# Algebraic number theory

In mathematics, algebraic number theory is the study of algebraic numbers and structures involving them, especially algebraic number fields.

# Introduction

Algebraic number theory is a branch of number theory that, in a nutshell, extends various properties of the integers to more general rings and fields. In doing so, many questions concerning Diophantine equations are resolved, including the celebrated quadratic reciprocity theorem. More recently, the field has been linked to the theory of elliptic curve, and its ideas are responsible for the successful attack on Fermat's Last Theorem. This field is extremely rich and advanced (indeed, prerequisites to courses in this topic often include first-year graduate school material), and this article gives no more than a brief introduction.

# The basics

## Motivation

Interest in the field was largely motivated by the desire to understand numbers of the form $x^2+ny^2$ for some fixed, squarefree $n$ (if $n=x^2y$ is not squarefree, the case is equivalent to $n=y$). More specifically, the question at hand was what numbers could be expressed in that form. Since $k=x^2+ny^2 \implies x^2 \equiv k \pmod n$, there was a natural connection to quadratic residues in play. In the ring of integers, this is a difficult question to analyze, but it becomes much easier when considering the field $Q[-\sqrt{n}]$ because then $x^2+ny^2=(x+y\sqrt{-n})(x-y\sqrt{-n})$ is the product of two elements - and the norm of one. The case of $n=1$, in particular, leads to factorization in the Gaussian integers.

This extension of the concept of factorization relates to many other problems as well, most notably Fermat's Last Theorem. In the $n=3$ case, a simple solution exists by taking $y^3=z^3-x^3=(z-x)(z^2+xz+z^2)$ and using an infinite descent argument. However, this does not extend well to larger $n$ because analogous factorizations contain terms of degree larger than 2. Early attempts at proof factored $y^p=z^p-x^p$ into $y^p=\prod_{i=0}^{p-1}(z-\omega^ix)$, where $\omega$ is a $p$th root of unity. However, though this was not well-understood at the time, the implicit assumption was that $\mathbb{Z}[\omega]$ was a principal ideal domain, later shown to be false.

## Algebraic Numbers

An algebraic number is a number that is the root of some nonzero integer polynomial (i.e. a polynomial with integer coefficients). When that polynomial is monic, the number is said to be an algebraic integer. For example, all rational numbers are algebraic integers (and thus an algebraic number as well), as the linear polynomial $nx-m$ has root $\frac{m}{n}$ for any integers $m,n$ (with $n \neq 0$). When such a polynomial exists, it is called the minimal polynomial of the algebraic number in question. It can be shown that a number is an algebraic integer if and only if its minimal polynomial has integer coefficients.

The sum, difference, product, and quotient of any two algebraic numbers is itself an algebraic number; as a result, the algebraic numbers form a field. In this article, $K$ will denote an arbitrary algebraic number field; for example, $\mathbb{Q}[\sqrt{-5}]$, which consists of the numbers of the form $a+b\sqrt{-5}$ where $a,b$ are rational. As a sidenote, this shows that the sum of any two algebraic integers is itself an algebraic integer, and furthermore any rational algebraic integer is obviously also an integer. This gives an easy way to show that sums similar to $\sqrt{2}+\sqrt{3}+\sqrt{5}+\sqrt{7}$ are irrational - as all of these terms are algebraic integers (they are roots of $x^2-k$ for $k=2,3,5,7$), the sum is an algebraic integer as well, and so must be an integer if rational. But any old approximation is sufficient to determine that this sum is not an integer, hence it is irrational.

## Unique factorization

In the ring of integers, all numbers have unique factorizations by the Fundamental theorem of arithmetic, up to multiplication by the units 1 and -1. We will extend this notion to an integral domain $A$; in other words, a commutative ring in which the product of two nonzero elements is nonzero. An element $a \in A$ is a unit if $a$ is invertible in $A$; i.e. there exists an inverse $b \in A$ such that $ab=ba=1$ (where 1 is the multiplicative identity). An element $p \in A$ is prime if it is not zero, not a unit, and $p \mid ab \implies p \mid a$ or $p \mid b$. In a principal ideal domain, any element $a \in A$ can be unique factored as the product of primes, up to order and multiplication by units. The first order of business is to explore when unique factorization holds.

Of course, we first need to define the term factorization. An element $a \in A$ is irreducible if it is not a unit and cannot be written as the product of two nonunits; obviously, primes are irreducible (but not necessarily vice versa). A factorization of an element is its expression as a product of irreducible elements, and a ring is a UFD if this factorization is unique (up to order and multiplication by units).

As previously mentioned, $K$ is an arbitrary algebraic number field; as a field, factorization only makes sense in the presence of a subring. Fortunately, can be shown that the algebraic integers form a subring of $K$; however, though it can be shown that algebraic integers can always be factored, they do not generally form a UFD. For example, if $K=\mathbb{Q}[\sqrt{-5}]$, we have $$6=2 \cdot 3=(1+\sqrt{-5})(1-\sqrt{-5})$$ and so the factorization is not unique.

It can be shown that unique factorization occurs when irreducibles are necessary primes. When this is not the case, the concept can be largely recovered through the use of ideals. The general idea is to consider entities that "divide" irreducibles, for the purpose of recovering unique factorization. This may sound contrived, but in fact it is a very important idea. In the above example, $$6=2 \cdot 3=(1+\sqrt{-5})(1-\sqrt{-5})$$ we define ideals $\mathfrak{a}, \mathfrak{b}, \mathfrak{c}, \mathfrak{d}$ such that $$6=(\mathfrak{a} \cdot \mathfrak{b}) \cdot (\mathfrak{c} \cdot \mathfrak{d})=(\mathfrak{a} \cdot \mathfrak{c}) \cdot (\mathfrak{b} \cdot \mathfrak{d})$$ which is not unlike writing $210=6 \cdot 35=14 \cdot 15$ (or $210=(2 \cdot 3) \cdot (5 \cdot 7)=(2 \cdot 7) \cdot (3 \cdot 5)$). The usual divisibility rules - namely $\mathfrak{a} \mid a \implies \mathfrak{a} \mid ab$, $\mathfrak{a} \mid a, \mathfrak{a} \mid b \implies \mathfrak{a} \mid a+b$, $\mathfrak{a} \mid 0$ - still hold for ideals. When we define $\mathfrak{a}, \mathfrak{b}$ by the set of irreducibles they divide, we can extend the notion to multiplication as well: $$\mathfrak{a}\mathfrak{b}=\{a_nb_m : \mathfrak{a}\mid a_n, \mathfrak{b} \mid b_m\}$$ this recovers unique factorization, as we can write $a=\mathfrak{a}\mathfrak{b}\mathfrak{c}\hdots$ for all $a \in A$. When we have to resort to this definition, $K$ is a principal ideal domain.

## The norm map

We skated over a key detail in the above section: how do we know that $2, 3, 1+\sqrt{-5}$, and $1-\sqrt{-5}$ are themselves irreducibles? To show this, we use the norm map $N:\mathbb{Q}[\sqrt{-5}] \rightarrow \mathbb{Q}$, $N(a+b\sqrt{-5})=a^2+5b^2$. It can be shown that the norm is multiplicative, and so if $1+\sqrt{-5}=ab$ for some $a,b \in K$, we have $N(ab)=N(a)N(b)=6$. We can immediately discard the cases where $N(a)=1$, as this implies $a\overline{a}=1 \implies a$ is a unit, and we can easily verify that the other two cases do not occur. There is one final detail to process: we must show that no two of these irreducibles are associates; i.e. they do not differ only by units. Fortunately, it is easy to verify that $1+\sqrt{-5}=(a+b\sqrt{-5})(1-\sqrt{-5})$ has no solutions in $a,b \in \mathbb{Z}$.

# Major results

The discovery of Quadratic reciprocity was an early success of this research, stating the impressive formula $$\left(\frac{p}{q}\right)\left(\frac{q}{p}\right)=(-1)^{\frac{p-1}{2}\frac{q-1}{2}}$$ where $\left(\frac{p}{q}\right)$ is the Legendre symbol, equal to 1 if $p$ is a quadratic residue modulo $q$ and -1 otherwise (or 0 when $p \equiv 0 \pmod q$, but this clearly does not occur for primes). In other words, unless $p \equiv q \equiv 3 \pmod 4$, $p$ is a quadratic residue modulo $q$ if and only if $q$ is a quadratic residue modulo $p$.