fefes blog led me to a wonderful article dealing with cognitive dissonance:  Das Experiment — Rauchen ist gesund by Reto U. Schneider published in NZZ Folio 01/2007.

The uniqueness of factorization in the integeres is a well-known fact from high school calculus. Many people find it quite surprising that there might be situations, where unique factorization does not hold.

The standard example for a domain where unique factorization fails is . Rico told me about a more simple structure where non-uniqueness of factorization can already be observed: Consider all positive integers of the form , i.e. the set . Obviously, multiplication is a well-defined binary composition on this set. It also quite accessible, when to call a number prime (although irreducible would be more appropriate): any number besides one is prime, if there are no divisors besides and the number itself.

With this definition, it is easy to check, that is prime and so is ! Since the only non-trivial divisor is not an element of our set. Realizing this, we can easily construct an example for a number with non-unique prime decomposition. Take .

Christian Hesse wrote in Mitteilungen der DMV 17 (2009) on Mathematik und Schach und Schönheit.  The article is available as pdf.

The goal is to give a quick proof of the well-known theorem quoted in the title.

We will use two facts:

Fact 1:  The order of a group element divides the order of the group.

Fact 2: If is of order , then is of order .

Fact 3:  In any field, the equation has at most solutions .

Let be a commutative group with elements.  For every divisor of we define as the set of all elements in of order .

Remark:  If , then .

By Fact 1, the form a partition of :

.

Example:  Let .  For every the solutions of are given by .  Furthermore .

Lemma:  Let be a commutative group with elements, where for every divisor of the equation has at most solutions.

1. If is non-empty, then .
2. is never empty.

Proof:

1. Let .  Then and the elements of are all the solutions to .
   So, if $x$ solves $x^m = 1$, then $x \in \langle g \rangle$.  By the remark above, this shows .
   Furthermore $g^i$ is of order $m$ if and only if $gcd(i,m)=1$.  Therefore
   .
2. For we have $n/m \in A_m$ and therefore no is empty.  For general satisfying the assumptions of the Lemma we have
   
   
   .
   And equality only if all are nonempty.  But since equality is forced by the first and last term being equal this holds.

Corollary:  Units in a finite field form a cyclic group

Proof:  Units in a finite field satisfy the assumptions of the Lemma by Fact 3.  Any element of the nonempty set is a generator.

The Computeralgebra-Rundbrief 44 (2009) featured Thomas Markwig, Tropische Geometrie. An english version of the article with an extended list of references is available as pdf.

The Computeralgebra-Rundbrief 45 (2009) featured Andreas Enge, Komplexe Multiplikation: von numerisch bis symbolisch.  The article is available as pdf.

I really like the problems Tanya Khovanova presents on her blog.  It would be interesting to know how high school kids perform on the Subtraction and Multiplication Problems now compared to 50 years ago.

I particularly like the first one:

A stick has two ends. If you cut off one end, how many ends will the stick have left?

Bruce Schneier pointed in his blog to the following article on the base rate fallacy: A scanner to detect terrorists by Michael Blastland.  It includes a nicely worked example and an even better graphic.

After reading Tanya Khovanova’s Remember your Primes, I decided to do so.  The description is due to John Conway.

Instead of memorizing all primes below 1000, it is easier to recognize composites:

1.  Multiples of 2, 3, 5, 11 are easily detected.

2.  Squares are known.

3.  All that remains is to memorize 70 “non-trivial” composites (opposed to 168 primes in that range):

91, 119, 133, 161, 203, 217, 221, 247, 259, 287, 299, 301, 323, 329, 343, 371, 377, 391, 403, 413, 427, 437, 469, 481, 493, 497, 511, 527, 533, 551, 553, 559, 581, 589, 611, 623, 629, 637, 667, 679, 689, 697, 703, 707, 713, 721, 731, 749, 763, 767, 779, 791, 793, 799, 817, 833, 851, 871, 889, 893, 899, 901, 917, 923, 931, 943, 949, 959, 973, 989.

1. Remark:

If you are lazy, you can learn primes only up to 100. [...] [Y]ou need to remember only one number: 91.

2. Remark:

If you are very ambitious and plan to learn the primes up to 50,000, then the trick of learning non-trivial composites instead of primes is of no use to you. [...] The turning point is around 11,625: the number of primes below 11,625 equals the number of non-trivial composites below it.

3. As soon as you feel comfortable in the range below 1000 two more classes of composites become trivial to detect: Multiples of 7 and 13, since you can find the residue of a given number modulo 1001 by taking the alternating sum of three-digits.

Udo Vetter’s law blog cites under the title Verschlüsseln macht verdächtig from an interrogation transcript. [German]