Posted by Jason Polak on 28. January 2015 · Write a comment · Categories: math

Sometimes there are wild beasts hiding in the tamest of rings. Let’s flush one out! Let $k$ be a field. Consider the following subring $R$ of $k[x,y]$ consisting of polynomials all of whose monomials are never of the form $x^i$ for $i\geq 1$. So $f = xy + y^2 + x^2y\in R$ but $f = xy + x^2$ is not. True or false: $R/(xy) \cong K[y]$?

If you thought the answer is ‘true’, think again! We cannot just “set $xy= 0$” — in fact, $x^2y\not\in (xy)$, since $x\not\in R$. This curious example of Graham Evans provides all sorts of interesting phenomenon, and is a good example to keep in mind. Do you think you understand what makes Krull’s principal ideal theorem tick? Let $M$ be the prime ideal consisting of all polynomials with constant term zero. It is in fact maximal, and is minimal over the ideal $(y)$. Yet, ${\rm rank}(M) \geq 2$! If $R$ were Noetherian, Krull’s principal ideal theorem would state ${\rm rank}(M)\leq 1$. So, obviously $R$ is not Noetherian. Can you find an infinitely generated ideal? Yet, it’s easy to see that $R$ still satisfies the ascending chain condition on principal ideals, so the ascending chain condition on principal ideals is insufficient for the principal ideal theorem to hold.

Posted by Jason Polak on 05. January 2015 · Write a comment · Categories: commutative-algebra · Tags:
Problem. Let $R$ be a (commutative!) Noetherian local ring, $M\subset R$ its maximal ideal, and $A$ a finitely generated $R$-module. If ${\rm Ext}^1(A,R/M) = 0$ then $A$ is a free $R$-module.

This problem will be a stepping stone to showing that a Noetherian local ring is regular if and only if the injective dimension of $R/M$ is finite. A closely related variation of this statement is that a Noetherian local ring is regular if and only if it has finite global dimension. This characterisation will then allow us to prove swiftly that a regular local ring is a unique factorisation domain.

Solution. Fix an exact sequence $0\to K\to F\to A\to 0$ with the rank of $F$ minimal. We will show that $K = 0$, which will finish the proof.

To do this, apply the functor ${\rm Hom}_R(-,R/M)$ to this exact sequence. The hypothesis that ${\rm Ext}^1(A,R/M) = 0$ shows that every homomorphism $f:K\to R/M$ can be factored as $K\subseteq F\to R/M$. Since we chose the rank of $F$ to be minimal, $K\subseteq MF$, and hence ${\rm Hom}_R(K,R/M) = 0$.

Now, $K/MK$ is an $R/M$ vector space, so if $K/MK$ is nonzero, there exists a nontrivial $R/M$-module homomorphism $\varphi:K/MK\to R/M$, which is also an $R$-module homomorphism since the action of $R$ on both is through $R\to R/M$. Composing with $K\to K/MK$ gives a nontrivial homomorphism $K\to R/M$, which is a contradiction. Hence $K/MK = 0$, or equivalently, $K = MK$. Since $R$ is Noetherian, $K$ is finitely generated and so by Nakayama’s lemma, $K = 0$.

And, the short version of this proof: applying ${\rm Hom}_R(-,R/M)$ to a minimal short exact sequence $0\to K\to F\to A\to 0$ with $F$ free shows that ${\rm Hom}_R(K,R/M) = 0$ and hence $K = MK$ whence $K = 0$.

Here are nine examples of projective modules that are not free, some of which are finitely generated.

Direct Products

Consider the ring $R= \Z/2\times\Z/2$ and the submodule $\Z/2\times \{0\}$. It is by construction a direct summand of $R$ but certainly not free. And it’s finitely generated! Another example is the submodule $\Z/2\subset \Z/6$, though this is the same kind of thing because $\Z/6\cong\Z/2\times\Z/3$. This was the first example I ever saw of a nonfree projective module.

Infinite Direct Products

One can modify the above construction for infinite direct products of rings, too. For instance, $R = \prod_{i=1}^\infty \Z$ contains $\Z$ as a direct summand. Hence $\oplus_{i=1}^\infty\Z$ is a projective $R$ module, yet cannot be free since nonzero free modules are uncountable.

Ideals in Dedekind Domains

In a Dedekind domain $R$, take an ideal representing a nontrivial element in the class group. It will then be projective. As an example, the class number of $\Z[\sqrt{5}]$ is two, and the ideal $(2,1+\sqrt{5})$ represents the nontrivial element in the class group. It is not free since it is not principal, and it is finitely generated projective since it is invertible.

More generally, for any ring extension of commutative rings $R\subseteq S$, one may define invertible $R$-submodules of $S$ as it is done for Dedekind domains. Then any invertible $R$-submodule of $S$ will be finitely-generated and projective. For more details and a further example, see Lam’s ‘Lectures on Modules and Rings’, Sections 2B-2C.

Rings of Continuous Functions

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Posted by Jason Polak on 19. December 2014 · Write a comment · Categories: homological-algebra, modules · Tags:

There are many ways to define the propery of semisimple for a ring $R$. My favourite is the “left global dimension zero approach”: a ring $R$ is left semisimple if every left $R$-module is projective, which is just the same thing as saying that every left $R$-module is injective. In particular, ideals are direct summands, and an easy application of Zorn’s lemma shows that $R$ can be written as a direct product of minimal left ideals, which is actually a finite sum because $R$ contains $1$.

An attack of Schur’s lemma yields the famous Wedderburn-Artin theorem: a ring $R$ is semisimple if and only if it is the finite direct product of matrix rings over division rings.

Since $R$ can be written as a finite direct product of minimal left ideals, we see that $R$ must be Noetherian and Artinian. Is the converse true?

Of course not! Here is a minimal counterexample: $\Z/4$. This ring cannot be semisimple. Indeed if it were, by the Wedderburn-Artin theorem, it would be a direct product of fields since it is commutative. It is not a field so it is not $\F_4$, and the only other possibility is $\Z/2\times\Z/2$, which it is also not isomorphic to since $\Z/4$ is cyclic.

We don’t have to appeal to the Wedderburn-Artin theorem however: the reduction map $\Z/4\to\Z/2$ makes $\Z/2$ into a $\Z/4$-module. If $\Z/4$ were semisimple, then $\Z/2$ would be a projective $\Z/4$-module, and hence at the very least as abelian groups, $\Z/2$ would be a direct summand of $\Z/4$, which is also nonsensical.

Can you think of a noncommutative example?

If $R$ is a commutative ring and $M$ an $R$-module, a regular sequence on $M$ is a sequence $x_1,\dots,x_n\in R$ such that $(x_1,\dots,x_n)M \not=M$ and for each $i$, the element $x_{i+1}$ is not a zero divisor on the module $M/(x_1,\dots,x_{i})$. The term regular sequence in $R$ just refers to a regular sequence on $R$ as an $R$-module over itself. The length of any regular sequence is the number of elements in the sequence.

The projective dimension of an $R$-module $M$ is the infimum over all lengths of projective resolutions of $M$, and hence is either a nonnegative integer or infinity. We write ${\rm pd}_R(M)$ for the projective dimension of an $R$-module $M$. (Clearly, this definition also makes sense for noncommutative rings.)

In particular, any ideal of $R$ is an $R$-module by definition, so we can look at the projective dimension of ideals. For instance, if $x\in R$ is a nonzerodivisor (=an element that is not a zero divisor), then the ideal $Rx$ is a left $R$-module that is free because $x$ is not a zero divisor. The following is a generalisation of this remark:

Problem. If $I$ is an ideal in a commutative ring generated by a regular sequence of length $n > 0$ then ${\rm pd}_R(I) = n-1$.

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Suppose we have a $2\times 2$ matrix
M = \begin{pmatrix}
x_{11} & x_{12}\\
x_{21} & x_{22}
with entries in a field $F$. The characteristic polynomial of this matrix is $p(t) := {\rm det}(tI_2 – M) = t^2 – (x_{11} + x_{22})t + x_{11}x_{22} – x_{21}x_{12}$. One might ask: how can we produce a matrix with a given characteristic polynomial? This can be accomplished using the rational canonical form:
t^2 + at + b\mapsto
0 & -b\\
1 & -a
We can calculate that the characteristic polynomial of this matrix to be $t^2 + at + b$. This map gives a bijection between quadratic monic polynomials in $F[t]$ and matrices of the above form. One way to understand this phenomenon is through algebraic groups. To explain, let’s stick with $F$ having characteristic zero, though much of what we do can be done in characteristic $p$ for $p$ sufficiently large as well using very different techniques.
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Posted by Jason Polak on 14. November 2014 · Write a comment · Categories: math


Today Alexander Grothendieck, probably best known for his significant development of the theory of schemes, died this morning at the Saint-Girons hospital in Ariège. He was 86 years old. Let me just list my four favourite Grothendieck inventions as a personal tribute:

  1. The functor of points is the perspective that a scheme (also invented by Grothendieck) can be viewed as a contravariant functor from affine schemes to sets. This perspective is immensely useful and clarifying in the theory of algebraic groups especially, and the representability of such functors is a fascinating field of study.
  2. The definition of $K_0$ was invented for his Riemann-Roch theorem, Grothendieck defined this simple functor of an essentially small abelian category to be the free abelian group whose generators $[A]$ correspond to isomorphism classes of objects $A$ and with relations $[B] = [A] + [C]$ for every short exact sequence $0\to A\to B\to C\to 0$. It’s such a simple and elegant definition and today it forms part of the basis for algebraic $K$-theory.
  3. Grothendieck topologies are generalisations of open covers where the open inclusions are replaced with more general morphisms, so that things like etale cohomology can be defined. It’s a fairly simple idea with enormous applications.
  4. Universal homological $\delta$-functors are a clean and crisp way to talk about derived functors and constructing natural transformations between them and they are used in homological algebra everywhere.
Posted by Jason Polak on 13. November 2014 · Write a comment · Categories: math · Tags:

There are two StackExchange sites I regularly visit: MathOverflow and Mathematics. Roughly, the former is for research level questions and the latter is for almost any kind of mathematics question. Mathematics is very popular at 216 questions per day at this time of posting (I don’t know how this average is calculated). Unfortunately, that means most of the questions aren’t interesting to any given user. To help find the questions I might like, I used to use the ‘Favourite Tags’ and ‘Ignored Tags’ features, with ‘Ignored Tags’ set to not display. Unfortunately, this only works so well, since there are so many tags, and my ‘Ignored Tags’ list got very long:

calculus homework integration complex-numbers summation combinatorics statistics real-analysis notation education graph-theory matrices limits probability probability-theory induction differential-equations stochastic-processes differential-geometry logic inequality general-topology trigonometry svd pde computational-mathematics vector-spaces numerical-methods algebra-precalculus terminology arithmetic algorithms functions probability-distributions polynomials linear-algebra analytic-geometry logarithms sequences-and-series multivariable-calculus gradient-flows geometry game-theory elementary-number-theory functional-analysis systems-of-equations fourier-analysis optimization discrete-mathematics analysis finance matlab uniform continuity statistical-mechanics elementary-set-theory differential-topology limits-without-lhospital number-theory uniform-continuity group-theory trees metric-spaces absolute-value definition galois-theory rotations complex-geometry puzzle complex-analysis

To be clear, I like many of these topics, but most of the questions with these tags on math.stackexchange are not very interesting. However, I recently discovered the feature ‘Advanced Tag Subscriptions’ that can be accessed by clicking edit>advanced tag subscriptions on the right-hand side of any StackExchange site. There you can create various filters (think of them as custom StackExchange sites) that have options like this:
The effect is that you can make a whitelist, showing only the tags you like. Moreover these tags can be from different sites so your filter can display a questions listing from as many sites as you like. Unfortunately, there are still a few problems with this approach:

  • You can only add tags and not more advanced filtering (like answers:0, score:3, etc.)
  • You cannot search through the filtered questions, which would ameliorate the first point
  • The page displaying the filtered questions does not display the score of the question, only the number of answers
Posted by Jason Polak on 09. November 2014 · Write a comment · Categories: math · Tags: , ,

I keep a fair amount of notes on things I want to learn and results that I am working on. Usually, I keep these in a LaTeX file on my computer in a neatly arranged directory structure. Unfortunately, for some time I’ve been dissatisfied with this approach. For one, LaTeX compiles into a isolated document that can’t be linked easily to other documents. For another, LaTeX makes good looking papers and books but is mediocre at formatting small little notes that might be linked in strange ways.

A typical example is conference notes: I always take notes at conferences at varying degrees of completeness, usually more on the side of a jumbled mess. Sometimes I just jot down one or two things that I’d like to keep in my mind, such as a fact I didn’t know or something to look up later. In typing these notes, perhaps I’d like to highlight these things somehow, perhaps by putting them in a pretty coloured box. With LaTeX, although it’s possible to use coloured boxes for text (and I do), it requires some macro with weird syntax that I keep forgetting.
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Posted by Jason Polak on 05. November 2014 · Write a comment · Categories: math

Unfortunately for the subject of K-Theory, one of its major journals “The Journal of K-Theory”, has been shrouded in controversy for the past few years, and as of today been deserted by all but two of its editors. Originally, this Cambridge University Press journal was meant to take the torch from Springer’s “K-Theory”, which also ended under strange circumstances (more details here).

Although published by CUP, “The Journal of K-Theory” is owned by A. Bak’s company which has yet to hand over the journal to the four-year old K-Theory Foundation as promised. The editors, except for Bak and one other, have resigned in protest, and issued a final announcement today. The now-resigned editors have promised to help authors with pending submissions.

Luckily for the K-Theory world, the K-Theory Foundation will launch the new “Annals of K-Theory” next year, and it is already accepting submissions. The resignation announcement in particular encourages libraries to cancel their subscriptions to the old journal in the fear that, in the words of the editors, “a phantom of JKT might actually still exist” by the time the new journal is published.

Here’s a wish of good luck to the new “Annals of K-Theory”.