, k_1, \ldots , k_n \geq 0 and a_0, a_1, \ldots ,a_n \in R such that

x_n^{k_n}( \ldots (x_2^{k_2}(x_1^{k_1}(x_0^{k_0}(1+a_0x_0)+a_1x_1) + a_2x_2) + \ldots) + a_nx_n)=0.

Solution. By Problem 3, \text{K.dim}(R) \leq n if and only if \text{K.dim}((\ldots ((R_{x_0})_{x_1})_{x_2} \ldots)_{x_n})=-1. Therefore \text{K.dim}(R) \leq n if and only if (\ldots ((R_{x_0})_{x_1})_{x_2} \ldots)_{x_n}=\{0\}. \ \Box

Advertisements

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s