The Krull dimension of a commutative ring; another view

Posted: November 3, 2011 in Elementary Algebra; Problems & Solutions, Rings and Modules
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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 $(\ldots ((R_{x_0})_{x_1})_{x_2} \ldots)_{x_n}=\{0\}. \ \Box$