Krull Dimension

March 16, 2009 by hilbertthm90 | 3 Comments

I didn’t actually want to take that long of a break before this post, but I had to do a final exam and give/grade a final, so that ate up lots of time. The next natural thing to move on to is something called Krull dimension. This is sort of annoying to define, but highly useful. I’ve also decided I’m going to stop “fraking” my $\frak{p}$ ‘s, since it is annoying to type and just use capital P’s for prime ideals.

First we need to define something I’ll call “height.” A prime chain is a strictly decreasing chain of prime ideals: $P_0\supsetneq P_1 \supsetneq \cdots \supsetneq P_n$ . Now we define the height of a prime ideal P, ht(P), to be the length of the longest prime chain with $P=P_0$ .

Some quick examples: It is easy to check that ht(P)=0 if and only if P is minimal, and hence in an integral domain ht(P)=0 if and only if $P=\{0\}$ . Let $R=k[x_1, x_2, \ldots]$ where k is a field. Then let $P_i=(x_i, x_{i+1}, \ldots)$ be the prime ideal generated by those indeterminants (check that it is prime easily by noting $R/P_i\cong k[x_1, \ldots , x_{i-1}]$ which is clearly an integral domain). Then for any n, we can make a chain $P_1\supsetneq P_2\supsetneq \cdots \supsetneq P_{n+1}$ . Thus $ht(P_1)=\infty$ .

Now for the actual definition we want to work with. I’ll denote the Krull dimension simply by “dim” rather than “Krulldim”. Then we define: $dim(R)=\sup\{ht(P) : P\in Spec(R)\}$ . So our quick example here is that for integral domains, $dim(R)=0$ if and only if $R$ is a field.

My goal for the day is to characterize all Noetherian rings of dim 0. The claim is that dim(R)=0 if and only if every finitely generated R-module M has a composition series. Since R is Noetherian, there are only finitely many minimal prime ideals. Since dim(R)=0, every prime ideal is minimal and hence there are only finitely many. Let’s call them $P_1, \ldots, P_n$ .

Let’s look at the nilradical: $\sqrt{R}=\cap P_i$ . Since the radical is nilpotent, there is some m such that $(\sqrt{R})^m=\{0\}$ . So we define $N=P_1\cdots P_n\subset P_1\cap\cdots\cap P_n=\sqrt{R}$ , so $N^m=\{0\}$ .

Let M be a finitely generated R-module. Then we have the chain $M\supset P_1 M\supset P_1P_2M\supset\cdots\supset NM$ . Now note that as a module $\frac{P_1\cdots P_{i-1}M}{P_1\cdots P_i M}$ is an $R/P_i$ -module. But $P_i$ is maximal and so $R/P_i$ a field, so it is a vector space. But M is finitely generated, so finite-dimensional, thus we can refine the chain so that all factors are simple.

Now we do this same trick on each of the chains $j=1, \ldots, m$ : $N^jM\supset P_1N^jM\supset\cdots \supset N^{j+1}M$ . Since at the m stage we get $N^m=\{0\}$ , we have a composition series for M.

For the converse suppose every finitely generated R-module has a composition series. Dimension zero is equivalent to showing that R has no prime ideals P, and Q such that $P\supsetneq Q$ . Suppose such exist. Let’s pass to the quotient, $R/Q$ , and reinterpret our hypothesis. Then R is an integral domain that has a nonzero prime ideal and a composition series $R\supset I_1\supset \cdots \supset I_d\neq \{0\}$ . So $I_d$ is minimal. Let $x\in I_d$ be any nonzero element. Then since $x I_d\subset I_d$ and $xI_d\neq \{0\}$ (we’re in a domain), then by minimality we have $xI_d=I_d$ . So there is a $y\in I_d$ such that $xy=x$ , i.e. $y=1\in I_d$ . And hence $I_d=R$ . Thus R is a field which contradicts our having a nonzero prime ideal.

Well, I think that is enough fun for one day. I may post again tomorrow, since my final is Wed.

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Author: hilbertthm90

I am a mathematics graduate student fascinated in how all my interests fit together.

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Krull Dimension

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