Hilbert’s Theorem 90

That’s right. Today I’m going to talk about where my blogging name (and basically my online name everywhere) came from.

If you go up to a mathematician and ask if they know who David Hilbert is, chances are they will say, “Yes! He is one of the most famous mathematicians of all time.” If they are problem oriented, then they will probably go off on his 23 problems that were presented at the International Congress of Mathematics in 1900. They may even say that this is what the current “Millenium problems” are based on.

On the other hand, David Hilbert is quite well-known to philosophers of math and science as the great formalist. He wanted to completely axiomatize mathematics. He was concerned with constructive proofs from these foundations. People on this side of the fence probably are familiar with his quote, “Wir mussen wissen. Wir werden wissen.” (or “We must know. We will know.”)

I want to present some history that is lesser known. There are many theorems associated with Hilbert. People familiar with algebraic number theory are used to seeing things like “HIlbert’s Theorem 90” or “Hilbert’s Theorem 92” (often also called Hilbert Satz 90 …) etc. But what are these referring to? It can’t possibly be his 90th theorem.

These theorems are in reference to Hilbert’s book Zahlbericht. He wrote the text in 1897, and it was the number theory text that many famous mathematicians such as Artin, Hasse, Hecke, and Weyl used (it was basically the only modern treatment of algebraic number theory available). One should note that this is remarkable considering “bericht” means “report,” and Hilbert literally wrote this text as a report of the state of algebraic number theory for the German Mathematical Society.

When Hilbert was comissioned to write the report, he was to work with Minkowski to write the state of all of number theory. Minkowski’s half was never written.

It is also rather interesting to note that although Hilbert had done work in algebraic number theory, this was a report on all of number theory, so the intent was not original work. In fact, the work that Hilbert is most known for in the field had not even been done yet. The report was designed to give direction to the field. It is because of this that it received much criticism most notably from Kummer who blamed Hilbert for the delays in some of his publications. This is probably a valid argument considering Hilbert often criticized Kummer for using complicated computations and even said that he avoided a lot of his work in the report. Hilbert even went ahead and replaced many of Kummer’s proofs with his own (e.g. satz 166-171 are Kummer’s theorems but Hilbert’s proofs).

One interesting historical point is how our notation and terminology has evolved over the past 100 years. For instance, many simple ideas from algebra had not been formalized yet such as a quotient group. At one point Hilbert writes, “the members of G are obtained precisely once when we multiply the members of H by 1, g, g^2, \ldots, g^{n-1} where g is a suitably chosen member of G.” In our current terminology we would just say, “G/H is a finite cyclic group.”

On the other hand we now get to Hilbert’s Theorems 89-94, which are considered to have influenced the subject in a very positive way. The 90th theorem (my name!) is arguably the most famous. This actually is one of Kummer’s. I won’t go into the original, but in modern terminology it is: If K is a finite Galois extension of F with G=Gal(K/F) cyclic of order n (with generator \sigma), then if a\in K, then N_{K/F}(a)=1 if and only if a=b/(\sigma(b)) for some b\in K. A generalization in terms of cohomology was found by Emmy Noether.

So next time you run into a theorem labelled “Hilbert Theorem ___” or “Hilbert Satz ____” you know where it came from. You also know that it is very likely that it is not Hilbert’s theorem at all, but was just compiled by him for a report.

Citations: I must admit that little is published on this subject and everything I wrote here I got from the introduction to the english edition of the Zahlbericht by Lemmermeyer and Schappacher. The exact statement of the 90th satz was taken from Larry Grove’s Algebra.


Noetherian Rings

I promised this awhile back. It seems as if the Noetherian condition is really the last major thing I need before being able to move on.

A ring is Noetherian if every ascending chain of ideals stabilizes (or “terminates”). So, this means that given any collection of ideals \{I_n\}\subset R such that I_1\subset I_2\subset I_3 \subset \cdots we have that there exists some N so that I_n=I_{n+1}=\cdots for all n>N. This condition seems very strange at first. It is known as the Ascending Chain Condition, or ACC for short, but it turns out that it is equivalent to some other things and makes sure our rings are somewhat well-behaved.

Since for the purpose of this collection of posts we only care about commutative rings, the ACC is equivalent to the condition that every ideal is finitely generated.

Proof) Suppose every ideal is finitely generated. Then let I_n be an ascending chain of ideals. Since I=\cup I_n is an ideal, it is generated by say m elements: I=<a_1, \ldots, a_m>. But each one of these elements come from some specific ideal, so suppose a_1\in I_{n_1}, \ldots, a_m\in I_{n_m}. Then just take N=\max(n_1, \ldots, n_m) and we have that the chain stabilizes after that.

For the reverse we go by contrapositive. Let I\subset R be some ideal that is not finitely generated. Then we can find a_1\in I such that <a_1>\neq I. We can also find a_2\in I\setminus <a_1> such that <a_1, a_2>\neq I. We can continue this process without termination. If it terminated at some step then, the ideal would be finitely generated. Thus we now just note that we have an ascending chain that doesn’t terminate <a_1>\subset <a_1, a_2>\subset \cdots.

It is easily seen that every PID is Noetherian. Rings tend to stay Noetherian under new constructions. The ring of polynomials (in finitely many indeterminates) and ring of power series where the coefficients come from a Noetherian ring is Noetherian. The former is known as the Hilbert Basis Theorem. Both the quotient R/I and the ring of fractions S^{-1}R are Noetherian if R is Noetherian.

But remember we want to figure out how this works with prime ideals. It turns out that prime isn’t quite what we want to get the best results, but in order to not introduce yet another type of ideal, I’ll leave this out since it won’t appear in anything I do later. So it turns out that if I\subset R an ideal and R Noetherian, every prime ideal P\supset I contains a minimal-over-I prime ideal, say P_0\supset I. This is just a standard one step application of Zorn’s Lemma.

So I think I’ve beat primality to death. Next time I’ll do a sort of “history of math” type post on Hilbert’s Zahlbericht to put into the blog carnival. This will give me some time to think of where to go from here. I’m thinking the algebraic number theory side…I just don’t want to have to build Galois theory before I do it.

Synecdoche, NY

I never thought the day would come where The Fountain would have a rival for greatest work of art produced in my lifetime. Charlie Kaufman has done it with Synecdoche, New York. I don’t want to say too much except, “Go see it now!”

In an attempt to not give anything away, but to talk about why it rivals Darren Aronsofsky’s brilliant work, it is basically about the same thing. The only thing. Fear of death. But of course, it wouldn’t be great if that was it. It also pulls in all that I love about modern art. It really isn’t as confusing as most would make it out to be either. The Fountain is probably the harder of the two to follow. But Kaufman’s film is more complex. More subtle.

So there is the post-modern paradigm. The greats of Pynchon or DeLillo. They try create these sprawling novels that keep splitting off in different directions: following different characters. They try to incorporate the ideas from math and science that say formal systems are incomplete or that uncertainty always exists. They try to self-reference themselves without contradiction.

Then people like DFW (David Foster Wallace for the non-initiated) come along and use the post-modern paradigm to reject it. You can’t self-reference without paradox. A self-reference creates an infinite recursion, but the novel itself is finite, so it doesn’t work. The very attempt at showing the incompleteness of the system creates a meta-system more powerful than the one in which it is trying to live. It is a rejection of post-modernism.

Kaufman says, “No! DFW, you should know better. You have some mathematical training. I’ll show you a return to classic post-modernism without contradiction. You can have an infinite recursion take up finite space. It just has to converge somewhere.” This is the brilliance of it. Let’s combine the only theme in art worth exploring (fear of death (note that I have a fairly good argument based on The Fountain that all things stem from this)), with a return to the classic post-modern paradigm. Where does the infinite self-reference converge to? Death of course! It was so obvious all along.

Don’t worry. I’ve given nothing away. A few more words are necessary, though. This movie is without a doubt the most demanding I’ve ever seen. Hopefully you have a degree in literature to catch all the references (Hedda Gabler, Death of a Salesman, White Noise, post-modern philosophy, etc). The details are so intensely packed in as well. I’m pretty sure you could just randomly flip to any 5 second period of the movie and I could point out some important detail.

Lastly, what does synecdoche mean? [From wikipedia] Synecdoche is a figure of speech in which:

  • a term denoting a part of something is used to refer to the whole thing, or
  • a term denoting a thing (a “whole”) is used to refer to part of it, or
  • a term denoting a specific class of thing is used to refer to a larger, more general class, or
  • a term denoting a general class of thing is used to refer to a smaller, more specific class, or
  • a term denoting a material is used to refer to an object composed of that material.

The movie takes place in Schenectady, NY. What an amazing play on words. DFW would be proud.

How great! Go now!

More on Primality

I want to wrap up some loose ends on the greatness of prime ideals before moving on in the localization theme. So. Recall that we formed the ring of quotients just like you would form the field of quotients. Only this time your “denominator” can be an arbitrary multiplicative set and this construction only gets us a ring. Moreover, this ring is not necessarily local. If we do the construction on a ring R with and the multiplicative set R\setminus P where P is a prime ideal, then we do get a local ring and we call this the localization.

Definition. Unique Factorization Domain (UFD): An integral domain in which every non-zero non-unit element can be written as a product of primes. (Note that there are equivalent definitions other than this one).

Quick property: Every irreducible element is prime.

Thus, it is instructive to look at some properties of prime ideals. First off, let’s look at the special case of UFD’s. It turns out that if R is a UFD, then for a multiplicative set S, S^{-1}R is also a UFD. This mostly has to do with the fact that R\hookrightarrow S^{-1}R is an embedding and anything in S^{-1}R is associate to something in R. This makes a nice little exercise for the reader.

So what’s so special about prime ideals in UFD’s? Well every nonzero prime ideal contains a prime element.

Proof: Suppose P\neq 0 and P prime. Then there exists a\in P, a\neq 0 such that a=up_1\cdots p_n where u a unit and p_i prime. Thus u\notin P. But this means that p_1\cdots p_n\in P and since it is prime we have some p_j\in P.

Theorem: If R is not a PID, and P an ideal which is maximal with respect to the property of not being principal, then P is prime (and will always exist).

Sketch of existence: Zorn’s Lemma. The proof of this contains lots of nitty gritty element-wise computation and a weird trick, so I don’t see it as beneficial. What is beneficial is that we get this great corollary: A UFD is a PID if and only if every nonzero prime ideal is maximal.

I’ve been kind of stingy on the examples, so I’ll leave you with a pretty common example of a ring of fractions. These are usually called dyadic rational numbers. Take your ring to be \mathbb{Z}. Then take your multiplicative set to be S=\{1, 2, 2^2, 2^3, \ldots\}. Now S^{-1}\mathbb{Z} are just the rational numbers with denominator a power of 2.

More generally we can form the p-adic integers (although that term is laden with many meanings, so I hesitate to actually use it). Let R=\times_{i=1}^\infty \mathbb{Z}/p^i. Where we have the restriction a\in R iff a=(a_1, a_2, \ldots ) satisfies a_i\cong a_{i+1} \mod p^i. So  elements of the ring are sequences. (Note \mathbb{Z} embeds naturally since i\mapsto (i, i, i, \ldots) satisfies that relation). This is a ring with no zero divisors, so we can take it to be the multiplicative set and we get the field of fractions \mathbb{Q}_p. The multiplicative group has a nice breakdown as \mathbb{Q}_p^{\times}\cong p^{\mathbb{Z}}\times \mathbb{Z}_p^{\times}.

Next time: Why Noetherian is important. How primality relates to it. And possibly another example.

Localization 2

Let’s figure out what “local” means and see if our construction somehow makes a local ring, i.e. is a “localization.”

Local: A ring is called local if there is a unique maximal ideal. This seems like a rather silly term, but it actually makes sense when you look at how rings arise in algebraic geometry or manifold theory. We won’t go there, though.

Sadly, it turns out that S^{-1}R is not always a local ring. But this is where primality comes into play. If P\subset R is a prime ideal then S=R\setminus P is a multiplicative set. Suppose it weren’t, then there would be two elements x,y\in S such that xy\notin S, i.e. xy\in P, but this is impossible, since by definition either x\in P or y\in P. We now denote the localization of R at P, to be S^{-1}R=(R\setminus P)^{-1}R which we denote with the shorthand R_P. This does turn out to be local since by the property listed last time of the embedding \phi^{-1}(S^{-1}P)=P, so S^{-1}P=\{r/s : r\in P, \ s\notin P\} is the unique maximal ideal in R_P.

Proof: Suppose x\in R_P, then x=r/s with r\in R and s\notin P. If r\notin P, then r/s is a unit in R_P. So all nonunits are in S^{-1}P. Now if I is any ideal in R_P that contains an element r/s with r\notin P, then I=R_P. Thus every proper ideal in R_P is contained in S^{-1}P. So R_P is local with unique max ideal S^{-1}P. For notational purposes outside of this blog, people usually write the prime ideal as \mathfrak{p} and the unique maximal ideal of R_\mathfrak{p} as \mathfrak{p}R_{\mathfrak{p}}.

I guess I’ve been rather sparse on the examples. The first one that comes to mind is surely to take \mathbb{Z}=R. Then our prime ideals are just the principal ideals generated by the primes, so take P=p\mathbb{Z} for some prime p. Then \mathbb{Z}_P=\mathbb{Z}_{(p)}.

I guess the importance of prime ideals leads us to explore some properties of prime ideals that could be useful.

Property 1: If S\subset R is any multiplicative set (not containing 0) and if P\subset R\setminus S is a maximal ideal, then P is prime. Also, any ideal I\subset R\setminus S is contained in such a P. I’ll omit proving this. The first part is fiddling with things until it works and the second statement uses Zorn’s Lemma.

OK, well I thought I had some other properties, but I can’t seem to find them/think of them now. I’m not sure where I’m going next. I’ll either move on to some related things to get at this better like the nilradical, or I’ll generalize this one more time to modules and do it using the categorical construction. If anyone has suggestions on which of these paths to take, just post. You probably have a few days as I’ll get busy again.

Localization 1

I’m calling this “Localization 1” since it is technically the beginning of the construction that we eventually want. But for now, I don’t even want to define what it means for a ring to be local, or why this may or may not be a (the) “localization.”

Last time we formed the field of fractions, but it came with some restrictions that we don’t need (as long as we don’t care that the end result isn’t a field). So today we will be forming the “ring of fractions.” Let’s think about what made that construction work. Essentially we didn’t want to “divide” by things that could end up as zero. This was the restriction that R be a domain. So let’s throw out that restriction, and just make sure there are no zero divisors. We also didn’t really care about the ring structure at all for the denominator. We only cared that it was closed under multiplication.

Let S be a multiplicative set with 1 (it is closed under mult.) with no zero divisors. Now do the same exact construction as before on \{(r,s)\in R\times S\}. We call the set of equivalence classes the ring of fractions and denote it S^{-1}R. It is straightforward computation to check that this is still a ring.

Now the only reason I included 1 in S was so that there is a natural embedding \phi :R\hookrightarrow S^{-1}R given by r\mapsto r/1. This embedding tends to be really nice for proving certain properties about the ring of fractions. One of these properties is exactly what we would hope: the embedding sends ideals to ideals, i.e. if I\subset R an ideal such that I\cap S=\emptyset, then \phi(I)\subset S^{-1}R is an ideal (with this being a bijection between prime ideals).

So the preimage under the embedding turns out to be fairly important. What is it on the element level? Well \phi^{-1}(S^{-1}I)=\{x\in R : \exists s\in S, \ sx\in I\}. And without giving too much away, we get the nice result that if I is prime and I\cap S=\emptyset, then \phi^{-1}(S^{-1}I)=I. Then we get that if P\subset R prime, then S^{-1}P\subset S^{-1}R is prime. So maybe there will be some importance to the notion of prime later.

Now let’s backtrack and try to do this construction one more time (OK, so it will come up again next time, but…). We really, really don’t want to have to restrict the set S to have no zero divisors. This is just too pesky of a condition. Can we somehow get around it?

This time let’s suppose S is multiplicative with 1, and let’s only assume 0\notin S. There can be zero divisors, though. Let I=\{x\in R: \exists s\in S, \ sx=0\}. Then I is an ideal. Clearly if x,y\in I then pick s,t\in S such that sx=0 and ty=0. Then st\in S and st(x-y)=stx-sty=0, so x-y\in I. Also x\in I and r\in R, (since we’ve assumed commutativity) we clearly have rx\in I.

Since we have a perfectly good ideal, and this ideal in a sense mimics the property of zero divisors, we do the natural thing and mod out by it. Thus form R\to R/I=\overline{R}, and S\to \overline{S}. Then \overline{S} has no zero divisors in \overline{R}. Suppose there was one, say a\in \overline{S}. Then there exists r\in \overline{R} such that ra=0. By definition of the “bar” we have that a=s+I for some s\in S and r=r_0+I for some r_0\in R (and 0=I), so (s+I)(r_0+I)=sr_0+I=I so sr_0=0 which means that s itself was a zero divisor. But trivially all zero divisors are sent to 0 under the mapping, so a=0\in \overline{S}.

So we form S^{-1}R by first canonically factoring through \overline{S}^{-1}\overline{R}. Why all this trouble? In essence what we are trying to do is form a new minimal ring that contains the old ring such that every element is a unit. If the kernel of the mapping is not {0} (i.e. there were zero divisors in the set), then the mapping is not an embedding and so the new ring doesn’t contain the old ring. The astute reader will note that this almost seems like a universal construction. This is what we’ll do when I actually get to localization for real.