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Analytic number theory is only one of many different approaches to number theory. Another important branch of the subject is algebraic number theory, which studies algebraic structures (e.g. groups, rings, and fields) of number-theoretic interest. With this perspective, the classical field of rationals ${{\bf Q}}$, and the classical ring of integers ${{\bf Z}}$, are placed inside the much larger field ${\overline{{\bf Q}}}$ of algebraic numbers, and the much larger ring ${{\mathcal A}}$ of algebraic integers, respectively. Recall that an algebraic number is a root of a polynomial with integer coefficients, and an algebraic integer is a root of a monic polynomial with integer coefficients; thus for instance ${\sqrt{2}}$ is an algebraic integer (a root of ${x^2-2}$), while ${\sqrt{2}/2}$ is merely an algebraic number (a root of ${4x^2-2}$). For the purposes of this post, we will adopt the concrete (but somewhat artificial) perspective of viewing algebraic numbers and integers as lying inside the complex numbers ${{\bf C}}$, thus ${{\mathcal A} \subset \overline{{\bf Q}} \subset {\bf C}}$. (From a modern algebraic perspective, it is better to think of ${\overline{{\bf Q}}}$ as existing as an abstract field separate from ${{\bf C}}$, but which has a number of embeddings into ${{\bf C}}$ (as well as into other fields, such as the completed p-adics ${{\bf C}_p}$), no one of which should be considered favoured over any other; cf. this mathOverflow post. But for the rudimentary algebraic number theory in this post, we will not need to work at this level of abstraction.) In particular, we identify the algebraic integer ${\sqrt{-d}}$ with the complex number ${\sqrt{d} i}$ for any natural number ${d}$.

Exercise 1 Show that the field of algebraic numbers ${\overline{{\bf Q}}}$ is indeed a field, and that the ring of algebraic integers ${{\mathcal A}}$ is indeed a ring, and is in fact an integral domain. Also, show that ${{\bf Z} = {\mathcal A} \cap {\bf Q}}$, that is to say the ordinary integers are precisely the algebraic integers that are also rational. Because of this, we will sometimes refer to elements of ${{\bf Z}}$ as rational integers.

In practice, the field ${\overline{{\bf Q}}}$ is too big to conveniently work with directly, having infinite dimension (as a vector space) over ${{\bf Q}}$. Thus, algebraic number theory generally restricts attention to intermediate fields ${{\bf Q} \subset F \subset \overline{{\bf Q}}}$ between ${{\bf Q}}$ and ${\overline{{\bf Q}}}$, which are of finite dimension over ${{\bf Q}}$; that is to say, finite degree extensions of ${{\bf Q}}$. Such fields are known as algebraic number fields, or number fields for short. Apart from ${{\bf Q}}$ itself, the simplest examples of such number fields are the quadratic fields, which have dimension exactly two over ${{\bf Q}}$.

Exercise 2 Show that if ${\alpha}$ is a rational number that is not a perfect square, then the field ${{\bf Q}(\sqrt{\alpha})}$ generated by ${{\bf Q}}$ and either of the square roots of ${\alpha}$ is a quadratic field. Conversely, show that all quadratic fields arise in this fashion. (Hint: show that every element of a quadratic field is a root of a quadratic polynomial over the rationals.)

The ring of algebraic integers ${{\mathcal A}}$ is similarly too large to conveniently work with directly, so in algebraic number theory one usually works with the rings ${{\mathcal O}_F := {\mathcal A} \cap F}$ of algebraic integers inside a given number field ${F}$. One can (and does) study this situation in great generality, but for the purposes of this post we shall restrict attention to a simple but illustrative special case, namely the quadratic fields with a certain type of negative discriminant. (The positive discriminant case will be briefly discussed in Remark 42 below.)

Exercise 3 Let ${d}$ be a square-free natural number with ${d=1\ (4)}$ or ${d=2\ (4)}$. Show that the ring ${{\mathcal O} = {\mathcal O}_{{\bf Q}(\sqrt{-d})}}$ of algebraic integers in ${{\bf Q}(\sqrt{-d})}$ is given by

$\displaystyle {\mathcal O} = {\bf Z}[\sqrt{-d}] = \{ a + b \sqrt{-d}: a,b \in {\bf Z} \}.$

If instead ${d}$ is square-free with ${d=3\ (4)}$, show that the ring ${{\mathcal O} = {\mathcal O}_{{\bf Q}(\sqrt{-d})}}$ is instead given by

$\displaystyle {\mathcal O} = {\bf Z}[\frac{1+\sqrt{-d}}{2}] = \{ a + b \frac{1+\sqrt{-d}}{2}: a,b \in {\bf Z} \}.$

What happens if ${d}$ is not square-free, or negative?

Remark 4 In the case ${d=3\ (4)}$, it may naively appear more natural to work with the ring ${{\bf Z}[\sqrt{-d}]}$, which is an index two subring of ${{\mathcal O}}$. However, because this ring only captures some of the algebraic integers in ${{\bf Q}(\sqrt{-d})}$ rather than all of them, the algebraic properties of these rings are somewhat worse than those of ${{\mathcal O}}$ (in particular, they generally fail to be Dedekind domains) and so are not convenient to work with in algebraic number theory.

We refer to fields of the form ${{\bf Q}(\sqrt{-d})}$ for natural square-free numbers ${d}$ as quadratic fields of negative discriminant, and similarly refer to ${{\mathcal O}_{{\bf Q}(\sqrt{-d})}}$ as a ring of quadratic integers of negative discriminant. Quadratic fields and quadratic integers of positive discriminant are just as important to analytic number theory as their negative discriminant counterparts, but we will restrict attention to the latter here for simplicity of discussion.
Thus, for instance, when ${d=1}$, the ring of integers in ${{\bf Q}(\sqrt{-1})}$ is the ring of Gaussian integers

$\displaystyle {\bf Z}[\sqrt{-1}] = \{ x + y \sqrt{-1}: x,y \in {\bf Z} \}$

and when ${d=3}$, the ring of integers in ${{\bf Q}(\sqrt{-3})}$ is the ring of Eisenstein integers

$\displaystyle {\bf Z}[\omega] := \{ x + y \omega: x,y \in {\bf Z} \}$

where ${\omega := e^{2\pi i /3}}$ is a cube root of unity.
As these examples illustrate, the additive structure of a ring ${{\mathcal O} = {\mathcal O}_{{\bf Q}(\sqrt{-d})}}$ of quadratic integers is that of a two-dimensional lattice in ${{\bf C}}$, which is isomorphic as an additive group to ${{\bf Z}^2}$. Thus, from an additive viewpoint, one can view quadratic integers as “two-dimensional” analogues of rational integers. From a multiplicative viewpoint, however, the quadratic integers (and more generally, integers in a number field) behave very similarly to the rational integers (as opposed to being some sort of “higher-dimensional” version of such integers). Indeed, a large part of basic algebraic number theory is devoted to treating the multiplicative theory of integers in number fields in a unified fashion, that naturally generalises the classical multiplicative theory of the rational integers.
For instance, every rational integer ${n \in {\bf Z}}$ has an absolute value ${|n| \in {\bf N} \cup \{0\}}$, with the multiplicativity property ${|nm| = |n| |m|}$ for ${n,m \in {\bf Z}}$, and the positivity property ${|n| > 0}$ for all ${n \neq 0}$. Among other things, the absolute value detects units: ${|n| = 1}$ if and only if ${n}$ is a unit in ${{\bf Z}}$ (that is to say, it is multiplicatively invertible in ${{\bf Z}}$). Similarly, in any ring of quadratic integers ${{\mathcal O} = {\mathcal O}_{{\bf Q}(\sqrt{-d})}}$ with negative discriminant, we can assign a norm ${N(n) \in {\bf N} \cup \{0\}}$ to any quadratic integer ${n \in {\mathcal O}_{{\bf Q}(\sqrt{-d})}}$ by the formula

$\displaystyle N(n) = n \overline{n}$

where ${\overline{n}}$ is the complex conjugate of ${n}$. (When working with other number fields than quadratic fields of negative discriminant, one instead defines ${N(n)}$ to be the product of all the Galois conjugates of ${n}$.) Thus for instance, when ${d=1,2\ (4)}$ one has

$\displaystyle N(x + y \sqrt{-d}) = x^2 + dy^2 \ \ \ \ \ (1)$

and when ${d=3\ (4)}$ one has

$\displaystyle N(x + y \frac{1+\sqrt{-d}}{2}) = x^2 + xy + \frac{d+1}{4} y^2. \ \ \ \ \ (2)$

Analogously to the rational integers, we have the multiplicativity property ${N(nm) = N(n) N(m)}$ for ${n,m \in {\mathcal O}}$ and the positivity property ${N(n) > 0}$ for ${n \neq 0}$, and the units in ${{\mathcal O}}$ are precisely the elements of norm one.

Exercise 5 Establish the three claims of the previous paragraph. Conclude that the units (invertible elements) of ${{\mathcal O}}$ consist of the four elements ${\pm 1, \pm i}$ if ${d=1}$, the six elements ${\pm 1, \pm \omega, \pm \omega^2}$ if ${d=3}$, and the two elements ${\pm 1}$ if ${d \neq 1,3}$.

For the rational integers, we of course have the fundamental theorem of arithmetic, which asserts that every non-zero rational integer can be uniquely factored (up to permutation and units) as the product of irreducible integers, that is to say non-zero, non-unit integers that cannot be factored into the product of integers of strictly smaller norm. As it turns out, the same claim is true for a few additional rings of quadratic integers, such as the Gaussian integers and Eisenstein integers, but fails in general; for instance, in the ring ${{\bf Z}[\sqrt{-5}]}$, we have the famous counterexample

$\displaystyle 6 = 2 \times 3 = (1+\sqrt{-5}) (1-\sqrt{-5})$

that decomposes ${6}$ non-uniquely into the product of irreducibles in ${{\bf Z}[\sqrt{-5}]}$. Nevertheless, it is an important fact that the fundamental theorem of arithmetic can be salvaged if one uses an “idealised” notion of a number in a ring of integers ${{\mathcal O}}$, now known in modern language as an ideal of that ring. For instance, in ${{\bf Z}[\sqrt{-5}]}$, the principal ideal ${(6)}$ turns out to uniquely factor into the product of (non-principal) ideals ${(2) + (1+\sqrt{-5}), (2) + (1-\sqrt{-5}), (3) + (1+\sqrt{-5}), (3) + (1-\sqrt{-5})}$; see Exercise 27. We will review the basic theory of ideals in number fields (focusing primarily on quadratic fields of negative discriminant) below the fold.
The norm forms (1), (2) can be viewed as examples of positive definite quadratic forms ${Q: {\bf Z}^2 \rightarrow {\bf Z}}$ over the integers, by which we mean a polynomial of the form

$\displaystyle Q(x,y) = ax^2 + bxy + cy^2$

for some integer coefficients ${a,b,c}$. One can declare two quadratic forms ${Q, Q': {\bf Z}^2 \rightarrow {\bf Z}}$ to be equivalent if one can transform one to the other by an invertible linear transformation ${T: {\bf Z}^2 \rightarrow {\bf Z}^2}$, so that ${Q' = Q \circ T}$. For example, the quadratic forms ${(x,y) \mapsto x^2 + y^2}$ and ${(x',y') \mapsto 2 (x')^2 + 2 x' y' + (y')^2}$ are equivalent, as can be seen by using the invertible linear transformation ${(x,y) = (x',x'+y')}$. Such equivalences correspond to the different choices of basis available when expressing a ring such as ${{\mathcal O}}$ (or an ideal thereof) additively as a copy of ${{\bf Z}^2}$.
There is an important and classical invariant of a quadratic form ${(x,y) \mapsto ax^2 + bxy + c y^2}$, namely the discriminant ${\Delta := b^2 - 4ac}$, which will of course be familiar to most readers via the quadratic formula, which among other things tells us that a quadratic form will be positive definite precisely when its discriminant is negative. It is not difficult (particularly if one exploits the multiplicativity of the determinant of ${2 \times 2}$ matrices) to show that two equivalent quadratic forms have the same discriminant. Thus for instance any quadratic form equivalent to (1) has discriminant ${-4d}$, while any quadratic form equivalent to (2) has discriminant ${-d}$. Thus we see that each ring ${{\mathcal O}[\sqrt{-d}]}$ of quadratic integers is associated with a certain negative discriminant ${D}$, defined to equal ${-4d}$ when ${d=1,2\ (4)}$ and ${-d}$ when ${d=3\ (4)}$.

Exercise 6 (Geometric interpretation of discriminant) Let ${Q: {\bf Z}^2 \rightarrow {\bf Z}}$ be a quadratic form of negative discriminant ${D}$, and extend it to a real form ${Q: {\bf R}^2 \rightarrow {\bf R}}$ in the obvious fashion. Show that for any ${X>0}$, the set ${\{ (x,y) \in {\bf R}^2: Q(x,y) \leq X \}}$ is an ellipse of area ${2\pi X / \sqrt{|D|}}$.

It is natural to ask the converse question: if two quadratic forms have the same discriminant, are they necessarily equivalent? For certain choices of discriminant, this is the case:

Exercise 7 Show that any quadratic form ${ax^2+bxy+cy^2}$ of discriminant ${-4}$ is equivalent to the form ${x^2+y^2}$, and any quadratic form of discriminant ${-3}$ is equivalent to ${x^2+xy+y^2}$. (Hint: use elementary transformations to try to make ${|b|}$ as small as possible, to the point where one only has to check a finite number of cases; this argument is due to Legendre.) More generally, show that for any negative discriminant ${D}$, there are only finitely many quadratic forms of that discriminant up to equivalence (a result first established by Gauss).

Unfortunately, for most choices of discriminant, the converse question fails; for instance, the quadratic forms ${x^2+5y^2}$ and ${2x^2+2xy+3y^2}$ both have discriminant ${-20}$, but are not equivalent (Exercise 38). This particular failure of equivalence turns out to be intimately related to the failure of unique factorisation in the ring ${{\bf Z}[\sqrt{-5}]}$.
It turns out that there is a fundamental connection between quadratic fields, equivalence classes of quadratic forms of a given discriminant, and real Dirichlet characters, thus connecting the material discussed above with the last section of the previous set of notes. Here is a typical instance of this connection:

Proposition 8 Let ${\chi_4: {\bf N} \rightarrow {\bf R}}$ be the real non-principal Dirichlet character of modulus ${4}$, or more explicitly ${\chi_4(n)}$ is equal to ${+1}$ when ${n = 1\ (4)}$, ${-1}$ when ${n = 3\ (4)}$, and ${0}$ when ${n = 0,2\ (4)}$.

• (i) For any natural number ${n}$, the number of Gaussian integers ${m \in {\bf Z}[\sqrt{-1}]}$ with norm ${N(m)=n}$ is equal to ${4(1 * \chi_4)(n)}$. Equivalently, the number of solutions to the equation ${n = x^2+y^2}$ with ${x,y \in{\bf Z}}$ is ${4(1*\chi_4)(n)}$. (Here, as in the previous post, the symbol ${*}$ denotes Dirichlet convolution.)
• (ii) For any natural number ${n}$, the number of Gaussian integers ${m \in {\bf Z}[\sqrt{-1}]}$ that divide ${n}$ (thus ${n = dm}$ for some ${d \in {\bf Z}[\sqrt{-1}]}$) is ${4(1*1*1*\mu\chi_4)(n)}$.

We will prove this proposition later in these notes. We observe that as a special case of part (i) of this proposition, we recover the Fermat two-square theorem: an odd prime ${p}$ is expressible as the sum of two squares if and only if ${p = 1\ (4)}$. This proposition should also be compared with the fact, used crucially in the previous post to prove Dirichlet’s theorem, that ${1*\chi(n)}$ is non-negative for any ${n}$, and at least one when ${n}$ is a square, for any quadratic character ${\chi}$.
As an illustration of the relevance of such connections to analytic number theory, let us now explicitly compute ${L(1,\chi_4)}$.

Corollary 9 ${L(1,\chi_4) = \frac{\pi}{4}}$.

This particular identity is also known as the Leibniz formula.
Proof: For a large number ${x}$, consider the quantity

$\displaystyle \sum_{n \in {\bf Z}[\sqrt{-1}]: N(n) \leq x} 1$

of all the Gaussian integers of norm less than ${x}$. On the one hand, this is the same as the number of lattice points of ${{\bf Z}^2}$ in the disk ${\{ (a,b) \in {\bf R}^2: a^2+b^2 \leq x \}}$ of radius ${\sqrt{x}}$. Placing a unit square centred at each such lattice point, we obtain a region which differs from the disk by a region contained in an annulus of area ${O(\sqrt{x})}$. As the area of the disk is ${\pi x}$, we conclude the Gauss bound

$\displaystyle \sum_{n \in {\bf Z}[\sqrt{-1}]: N(n) \leq x} 1 = \pi x + O(\sqrt{x}).$

On the other hand, by Proposition 8(i) (and removing the ${n=0}$ contribution), we see that

$\displaystyle \sum_{n \in {\bf Z}[\sqrt{-1}]: N(n) \leq x} 1 = 1 + 4 \sum_{n \leq x} 1 * \chi_4(n).$

Now we use the Dirichlet hyperbola method to expand the right-hand side sum, first expressing

$\displaystyle \sum_{n \leq x} 1 * \chi_4(n) = \sum_{d \leq \sqrt{x}} \chi_4(d) \sum_{m \leq x/d} 1 + \sum_{m \leq \sqrt{x}} \sum_{d \leq x/m} \chi_4(d)$

$\displaystyle - (\sum_{d \leq \sqrt{x}} \chi_4(d)) (\sum_{m \leq \sqrt{x}} 1)$

and then using the bounds ${\sum_{d \leq y} \chi_4(d) = O(1)}$, ${\sum_{m \leq y} 1 = y + O(1)}$, ${\sum_{d \leq \sqrt{x}} \frac{\chi_4(d)}{d} = L(1,\chi_4) + O(\frac{1}{\sqrt{x}})}$ from the previous set of notes to conclude that

$\displaystyle \sum_{n \leq x} 1 * \chi_4(n) = x L(1,\chi_4) + O(\sqrt{x}).$

Comparing the two formulae for ${\sum_{n \in {\bf Z}[\sqrt{-1}]: N(n) \leq x} 1}$ and sending ${x \rightarrow \infty}$, we obtain the claim. $\Box$

Exercise 10 Give an alternate proof of Corollary 9 that relies on obtaining asymptotics for the Dirichlet series ${\sum_{n \in {\bf Z}} \frac{1 * \chi_4(n)}{n^s}}$ as ${s \rightarrow 1^+}$, rather than using the Dirichlet hyperbola method.

Exercise 11 Give a direct proof of Corollary 9 that does not use Proposition 8, instead using Taylor expansion of the complex logarithm ${\log(1+z)}$. (One can also use Taylor expansions of some other functions related to the complex logarithm here, such as the arctangent function.)

More generally, one can relate ${L(1,\chi)}$ for a real Dirichlet character ${\chi}$ with the number of inequivalent quadratic forms of a certain discriminant, via the famous class number formula; we will give a special case of this formula below the fold.
The material here is only a very rudimentary introduction to algebraic number theory, and is not essential to the rest of the course. A slightly expanded version of the material here, from the perspective of analytic number theory, may be found in Sections 5 and 6 of Davenport’s book. A more in-depth treatment of algebraic number theory may be found in a number of texts, e.g. Fröhlich and Taylor.
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