Posted by Jason Polak on 23. August 2014 · Write a comment · Categories: homological-algebra

Suppose $k$ is a ring and $R$ is the ring of $2\times 2$ upper triangular matrices with entries in $k$. Via left matrix multiplication, $k^2$ viewed as column vectors becomes a left $R$-module. Now it turns out that $k^2$ is projective as a left $R$-module. Today we’ll see one methods to show this. Before we do this, let us briefly outline another solution to this problem that will make the reader appreciate the second method. The most straightforward approach to this problem is perhaps using the lifting criterion, or at least the following variation of it: define a surjection $R\to k^2$ by $1\mapsto [0,1]^t$. Now, one just has find a splitting of this surjection. Exercise: find a map that’s a splitting and prove carefully that it is actually an $R$-module homomorphism.

Now let’s see another way to do this. Instead of constructing a splitting, we could continue the surjection $R\to k^2$ defined by $1\mapsto [0,1]^t$ above into a projective resolution. Since maps out of free $R$-modules are easy to construct, if the resoluting projective resolution is small, then it will be easy to see by inspecting the resolution whether $k^2$ is projective (or so we hope).

To continue the projective resolution, we just find the kernel of $R\to k^2$ and map onto it with another free module. The reader may wish to write this process down before reading on.
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Posted by Jason Polak on 23. August 2014 · Write a comment · Categories: algebraic-geometry, group-theory · Tags:

Suppose one day you run into the following algebraic group, defined on $\mathbb{Z}$-algebras $R$ by
$$G(R) = \left\{ \begin{pmatrix}
a_{11} & a_{12} & a_{13} & a_{14} \\
a_{12} & a_{11} & -a_{14} & -a_{13} \\
a_{13} & -a_{14} & a_{11} & -a_{12} \\
a_{14} & -a_{13} & -a_{12} & a_{11}
\end{pmatrix} \in\mathrm{GL}_4(R) \right\}$$
Can you figure out what this group is? I actually did run into this group one day, but luckily I discovered its true identity. Today we’ll see one way to do so.

One thing to try to identify unknown algebraic groups in practice is to check out their matrix multiplication. If $A = (a_{ij}), B = (b_{ij})$ have the above form then $C = AB$ has first row
c_{11} = a_{11} b_{11} + a_{12} b_{12} + a_{13} b_{13} + a_{14} b_{14} \\c_{12}= a_{12} b_{11} + a_{11} b_{12} – a_{14} b_{13} – a_{13} b_{14} \\c_{13}= a_{13} b_{11} – a_{14} b_{12} + a_{11} b_{13} – a_{12} b_{14} \\c_{14}= a_{14} b_{11} – a_{13} b_{12} – a_{12} b_{13} + a_{11} b_{14}
(We only need to specify the first row to get the whole matrix in this group). Can you tell what the group is yet? At first, I couldn’t, so let’s try something else. One can check now that this group is actually commutative, and this would be enough to determine what it is given the context of where it came from, but let’s assume we don’t know that. Instead, let’s take the determinant:
\mathrm{det}(A) = (a_{11} + a_{12} + a_{13} – a_{14})(a_{11} + a_{12} – a_{13} + a_{14})\\\times(a_{11} – a_{12} + a_{13} + a_{14})(a_{11} – a_{12} – a_{13} – a_{14})$$
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Posted by Jason Polak on 03. May 2014 · Write a comment · Categories: modules · Tags:

Let \(R\) be a finite ring. The example we’ll have in mind at the end is the ring of \(2\times 2\) matrices over a finite field, and subrings. A. Kuku proved that \(K_i(R)\) for \(i\geq 1\) are finite abelian groups. Here, \(K_i(R)\) denotes Quillen’s \(i\)th \(K\)-group of the ring \(R\). In this post we will look at an example, slightly less simple than \(K_1\) of finite fields, showing that these groups can be arbitrarily large. Before we do this, let us briefly go over why this is true

But even before this, can you think of an example showing why this is false for \(i=0\)?
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Posted by Jason Polak on 14. April 2014 · Write a comment · Categories: number-theory

Here’s a cool picture I made with Sage and R of class numbers of $\mathbb{Q}(\sqrt{-D})$ where $D$ is squarefree. It consists of the first five thousand such $D$ (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.

Let $ R$ be an integral domain and $ K$ is fraction field. If $ K$ is finitely generated over $ R$ then we say that $ R$ is a $ G$-domain, named after Oscar Goldman. This innocuous-looking definition is actually an extremely useful device in commutative algebra that pops up all over the place. In fact, the $ G$-domain usually comes up in the context of quotienting by a prime ideal, so let’s call a prime $ P$ in $ R$ a $ G$-ideal if $ R/P$ is a $ G$-domain. In this post, we shall see a few applications of this concept, following Kaplansky’s book “Commutative Rings” for the theory and some standard examples for illustrations. At the end, we shall see a short paragraph proof of the Nullstellentsatz assuming the theory of $ G$-ideals.

Why is this concept so useful? Perhaps because of the following result: an ideal $ I$ in $ R$ is a $ G$-ideal if and only if it is the contraction of a maximal ideal in $ R[x]$ (although we won’t use it, it’s interesting to note that the nilradical is actually the intersection of all the $ G$-ideals). It’s worth looking at one direction of a proof of this result since it’s so short. First, the reader should prove that $ K$ can be generated by one element over $ R$ if and only if it is finitely generated, as an exercise.
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Posted by Jason Polak on 31. January 2014 · Write a comment · Categories: commutative-algebra · Tags: ,

P1000536It 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 $ R$, a nonzero element $ r\in R$ is called a zero divisor if $ rs = 0$ for some nonzero $ s\in R$. 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 $ R$ is commutative. The rings probably easiest to understand, at least if we’re not considering relations to other rings, are fields. If $ R$ is not a field, then it has an ideal $ I$ that is not prime so $ R/I$ already has zero divisors. So even if $ R$ is a domain, some of its quotients will not be as long as $ R$ is not a field.
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A commutative Noetherian local ring $ R$ with maximal ideal $ M$ is called a regular local ring if the Krull dimension of $ R$ is the same as the dimension of $ M/M^2$ as a $ R/M$-vector space.

In studying regular local rings one often uses the following lemma in inductive arguments: if $ R$ is an arbitrary commutative Noetherian local ring with maximal ideal $ M$, and $ M$ consists entirely of zero divisors, then the projective dimension $ \mathrm{pd}_R(A)$ is either zero or infinity for any finitely-generated $ R$-module $ A$. In other words, the only $ R$-modules of finite projective dimension are the projective (hence free) modules.

This is a neat little result that has a fun More »

It’s time for another installment of Wild Spectral Sequences! We shall start our investigations with a classic theorem useful in many applications of homological algebra called Schanuel’s lemma, named after Stephen Hoel Schanuel who first proved it.

Consider for a ring $ R$ the category of left $ R$-modules, and let $ A$ be any $ R$-module. Schanuel’s lemma states: if $ 0\to K_1\to P_1\to A\to 0$ and $ 0\to K_2\to P_2\to A\to 0$ are exact sequences of $ R$-modules with $ P$ projective, then $ K_1\oplus P_2\cong K_2\oplus P_1$.

We shall prove this using spectral sequences. I came up with this proof while trying to remember the “usual” proof of Schanuel’s lemma and I thought that this would be a good illustration of how spectral sequences can be used to eliminate the dearth of clarity in the dangerous world of diagram chasing.

Before I start, I’d like to review a pretty cool fact I which I think of as expanding the kernel, which is pretty useful More »

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 $ 1 – 1 = 0$. For the third row we have $ 1 – 3 + 3 – 1 = 0$. For the fourth row we have $ 1 -4 + 6 – 4 + 1 = 0$, etc. So it seems that we have the following for $ n > 0$:

$ \sum_{i=0}^n (-1)^n\binom{n}{i} = 0$

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.
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Take yourself away from this cold day in December and transport yourself to the world of commutative rings with identity. In this land there is a wonderful tool called the theory of regular sequences, which we will examine in this post. Our aim will be to get a quick idea of what regular sequences are, without going into too much tedious detail, with the hope that everyone reading this will think regular sequences are cool.

Now before I even define regular sequences, let us look at some examples of regular sequences:

  1. In the ring $ k[x,y,z]$, the sequence $ x,y,z$.
  2. In the ring $ \mathbb{Z}[x]$, the sequence $ 2,x$.
  3. In the ring $ \mathbb{Z}$, the sequence $ 4$

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