Spring Break

I feel I owe a little explanation, since my original plan was to pick something cool like Morse theory and do a series of posts on it for a spring break project. Instead, my spring break became a rather intense experience.

My parents came to visit me, which was in the original plans. We did a few basic touristy things through Tues., then disaster struck. My dad we’ll just say started losing a lot of blood and ended up in the hospital. You should note that this is literally on the other side of the continent from where my parents actually live. So it is not a good place for this type of thing to happen.

He ended up receiving 4 units of blood over a course of 2 transfusions. They were due to leave extremely early on Friday morning. But in order to figure out what caused the bleeding, the doctors needed to do a colonoscopy (scheduled, you guessed it, Thursday afternoon). Thus, they had to extend all their reservations (rental car, hotel, plane tickets, etc) for an unspecified amount of time. They guessed it would be over by this morning, and it was luckily.

So without going into much detail, that is the general story of my spring break (and hence for any of you that read this that wanted me to do things with you over break and I promised I would, that is why I didn’t. I wasn’t just ignoring you). On a more positive note, I’ve made a lot of progress over this break on one of my life-long goals: learning to play the banjo. Really, do instruments get much cooler than the banjo?

What Day is It?

Wow. Has it really been seven days since I last posted? I studied and did finals for the first half of the week. Then I went into recovery mode for the rest of the week, and now my parents are visiting me. Anyway. My most recent project is that I’m giving a lecture on David Foster Wallace on Tuesday. I don’t feel like posting the details, since I’ve basically talked about most of it in previous posts.

Here is my tentative outline for the lecture.

Title: DFW: A New Era of Fiction
I. The Literary Landscape
–Talk about Barth’s “The Literature of Exhaustion”

II. Technical Virtuosity
a) Use of endnotes/footnotes:
–Disrupts linearity and how consciousness works
–Novel is an external construct
–Relationship with author = work
b) Philosophical illumination
–Open vs closed systems and use of abbreviations
–Language, Wittgenstein, and Derrida
–Lacan: “In the unconscious is the whole structure of language.”
c) Metaphos, etc.
–Fiction within fiction and “Westward”
–Referencing and mocking Barth

—haven’t figured this part out yet.

I really like how cryptic and hard to follow this post is. There is just no way most people could have a clue what this talk is actually going to be like. Oh well.

Krull Dimension

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.

Lying Over and Going Up Part II

I realized there was one more result I probably should have included last time. Oh well. Here goes:

Let $R^*/R$ be integral, $\frak{p}$ a prime ideal in R and $\frak{p}^*, \frak{q}^*$ prime ideals in $R^*$ lying over $\frak{p}$. If $\frak{p}^*\subset \frak{q}^*$, then $\frak{p}^*=\frak{q}^*$.

Proof: Recall that $S^{-1}R^*$ is integral over $S^{-1}R$ by last time for any multiplicative set, and also that prime ideals are preserved in rings of fractions. Thus the hypotheses still hold if we localize at $\frak{p}$. Thus $R_\frak{p}^*$ is integral over $R_\frak{p}$, and $\frak{p}^*R_\frak{p}^*\subset \frak{q}^*R_\frak{p}^*$ are prime ideals. Thus we can WLOG replace $R^*$ and $R$ by their localizations and hence assume they are local. So now $\frak{p}$ is a maximal ideal in $R$. Thus by last time $\frak{p}^*$ is maximal. Since $\frak{p}^*\subset \frak{q}^*$, we have $\frak{p}^*=\frak{q}^*$.

Now we are ready for the two big theorems. Here is the “Lying Over” Theorem. Let $R^*/R$ be an integral extension. If $\frak{p}$ is a prime ideal in R, then there is a prime ideal $\frak{p}^*$ in $R^*$ lying over $\frak{p}$, i.e. $\frak{p}^*\cap R=\frak{p}$.

Proof: First note that $R \stackrel{i}{\longrightarrow} R^* \stackrel{h^*}{\longrightarrow} S^{-1}R^*$ and $R \stackrel{h}{\longrightarrow} R_\frak{p} \stackrel{j}{\longrightarrow} S^{-1}R^*$ form the two sides of a commutative diagram. By last time $S^{-1}R^*$ is integral over $R_\frak{p}$. Choose a maximal ideal $\frak{m}^*$ in $S^{-1}R^*$. Thus $\frak{m}^*\cap R_\frak{p}$ is maximal in $R_\frak{p}$. But $R_{\frak{p}}$ is local with unique max ideal $\frak{p}R_\frak{p}$, so $\frak{m}^*\cap R_\frak{p}=\frak{p} R_\frak{p}$. But the preimage of a prime ideal is prime, so $\frak{p}^8=(h^*)^{-1}(\frak{m}^*)$ is a prime ideal in $R^*$.

Now we just diagram chase: $(h^*i)^{-1}(\frak{m}^*)=i^{-1}(h^*)^{-1}(\frak{m}^*)=i^{-1}(\frak{p}^*)=\frak{p}^*\cap R$. And also: $(jh)^{-1}(\frak{m}^*)=h^{-1}j^{-1}(\frak{m}^*)=h^{-1}(\frak{m}^*\cap R_\frak{p})=h^{-1}(\frak{p}R_\frak{p})=\frak{p}$.

Thus $\frak{p}^*$ lies over $\frak{p}$.

Our other big theorem is the one about “Going Up”: If $R^*/R$ is an integral extension and $\frak{p}\subset \frak{q}$ are prime in R, and $\frak{p}^*$ lies over $\frak{p}$, then there is a prime ideal $\frak{q}^*$ lying over $\frak{q}$ with $\frak{p}^*\subset \frak{q}^*$.

Proof: By last time $(R^*/\frak{p}^*)/(R/\frak{p})$ is an integral extension where $R/\frak{p}$ is embedded in $R^*/\frak{p}^*$ as $(R+\frak{p}^*)/\frak{p}^*$. Now we just replace $R^*$ and R by these rings so that both $\frak{p}^*$ and $\frak{p}$ are $\{0\}$. Now we just apply the Lying Over Theorem to get our result.

So as we see here integral extensions behave extremely nicely. These theorems guarantee that se always have prime ideals lying over ones in the lower field. This has some important applications to the Krull dimension that we’ll start looking at next time.

Lying Over and Going Up

If you haven’t heard the terms in the title of this post, then you are probably bracing yourself for this to be some weird post on innuendos or something. Let’s first do some motivation (something I’m not often good at…remember that Jacobson radical series of posts? What is that even used for? Maybe at a later date we’ll return to such questions). We can do ring extensions just as we do field extensions, but they tend to be messier for obvious reasons. So we want some sort of property that will force an extension to be with respect to prime ideals. Two such properties are “lying over” and “going up.”

Let $R^*/R$ be a ring extension. Then we say it satisfies “lying over” if for every prime ideal $\mathfrak{p}\subset R$ in the base, there is a prime ideal $\mathfrak{p}^*\subset R^*$ in the extension such that $\mathfrak{p}^*\cap R=\mathfrak{p}$. We say that $R^*/R$ satisfies “going up” if in the base ring $\mathfrak{p}\subset\mathfrak{q}$ are prime ideals, and $\mathfrak{p}^*$ lies over $\mathfrak{p}$, then there is a prime ideal $\mathfrak{q}^*\supset \mathfrak{p}^*$ which lies over $\mathfrak{q}$. (Remember that Spec is a contravariant functor).

Note that if we are lucky a whole bunch of posts of mine will finally be tied together and this was completely unplanned (spec, primality, localization, even *gasp* the Jacobson radical). First, let’s lay down a Lemma we will need:

Let $R^*$ be an integral extension of R. Then
i) If $\mathfrak{p}$ a prime ideal of R and $\mathfrak{p}^*$ lies over $\mathfrak{p}$, then $R^*/\frak{p}^*$ is integral over $R/\mathfrak{p}$.
ii) If $S\subset R$, then $S^{-1}R^*$ is integral over $S^{-1}R$.

Proof: By the second iso theorem $R/\frak{p}=R/(\frak{p}^*\cap R)\cong (R+\frak{p}^*)/\frak{p}^*\subset R^*/\frak{p}^*$, so we can consider $R/\frak{p}$ as a subring of $R^*/\frak{p}^*$. Take any element $a+\frak{p}^*\in R^*/\frak{p}^*$. By integrality there is an equation $a^n+r_{n-1}a^{n-1}+\cdots + r_0=0$ with the $r_i\in R$. Now just take everything $\mod \frak{p}^*$ to get that $a+\frak{p}^*$ integral over $R/\frak{p}$. This yields part (i).

For part (ii), let $a^*\in S^{-1}R^*$, then $a^*=a/b$, where $a\in R^*$ and $b\in\overline{S}$. By integrality again we have that $a^n+r_{n-1}a^{n-1}+\cdots + r_0=0$, so we multiply through by $1/b^n$ in the ring of quotients to get $(a/b)^n+(r_{n-1}/b)(a/b)^{n-1}+\cdots +r_0/b^n=0$. Thus $a/b$ is integral over $S^{-1}R$.

I’ll do two quick results from here that will hopefully put us in a place to tackle the two big results of Cohen and Seidenberg next time.

First: If $R^*/R$ is an integral ring extension, then $R^*$ is a field if and only if $R$ is a field. If you want to prove this, there are no new techniques from what was done above, but you won’t explicitly use the above result, so I won’t go through it.

Second: If $R^*/R$ is an integral ring extension, ten if $\frak{p}$ is a prime ideal in R and $\frak{p}^*$ is a prime ideal lying over $\frak{p}$, then $\frak{p}$ is maximal if and only if $\frak{p}^*$ is maximal.

Proof: By part (i) of above, $R^*/\frak{p}^*$ is integral over $R/\frak{p}$ and so as a corollary to “First” we have one is a field if and only if the other is. This is precisely the statement that $\frak{p}$ is maximal iff $\frak{p}^*$ is maximal.