Posted by Jason Polak on 25. March 2015 · Write a comment · Categories: algebraic-topology, books

Doug Ravenel has made his book Nilpotence and periodicity in stable homotopy theory available for free download along with a list of errata, also available at the same page as the book.

Here is the official description from Princeton University Press:

Nilpotence and Periodicity in Stable Homotopy Theory describes some major advances made in algebraic topology in recent years, centering on the nilpotence and periodicity theorems, which were conjectured by the author in 1977 and proved by Devinatz, Hopkins, and Smith in 1985. During the last ten years a number of significant advances have been made in homotopy theory, and this book fills a real need for an up-to-date text on that topic.

Ravenel’s first few chapters are written with a general mathematical audience in mind. They survey both the ideas that lead up to the theorems and their applications to homotopy theory. The book begins with some elementary concepts of homotopy theory that are needed to state the problem. This includes such notions as homotopy, homotopy equivalence, CW-complex, and suspension. Next the machinery of complex cobordism, Morava K-theory, and formal group laws in characteristic p are introduced. The latter portion of the book provides specialists with a coherent and rigorous account of the proofs. It includes hitherto unpublished material on the smash product and chromatic convergence theorems and on modular representations of the symmetric group.

At the official page of the book, you can also buy a paperback copy.

Posted by Jason Polak on 18. March 2015 · Write a comment · Categories: math

It’s always a treat when a set of lecture notes for a workshop find their way to my arXiv RSS feeds. Today I found notes written by Samuele Anni on Vladimir Dokchitser’s lectures $l$-adic representations and their associated invariants (20pp). Here is the abstract:

These are notes from a 3-lecture course given by V. Dokchitser at the ICTP in Trieste, Italy, 1st–5th of September 2014, as part of a graduate summer school on “L-functions and modular forms”. The course is meant to serve as an introduction to l-adic Galois representations over local fields with “l not equal to p”, and has a slightly computational bent. It is worth mentioning that the course is not about varieties and their etale cohomology, but merely about the representation theory.

Here is a quotation from the introduction containing the prerequisites:

The prerequisites for the course are fairly modest: the theory of local fields and representation theory of finite groups, such as might be covered in a Masters level course, and, for the sake of examples and motivation, the theory of elliptic curves over local fields (Silverman’s book is more than sufficient). Some previous exposure to L-functions is desirable, as, after all, this was the main topic of the school.

Posted by Jason Polak on 18. March 2015 · Write a comment · Categories: math

Given two similar definitions, it is very valuable to have a counterexample distinguishing them. Here is one case where this arises: A Hurewicz fibration $p:E\to B$ of unbased spaces is a continuous map such that for every space $X$ and every commutative diagram
$$\begin{matrix} X & \longrightarrow & E\\ \downarrow & ~ & \downarrow \\ X\times I & \longrightarrow & B\end{matrix}$$
there exists a map $X\times I\to E$ making the resulting diagram commute. On the other hand, a Serre fibration $p:E\to B$ is a continuous map with the same property for every CW-complex $X$. From the definition it seems that there might be Serre fibrations that are not Hurewicz fibrations. In fact, there are! An example first appeared in R. Brown’s paper [1] in 1966. We define $E$ to be the subset of the Euclidean plane $\R^2$ defined by More »

Posted by Jason Polak on 10. March 2015 · Write a comment · Categories: math

The other day I wondered to myself, “am I using the best commutative diagram drawing package available?” Reading a little about Tikz on tex.stackexchange.com got me thinking that there might be improvements in store for my future commutative diagrams. Of course, being an ardent xy-pic fan, I was a bit hesitant. Nevertheless, I decided to try tikz-cd, by making two almost identical commutative diagrams with each package:

xypic tikz

Can you tell which one is which?
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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
$$
M = \begin{pmatrix}
x_{11} & x_{12}\\
x_{21} & x_{22}
\end{pmatrix}
$$
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:
$$
t^2 + at + b\mapsto
\begin{pmatrix}
0 & -b\\
1 & -a
\end{pmatrix}.
$$
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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