Book mites

Got mites. I saw them crawling around on my gameboy advance screen. It was a pretty nasty thing to find. Since they're so small they're really hard to see and they could be anywhere.. I think they came along with my book on minkowski's geometry of numbers.

It spurred me on to tidying which was good. They mites are called psocoptera or book lice. They are an all female species (!) and they feed on microscopic mold particles like the paste used for wallpaper or book binding. Of course they like powdered food too such as flour, sugar, semolina. I can only hope they don't reach my kitchen.

I think the solution is to dehumidify: if there is low humidity their food can't exist so they will disappear too.


A diophantine equation

In some number theory notes it was said Euler solved $$x^3 = y^2 + 2$$ by working in $\mathbb Z[\sqrt{-2}]$. I had a look in Mordell (Diophantine equations) and apparently Eulers proof was "faulty". I constructed most of the proof but missed out an important detail about coprimality.

First some annoying technical stuff: Suppose $x$ is even then we have $y^2 \equiv -2 \equiv 6 \pmod 8$ but this is impossible so $x$ is odd, so $y$ is odd too.

Factor the RHS $(y + \sqrt{-2})(y - \sqrt{-2})$ and consider the GCD $g$ of these two terms, we want to show they are coprime (by showing $g=1$). Because the form $A(y + \sqrt{-2}) + B(y - \sqrt{-2})$ takes on values $2y$, $2 \sqrt{-2}$, We know the GCD is odd so throw away the $2$, therefore $g$ divides both $y$ and $\sqrt{-2}$. If $\sqrt{-2} | g$ then $2 | y$ so we can conclude $g=1$.

Now the good bit: $(y + \sqrt{-2}) = (a + \sqrt{-2}b)^3$, multiply this out


? Mod(a + x*b, x^2 + 2)^3
%1 = Mod((3*b*a^2 - 2*b^3)*x + (a^3 - 6*b^2*a), x^2 + 2)

which clearly only holds when $b = \pm 1$ and $3a^2 - 2b^2 \mp 1$! The second equation is sort of like a generalized Pell equation, it has the trivial solution $a=b=1$ but infinitely many more - of course $b$ is large in all of them so they are irrelevant to the original equation.

The only solution of the equation is then $3^3 = 5^2 + 2$.


Algebraic Number Theory

Basic question in algebraic number: why/why is $\mathbb Q(\alpha) \simeq \mathbb Q[\alpha]$?

My knee-jerk reaction was to use the norm function, $N(\alpha) = \prod_i \sigma_i \alpha \in \mathbb Q$ so $\alpha^{-1} = \prod_{i \not = 1} \sigma_i / N(\alpha)$.

This works in a lot of cases but not always, it's not enough for a proof - the conjugates of $\alpha$ may not live in the field. The real proof is very simple: $\mathbb Q[\alpha]$ is a field since it's isomorphic to $\mathbb Q[X]/(p(X))$ for some irreducible polynomial $p$. The key is irreducibility.


Fact: the field of fractions of the ring of integers of $K$ is $K$. In fact any number in $K$ is $\alpha/d$ for some algebraic integer $\alpha$ and integer $d$. This is very easy, proved in J.S.Milne pg 29.


Function Fields

I read an interesting aside in the notes: The integers of a function field behave a lot like integers of a number field. There are some mentions of this on wikipedia:


Apparently they have their own notions of prime numbers, mobius function etc. It sounds very interesting and I hope to learn about it soon


Bilinear forms

The idea of a bilinear form is very general. The most basic example is the inner product in $\mathbb R^n$. Another example is a that quadratic forms have bilinear forms attached. It may seem natural enough to want to attack a bilinear form to a number field.

In general you can take the discriminant of any bilinear form. Why do we take the trace of multiplying two numbers together rather than the norm? It would not be bilinear if we used norm.

In this lecture note (BILINEAR FORMS AND THE TRACE. - IAN KIMING) <http://www.math.ku.dk/~kiming/lecture_notes/2003-2004-algebraic_number_theory_koch/bilinear_forms.pdf> I found a nice general treatment of bilinear forms and the dual basis in the case of a non-degenerate form. It then specific to the algebraic number field case and proves some important results about discriminant.


Finite fields

I originally thought $$\mathbb F_{p} \le \mathbb F_{p^2} \le \mathbb F_{p^3} \le \cdots,$$ it turns out finite fields are not linearly ordered like this! I should do the theory of finite fields proper for once.


Many impressive results

I enjoyed reading about Zhang: https://archive.is/nVuDn>. It would be nice to study his theorem one day.