## Basic notions pt 1 – Number fields and rings of integers

So yeah,  now that we have seen a little of what algebraic number theory is about, let’s get started properly.

Usually in number theory we  love to work with integers, or at a push rational numbers. This is not really enough to solve some problems.

A fact that you all know is that really each rational number can be written as a ratio of two integers (excluding $0$ on the denominator). More formally, $\mathbb{Q}$ is the field of fractions of the integral domain $\mathbb{Z}$.

The field $\mathbb{Q}$ is in a way the biggest thing we work with in elementary number theory.

However, in (basic) algebraic number theory we look at other fields of “numbers” obtained from $\mathbb{Q}$ by taking field extensions of finite degree (remember that the degree of a field extension $L/K$ is the dimension of $L$ when viewed as a $K$-vector space).

Such fields are called number fields.

Examples of number fields are; $\mathbb{Q}(\sqrt{2})$, $\mathbb{Q}(\sqrt[3]{5}, \sqrt{7})$ and $\mathbb{Q}(\zeta)$ where $\zeta$ is a primitive $n$th root of unity.

The degree of a number field is defined to be the degree of the corresponding extension of $\mathbb{Q}$.

Recall that a field extension $L/K$ is algebraic if every element of $L$ satisfies a polynomial with coefficients in $K$. It is known that every field extension of finite degree is algebraic.

So the reason that we define number fields to have this finite degree condition is so that we can guarantee that we get an algebraic extension of $\mathbb{Q}$ and that we have a finite basis over $\mathbb{Q}$.

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By Artin’s primitive element theorem, there exists a primitive element for any number field as an extension of $\mathbb{Q}$.

So we may always write a number field in the form $\mathbb{Q}(\theta)$ for some $\theta$ algebraic over $\mathbb{Q}$.

Not only does the proof of this theorem provide the existence of such a generator but it also gives a way of finding one. Generators are never unique.

See http://en.wikipedia.org/wiki/Primitive_element_theorem for some info, including details of the construction of a primitive element.

A simple example is that $\mathbb{Q}(\sqrt{2},\sqrt{3})=\mathbb{Q}(\sqrt{2}+\sqrt{3})$. One can check this by hand by showing that each side contains the generators for the other side but also this can be shown via the construction in Artin’s argument.

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One nice thing about number fields is that they each contain a very special ring, the ring of integers.

We already know that number fields are algebraic extensions of $\mathbb{Q}$, so that every element of a number field satisfies some monic polynomial over $\mathbb{Q}$.

We can always clear the denominators and find that every element actually satisfies a polynomial over $\mathbb{Z}$. The key thing to ask now is, which elements satisfy a monic polynomial over $\mathbb{Z}$?

These elements in a number field $K$ form the ring of integers, denoted $\mathfrak{O}_K$. Proving that we actually have a ring here is not so simple.

To give you some kind of idea why this is not so straight-forward, if I give you two completely random algebraic numbers say $\sqrt[57]{8217}$ and $2+57\sqrt{5}$, can you immediately tell me the polynomials that their sum and product satisfy?

The elements of $\mathfrak{O}_K$, in some sense, generalize the notion of “integer”. In fact when we take the number field $K=\mathbb{Q}$ then it is easily shown that $\mathfrak{O}_K=\mathbb{Z}$.

To see this, note that the minimal polynomial of any rational number $\frac{a}{b}$ over $\mathbb{Q}$ is $bx+a$. This can only be a monic polynomial over $\mathbb{Z}$ when $b=1$, i.e. when we started with an integer. Conversely, it is easy to see that any integer $n\in\mathbb{Z}$ satisfies the polynomial $x-n$ and so belongs to $\mathfrak{O}_K$.

The ring of integers is trivially an integral domain and conventionally its elements are called integers. Elements of $\mathbb{Z}$ become known as rational integers.

Here are a few examples of rings of integers. Now the ring of integers of $\mathbb{Q}$ is $\mathbb{Z}$. We saw this above. The ring of integers of the Gaussian integers $\mathbb{Q}(i)$ is $\latex mathbb{Z}(i)$…this isn’t hard to prove. The ring of integers of $\mathbb{Q}(\zeta_n)$, where $\zeta_n$ is a primitive $n$-th root of unity, is $\mathbb{Z}[\zeta_n]$. This is harder to prove.

We are starting to see a pattern here…to get the ring of integers it seems as though we are just replacing $\mathbb{Q}$ with $\mathbb{Z}$.

Unfortunately things are not that simple. The ring of integers of $\mathbb{Q}(\alpha)$ will always contain $\mathbb{Z}[\alpha]$ but we do not always have an equality. For example, in $\mathbb{Q}(\sqrt{5})$ we have extra integers…the element $\frac{1+\sqrt{5}}{2}$ is a root of the monic integer polynomial $x^2 - x - 1$, yet this element is not in $\mathbb{Z}[\sqrt{5}]$.

Fortunately for quadratic fields we don’t get much worse behaviour than this. If $n$ is a square-free integer then the ring of integers of $\mathbb{Q}(\sqrt{n})$ is $\mathbb{Z}\left[\frac{1+\sqrt{n}}{2}\right]$ if $n\equiv 1 \bmod 4$ and is $\mathbb{Z}[\sqrt{n}]$ if $n\equiv 3 \bmod 4$. So really this unusual behaviour only happens for $n$‘s that are $1 \bmod 4$.

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At the start of this post we remarked at the field of fractions relationship between between the integral domain $\mathbb{Z}$ and the field $\mathbb{Q}$. In general we have the same relationship between the integral domain $\mathfrak{O}_K$ and the field $K$.

It is not too difficult to show that every number field can be generated by an integer (not just an algebraic number).

Calculating the ring of integers is not an easy task in general…but there are algorithms to do it. See the book by Stewart/Tall.

In the next post we will investigate factorisation of things in the ring of integers. Unfortunately it isn’t always unique…but we will find a way to restore uniqueness (by moving to ideals). Then we will construct a finite group that measures the non-uniqueness, called the ideal class group.