Here’s a cool picture I made with Sage and R of class numbers of where is squarefree. It consists of the first five thousand such (click image to enlarge):
The even class numbers are shown in red +-signs and the odd class numbers are shown as blue disks. If you look carefully at the bottom left-hand corner where the points start, you’ll see a small conglomeration of blue dots that consist of the only nine imaginary quadratic fields that have class number one.
It was more than a year ago that I opened a package that I got in the mail, taking out this green ex-library hardcover in excellent condition. Now, I honestly can’t remember what prompted me to order it (perhaps it was the author’s name), but I remember reading the first few sections and feeling that it would be worthwhile to spend some time over its pages to learn more about those primes. Unfortunately, I didn’t manage to keep reading at the time, but a few months ago I decided to push through this volume with a little spare time I had, and this became my first real serious conversation with the zero divisor.
Certainly, no algebraist can ever escape the grasp of the zero divisor. In a ring , a nonzero element is called a zero divisor if for some nonzero . In even basic questions on ring theory, zero divisors are bound to be lurking. Our topic today is commutative rings, so we’ll assume from now on that is commutative. The rings probably easiest to understand, at least if we’re not considering relations to other rings, are fields. If is not a field, then it has an ideal that is not prime so already has zero divisors. So even if is a domain, some of its quotients will not be as long as is not a field.
Consider the good old Pascal’s triangle:
Except for the first row, take the alternating sum of the entries. So for the second row we have . For the third row we have . For the fourth row we have , etc. So it seems that we have the following for :
One can use the binomial theorem to prove this. In the process of reviewing some material on regular sequences, I came up with a slightly different proof that could be the most pretentious and ridiculous proof of this fact. (However, even more ridiculous proofs using heavy machinery would be welcome in the comments.) The reader may wish to consult the previous post describing the Koszul resolution before reading onwards.
As far as integral domains go, the UFD or unique factorisation domain is certainly one of the nicest. In this post we recall a few facts about UFDs and look at a wider class of rings called GCD-domains, which are domains in which every two elements have a greatest common divisor, and explain a proof following Gilmer and Parker that a polynomial extension of a GCD domain is a GCD domain. Along the way, we will see many cool algebraic facts!
Let’s recall that an integral domain is a UFD if every nonzero nonunit can be written uniquely (up to order and multiplication by a unit) as a product of irreducible (equivalently, prime) elements. The ultimate example of such a docile creature is the integral domain , the integers. Fields are UFDs as well, and a basic theorem of algebra is: if is a UFD then is a UFD, where is the polynomial ring in one variable.
Thus we get that for a field and are UFDs. given a UFD , however, it is not always true that , the power series ring in one variable, is a UFD. On the bright side, if is a principal ideal domain (hence a UFD), then is a UFD. The proof of this boils down to examining the evaluation map at zero, which would tell us after a bit of thought that if is a prime ideal in and is its image, then is principal if is principal and . We could then finish the proof by using the very useful fact that an integral domain is a UFD if and only if every prime ideal contains a principal prime ideal.
To reiterate, if is a principal ideal domain then is a unique factorisation domain. I wonder if the converse is true? But let us move on.