# 9 LaTeX typesetting errors that will hurt your eyes

Posted by Jason Polak on 18. May 2017 · Write a comment · Categories: math · Tags:

Here are some common LaTeX typesetting errors that you can avoid to make your document look good:

## The wrong inner product

Ever seen inner products written like this?

This horrible inner product typesetting was made using signs that should only be used as relation symbols such as less than or greater than. If you want to typeset a proper inner product using angle brackets, use \langle and \rangle.

## Using multiple fonts

When the ability to use other fonts are discovered, it leads to using them. Unfortunately, sometimes they get mixed when they shouldn’t:

This fragment shows a sans-serif font being used to emphasise a definition word in a sentence that is typeset with a serif font. Truly shocking, though it’s not as bad as using Times New Roman for the body text and Computer Modern Roman for the math font. Now that’s really bad. Take my advice: stick to one font.

## Not using the right-sized parentheses

This is one of the most common mistakes. Ever see something like this:

Yeah. This was acceptable when we were using typewriters. Not any more. For correctly sized parentheses and other delimters, use \left before the left delimiter and \right before the right delimiter.

## Keep weird fraktur letters to a minimum

Ever see a document with hundreds of fraktur letters?

Seriously? What’s wrong with $I$ and $J$?

## Use numbering that makes sense

What about those definitions, theorems, lemmas, and corollaries that all use separate numbering?

This makes for finding stuff in a document frustrating, and impossible in a long one. Use one numbering system to help everyone keep their sanity.
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# Where does convolution come from?

Posted by Jason Polak on 06. May 2017 · 2 comments · Categories: group-theory

When I first saw convolution I happily used it without thinking about why we should define it in the first place. Here’s a post that might tell you why it makes sense from an algebraic viewpoint. Let’s recall the convolution product, as it is for functions $f:\R\to\R$. If $f,g:\R\to\R$ are two such functions, then the convolution $f*g$ is defined as
$$(f*g)(y) := \int_\R f(y-x)g(x){\rm d}x$$
provided the integral exists. Why do we do this? For convolution, the explanation starts with polynomials. Multiplying polynomials is simplicity itself. For example, here is the product of two real-valued polynomials:
$$(1 + 2t + 5t^2)(1 + 3t) = 1 + 5t + 11t^2 + 15y^3$$
But a real-valued polynomial could also be thought of as a function $f:\Z\to \R$, where $f(x)$ is zero for all but finitely many $x$ and nonzero only if $x \geq 0$. In this setup, $f(x)$ represents the coefficient of $t^x$. Addition of polynomials becomes pointwise addition of functions. Now here’s the good part: polynomial multiplication becomes convolution: if $f,g:\Z\to\R$ are two polynomials then
$$(fg)(y) = \sum_{x\in \Z} f(y-x)g(x).$$
For a finite group $G$ then, complex representations of $G$ correspond to representations of the algebra of complex-valued functions on $G$ where the multiplication is the convolution product.

# Check out the new post series page

Posted by Jason Polak on 15. April 2017 · Write a comment · Categories: math

I’ve made a small organizational change to this blog. I realize it might not have been easy to find series of posts on one topic, or the “post series”. They used to be on a page called “features” but I think that was too obscure. Now there’s the page “post series”, which is found above on the header. Along with the current series I have going on Non-unique Factorisation, it also lists five posts on Waldhausen categories.

Perhaps the most exciting in my mind is the six posts of “Wild Spectral Sequences”, which give little toy problems that can be solved using spectral sequences. I came up with all the proofs myself, and some of them might be published nowhere else. Though I admit some of them are quite basic, I think they are quite instructive!

What are you waiting for? Visit the Post Series page today for these topics!

# Non-unique Factorisation: Part 2

Posted by Jason Polak on 15. April 2017 · Write a comment · Categories: commutative-algebra · Tags:

We are continuing the series on non-unique factorisation. For a handy table of contents, visit the Post Series directory.

In Part 1 of this series, we introduced for a commutative ring three types of relations:

1. Associaties: $a\sim b$ means that $(a) = (b)$
2. Strong associates: $a\approx b$ means that $a = ub$ for $u\in U(R)$
3. Very strong associates: $a\cong B$ means that $a\sim b$ and either $a = b=0$ or $a = rb$ implies that $r\in U(R)$

Here, $U(R)$ denotes the group of units of $R$. We have already seen in a ring with nontrivial idempotents like $\Z\times \Z$, a nontrivial idempotent $e$ will be satisfy $e\sim e$ and $e\approx e$, but $e\not\cong e$ because $e = ee$ and yet $e$ is not a unit and nonzero.

Therefore, $\cong$ is not an equivalence relation for all commutative rings. But it is symmetric:

Proof. Suppose $a\cong b$. Then $a\sim b$ and so $b\sim a$. If $a$ and $b$ are not both zero, write $a = sb, b = ta$. If $b = ra$ then $a = sra = s^2rb$. Since $a\cong b$, this implies that $s^2r$ is a unit and so $r$ is a unit. Hence $b\cong a$.

Guess what? The relation $\cong$ is also transitive. Since the proof is similarly short I’ll leave the proof to the reader. So, $\cong$ is just missing being reflexive for all rings to be an equivalence relation for all rings. If $\cong$ is an equivalence relation for a ring $R$, then we say that $R$ is presimplifiable. We introduced this type of ring last time.
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# Non-unique Factorisation: Part 1

Posted by Jason Polak on 11. April 2017 · Write a comment · Categories: commutative-algebra · Tags:

If $F$ is a field then the polynomial ring $F[x]$ is a unique factorisation domain: that is, every nonunit can be written uniquely as a product of irreducible elements up to a unit multiple. So in $\Q[x]$ for example, you can be sure that the polynomial $x^2 – 2 = (x-2)(x+2)$ can’t be factored any other way, and thus the only zeros of $x^2 – 2$ really are $\pm 2$.

If $F$ is not a field, then a polynomial might have a bunch of different factorisations. For example, in the ring $\Z/4[x]$ we can write $x^2 = (x)(x) = (x+2)(x+2)$. How can we make sense of factorisations in rings that are not unique factorisation domains? In order to do so, we first should make sure we understand what irreducible means in this context.

In the next several posts we’ll look at non-unique factorisation more closely, following a paper of Anderson and Valdes-Leon [1], but keeping the posts self-contained. We’ll start by looking at the concept of associates. One can in fact look at several different variations of associates. Here are three:
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# Switching the order of summation

Posted by Jason Polak on 21. March 2017 · Write a comment · Categories: elementary, exposition

Characteristic functions have magical properties. For example, consider a double summation:
$$\sum_{k=1}^M\sum_{r=1}^k a_{r,k}.$$
How do you switch the order of summation here? A geometric way to think of this is arrange the terms out and “see” that this sum must be equal to
$$\sum_{r=1}^M\sum_{k=r}^M a_{r,k}.$$
I find this unsatisfactory because the whole point of good notation is that you shouldn’t have to think about what it actually means. I do think it’s very important to understand what the notation means, but in doing formal manipulations it’s nice not to have to do that all the time.

A superior proof that these two sums are equal would be to write
$$\sum_{k=1}^M\sum_{r=1}^k a_{r,k} = \sum_{k=1}^M\sum_{r=1}^M a_{r,k}\delta(r\leq k)$$
where $\delta(r\leq k)$ is the function of the variables $r$ and $k$ that equals one exactly when $r\leq k$ and zero otherwise. Then we can happily switch the order of summation to get
$$\sum_{r=1}^M\sum_{k=1}^M a_{r,k}\delta(r\leq k).$$
Now, it’s trivial to get rid of the $\delta(r\leq k)$ function by writing
$$\sum_{r=1}^M\sum_{k=r}^M a_{r,k}.$$

# More about Ext Calculations with Regular Sequences

Posted by Jason Polak on 21. March 2017 · Write a comment · Categories: commutative-algebra · Tags:

This post is a continuation of this previous one, though I repeat the main definitions for convenience.

Let $R$ be a commutative ring and $A$ and $R$-module. We say that $x_1,\dots,x_n\in R$ is a regular sequence on $A$ if $(x_1,\dots,x_n)A\not = A$ and $x_i$ is not a zero divisor on $A/(x_1,\dots,x_{i-1})A$ for all $i$. Last time, we looked at the following theorem:

Theorem. Let $A$ and $B$ be $R$-modules and $x_1,\dots,x_n$ a regular sequence on $A$. If $(x_1,\dots,x_n)B = 0$ then
$${\rm Ext}_R^n(B,A) \cong {\rm Hom}_R(B,A/(x_1,\dots,x_n)A)$$

When $R$ is a Noetherian ring, $I$ a proper ideal of $R$, and $A$ a finitely-generated $R$-module, this theorem for $B = R/I$ says that the least integer $n$ such that ${\rm Ext}_R^n(R/I,A)\not= 0$ is exactly the length of a maximal regular sequence in $I$ on $A$.

The Noetherian and finitely generated hypotheses are crucial. Why is this? It’s because you need to have control over zero divisors. In fact you can see this by looking at the case $n = 0$:

Theorem. Let $R$ be a Noetherian ring, $I$ a proper ideal of $R$, and $A$ a finitely-generated $R$-module. Then every element of $I$ is a zero divisor on $A$ if and only if ${\rm Hom}_R(R/I,A)\not= 0$.
Proof. Since $A$ is a finitely generated $R$-module, that every element of $I$ is a zero divisor on $A$ is equivalent to $I$ being contained in the annihilator of a single nonzero element $a\in A$, which is in turn equivalent to every element of $I$ being sent to zero under the homomorphism
$$R\to A\\ 1\mapsto a.$$
Such homomorphisms are the same as nonzero homomorphisms $R/I\to A$. QED.

Here we are using this crucial fact:

Cool Theorem. For a finitely generated module $A$ over a Noetherian ring $R$, the zero divisors $Z(A)$ of $A$ in $R$ are a union of prime ideals of $R$, each of which are ideals maximal with respect to the property of being in $Z(A)$. Furthermore, each such prime is the annihilator of a single nonzero element of $A$.

In general, primes that are equal to the annihilator of a single element of a module $M$ are called the associated primes of $M$, and of course the theory of associated primes and primary decomposition is much more vast than this simple ‘Cool Theorem’, as is evident from Eisenbud’s 30-page treatment of them in his book Commutative Algebra. In practice however, I only ever seem to need this simple version of the ‘Cool Theorem’.

# Calculating Factorials in C

Posted by Jason Polak on 08. March 2017 · Write a comment · Categories: math

If $n \geq 0$ is an integer, define the number $n!$, pronounced “$n$ factorial”, as the number of bijections from an $n$-element set to itself. Therefore, $0! = 1$ and if $n > 0$ then
$$n! = 1\times \cdots\times n$$.
What’s the best way to calculate this quantity? The following “First Method” could possibly be the most straightforward method in the C language:

On other other hand, one might be tempted to reduce the number More »

# Paper Announcement: Separable Polynomials in Z/n[x]

Posted by Jason Polak on 05. March 2017 · Write a comment · Categories: commutative-algebra · Tags:

What is this paper about? It uses the theory of separable algebras to study separable polynomials in $\Z/n[x]$, which extends the usual definition of separability for polynomials over a field.

Let $d\geq 2$. A classical theorem of Leonard Carlitz says that for a prime $p$ with $q=p^k$, the number of monic separable polynomials of degree $d$ in $\F_q[x]$ is $q^d – q^{d-1}$. One can also define separable for polynomials in $\Z/n[x]$. In this case, since a polynomial cannot always be converted to a monic one by multiplying by a unit, it makes more sense count all separable polynomials. Deriving a formula for this number is exactly what my paper does.

Read the paper to see how it’s done! Although it talks about separable algebras, you can actually read it without knowing anything about this more advanced stuff as the interface between separable algebra theory and the concrete combinatorics is pretty clean. Or, you can just look at the final answer: the number of separable polynomials in $\Z/n[x]$ of degree at most $d$ for $d\geq 1$ is given by
$$\phi(n)n^d\prod_{i=1}^m(1 + p_i^d)$$
where $n = p_1^{k_1}\cdots p_m^{k_m}$ is the prime factorization of $n$ and $\phi(n) = |(\Z/n)^\times|$ is Euler’s phi function. The formulas in the paper have been checked mutliple times with Sage.

# Counting Bijective, Injective, and Surjective Functions

Posted by Jason Polak on 01. March 2017 · Write a comment · Categories: combinatorics

In this post we’ll give formulas for the number of bijective, injective, and surjective functions from one finite set to another. Although it’s not difficult, a formula for the number of surjective functions was one of the first problems I solved as an undergrad that got me interested in recurrence relations and combinatorics.

Let’s use the notation $[n] = \{ 1,2,\dots,n\}$ for an $n$-element set.
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