Let $R$ be a ring and $M$ an $R$-module with a finite free resolution (an “FFR module”). That is, there exists an exact sequence $0\to F_n\to F_{n-1}\to \cdots\to F_0\to M\to 0$ with each $F_i$ a finitely generated free $R$-module. If we denote by $r_i$ the rank of $F_i$, then the Euler characteristic of $M$ is defined to be $\chi(M) = r_0 – r_1 + \cdots + (-1)^nr_n$. One can easily prove that this is independent of the finite free resolution chosen and hence is a well-defined integer. In the post The Alternating Binomial Sum Vanishes we saw that it’s possible to prove that the alternating sum $\sum (-1)^i\binom{n}{i}$ vanishes by using some facts about Euler characteristics together with the Koszul complex.

Today’s problem shows that the Euler characteristic ins’t that interesting for ideals or homomorphic images of rings. First, here are some facts about the Euler characteristic, some of which might be useful for solving the problem:

  1. Any FFR module over a commutative ring has nonnegative Euler characteristic
  2. If an FFR module over a commutative ring has nonzero Euler characteristic, then the annihilator of $A$ is nil
  3. (Stallings) If $A$ is an FFR module with annihilator $I$ then $\chi(A)I = 0$.

We refer to these as Fact 1, Fact 2, and Fact 3.

Problem. If $I$ is a nonzero (!) ideal of a ring $R$ with a finite free resolution, show that $R/I$ also has a finite free resolution and that $\chi(I) = 1$ and $\chi(R/I) = 0$.

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Posted by Jason Polak on 15. October 2014 · Write a comment · Categories: math

If $R$ is a commutative ring and $A$ an $R$ module, let $Z(A)$ denote the zero divisors of $R$ on $A$. In other words, $Z(A) = \{ r\in R : \exists_a(ra = 0) \}$. If $I$ is the annihilator of $A$, then it is known that any prime minimal over $I$ is contained in $A$. In particular, any minimal prime of $R$ is contained in $Z(R)$. (Recall that a prime $P\subset R$ minimal if it does not properly contain any other prime, and $P$ is said to be minimal over $I$ if there is no inclusion $I\subseteq Q\subset P$ with $Q$ another prime.)

Problem. If $R$ is a ring with no nonzero nilpotent elements, then $Z(R) = \cup_P P$ where $P$ ranges over all the minimal primes of $R$.

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Posted by Jason Polak on 11. October 2014 · Write a comment · Categories: math

Let $R$ be a commutative ring. We say that $R$ is a valuation ring if for any $a,b\in R$, either $a | b$ or $b | a$. For an integral domain $R$, this is equivalent to say that for any nonzero $r$ in the fraction field of $R$, either $r\in R$ or $r^{-1}\in R$.

Examples of valuation rings include discrete valuation rings such as the ring of formal power series $k[[t]]$ for a field $k$ and the ring $\mathbb{Z}_p$ of $p$-adic integers for a prime $p$. Valuation rings also have unique maximal ideals (exercise). Can you think of a valuation ring that is not a discrete valuation ring and not a field? Anyways, here’s the main problem for today:

Problem. Show that every radical ideal in a valuation ring is a prime ideal.

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Posted by Jason Polak on 11. October 2014 · Write a comment · Categories: math, opensource · Tags: ,

Sage is a neat bundle of mathematical software that can be used to do anything from finding class numbers of number fields (like I did to make this graph) to testing whether a finitely presented group is trivial (sometimes…). Typically one works with Sage in a “worksheet” in a web browser, so that it’s easy to organise large numbers of calculations for later reference. Navigating large worksheets, however, might be cumbersome without a table of contents, just like long math papers without a table of contents are hard to read.

In this post, we’ll go over how to (manually) make a table of contents in your Sage worksheet with HTML, with the following example: More »

Posted by Jason Polak on 08. October 2014 · Write a comment · Categories: math

In an effort to keep this blog active, I’ve decided to post some old solutions to various problems I’ve done from books, starting with Kaplansky’s “Commutative Rings”. Hopefully they might be useful to someone learning the subject. I’ve done about half the problems, so it might take a while to post them all, and perhaps posting them will motivate me to do the other half. I won’t necessarily post them all in order, but I will start with the first problem in the book. I encourage readers to try the problem first, and post alternative solutions in the comments! The wording is changed slightly to avoid copyright issues, and I will attempt to provide some background comments to make the post self-contained. All the solutions will be linked to in the Solutions Page for easy reference.

Problem. If every proper ideal in a commutative ring $R$ is prime, show that $R$ is a field.

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Posted by Jason Polak on 01. October 2014 · Write a comment · Categories: math · Tags: ,

AZC has switched from WP-LaTeX to MathJax. There was no real problem with WP-LaTeX, but MathJax offered some improvements:

  1. Improved rendering
  2. Local installation: MathJax can be installed on your own webserver, so that you’re no longer reliant on external services
  3. Easier typing of LaTeX: in WP-LaTeX, you have to type the word LaTeX after the first dollar sign, and in MathJax you don’t

Here is a before-and-after image (you’ll have to click on it to view full size to see the difference properly):

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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.