You are currently browsing the tag archive for the ‘Dirichlet character’ tag.

A fundamental and recurring problem in analytic number theory is to demonstrate the presence of cancellation in an oscillating sum, a typical example of which might be a correlation

$\displaystyle \sum_{n} f(n) \overline{g(n)} \ \ \ \ \ (1)$

between two arithmetic functions ${f: {\bf N} \rightarrow {\bf C}}$ and ${g: {\bf N} \rightarrow {\bf C}}$, which to avoid technicalities we will assume to be finitely supported (or that the ${n}$ variable is localised to a finite range, such as ${\{ n: n \leq x \}}$). A key example to keep in mind for the purposes of this set of notes is the twisted von Mangoldt summatory function

$\displaystyle \sum_{n \leq x} \Lambda(n) \overline{\chi(n)} \ \ \ \ \ (2)$

that measures the correlation between the primes and a Dirichlet character ${\chi}$. One can get a “trivial” bound on such sums from the triangle inequality

$\displaystyle |\sum_{n} f(n) \overline{g(n)}| \leq \sum_{n} |f(n)| |g(n)|;$

for instance, from the triangle inequality and the prime number theorem we have

$\displaystyle |\sum_{n \leq x} \Lambda(n) \overline{\chi(n)}| \leq x + o(x) \ \ \ \ \ (3)$

as ${x \rightarrow \infty}$. But the triangle inequality is insensitive to the phase oscillations of the summands, and often we expect (e.g. from the probabilistic heuristics from Supplement 4) to be able to improve upon the trivial triangle inequality bound by a substantial amount; in the best case scenario, one typically expects a “square root cancellation” that gains a factor that is roughly the square root of the number of summands. (For instance, for Dirichlet characters ${\chi}$ of conductor ${O(x^{O(1)})}$, it is expected from probabilistic heuristics that the left-hand side of (3) should in fact be ${O_\varepsilon(x^{1/2+\varepsilon})}$ for any ${\varepsilon>0}$.)

It has proven surprisingly difficult, however, to establish significant cancellation in many of the sums of interest in analytic number theory, particularly if the sums do not have a strong amount of algebraic structure (e.g. multiplicative structure) which allow for the deployment of specialised techniques (such as multiplicative number theory techniques). In fact, we are forced to rely (to an embarrassingly large extent) on (many variations of) a single basic tool to capture at least some cancellation, namely the Cauchy-Schwarz inequality. In fact, in many cases the classical case

$\displaystyle |\sum_n f(n) \overline{g(n)}| \leq (\sum_n |f(n)|^2)^{1/2} (\sum_n |g(n)|^2)^{1/2}, \ \ \ \ \ (4)$

considered by Cauchy, where at least one of ${f, g: {\bf N} \rightarrow {\bf C}}$ is finitely supported, suffices for applications. Roughly speaking, the Cauchy-Schwarz inequality replaces the task of estimating a cross-correlation between two different functions ${f,g}$, to that of measuring self-correlations between ${f}$ and itself, or ${g}$ and itself, which are usually easier to compute (albeit at the cost of capturing less cancellation). Note that the Cauchy-Schwarz inequality requires almost no hypotheses on the functions ${f}$ or ${g}$, making it a very widely applicable tool.

There is however some skill required to decide exactly how to deploy the Cauchy-Schwarz inequality (and in particular, how to select ${f}$ and ${g}$); if applied blindly, one loses all cancellation and can even end up with a worse estimate than the trivial bound. For instance, if one tries to bound (2) directly by applying Cauchy-Schwarz with the functions ${\Lambda}$ and ${\chi}$, one obtains the bound

$\displaystyle |\sum_{n \leq x} \Lambda(n) \overline{\chi(n)}| \leq (\sum_{n \leq x} \Lambda(n)^2)^{1/2} (\sum_{n \leq x} |\chi(n)|^2)^{1/2}.$

The right-hand side may be bounded by ${\ll x \log^{1/2} x}$, but this is worse than the trivial bound (3) by a logarithmic factor. This can be “blamed” on the fact that ${\Lambda}$ and ${\chi}$ are concentrated on rather different sets (${\Lambda}$ is concentrated on primes, while ${\chi}$ is more or less uniformly distributed amongst the natural numbers); but even if one corrects for this (e.g. by weighting Cauchy-Schwarz with some suitable “sieve weight” that is more concentrated on primes), one still does not do any better than (3). Indeed, the Cauchy-Schwarz inequality suffers from the same key weakness as the triangle inequality: it is insensitive to the phase oscillation of the factors ${f, g}$.

While the Cauchy-Schwarz inequality can be poor at estimating a single correlation such as (1), its power improves when considering an average (or sum, or square sum) of multiple correlations. In this set of notes, we will focus on one such situation of this type, namely that of trying to estimate a square sum

$\displaystyle (\sum_{j=1}^J |\sum_{n} f(n) \overline{g_j(n)}|^2)^{1/2} \ \ \ \ \ (5)$

that measures the correlations of a single function ${f: {\bf N} \rightarrow {\bf C}}$ with multiple other functions ${g_j: {\bf N} \rightarrow {\bf C}}$. One should think of the situation in which ${f}$ is a “complicated” function, such as the von Mangoldt function ${\Lambda}$, but the ${g_j}$ are relatively “simple” functions, such as Dirichlet characters. In the case when the ${g_j}$ are orthonormal functions, we of course have the classical Bessel inequality:

Lemma 1 (Bessel inequality) Let ${g_1,\dots,g_J: {\bf N} \rightarrow {\bf C}}$ be finitely supported functions obeying the orthonormality relationship

$\displaystyle \sum_n g_j(n) \overline{g_{j'}(n)} = 1_{j=j'}$

for all ${1 \leq j,j' \leq J}$. Then for any function ${f: {\bf N} \rightarrow {\bf C}}$, we have

$\displaystyle (\sum_{j=1}^J |\sum_{n} f(n) \overline{g_j(n)}|^2)^{1/2} \leq (\sum_n |f(n)|^2)^{1/2}.$

For sake of comparison, if one were to apply the Cauchy-Schwarz inequality (4) separately to each summand in (5), one would obtain the bound of ${J^{1/2} (\sum_n |f(n)|^2)^{1/2}}$, which is significantly inferior to the Bessel bound when ${J}$ is large. Geometrically, what is going on is this: the Cauchy-Schwarz inequality (4) is only close to sharp when ${f}$ and ${g}$ are close to parallel in the Hilbert space ${\ell^2({\bf N})}$. But if ${g_1,\dots,g_J}$ are orthonormal, then it is not possible for any other vector ${f}$ to be simultaneously close to parallel to too many of these orthonormal vectors, and so the inner products of ${f}$ with most of the ${g_j}$ should be small. (See this previous blog post for more discussion of this principle.) One can view the Bessel inequality as formalising a repulsion principle: if ${f}$ correlates too much with some of the ${g_j}$, then it does not have enough “energy” to have large correlation with the rest of the ${g_j}$.

In analytic number theory applications, it is useful to generalise the Bessel inequality to the situation in which the ${g_j}$ are not necessarily orthonormal. This can be accomplished via the Cauchy-Schwarz inequality:

Proposition 2 (Generalised Bessel inequality) Let ${g_1,\dots,g_J: {\bf N} \rightarrow {\bf C}}$ be finitely supported functions, and let ${\nu: {\bf N} \rightarrow {\bf R}^+}$ be a non-negative function. Let ${f: {\bf N} \rightarrow {\bf C}}$ be such that ${f}$ vanishes whenever ${\nu}$ vanishes, we have

$\displaystyle (\sum_{j=1}^J |\sum_{n} f(n) \overline{g_j(n)}|^2)^{1/2} \leq (\sum_n |f(n)|^2 / \nu(n))^{1/2} \ \ \ \ \ (6)$

$\displaystyle \times ( \sum_{j=1}^J \sum_{j'=1}^J c_j \overline{c_{j'}} \sum_n \nu(n) g_j(n) \overline{g_{j'}(n)} )^{1/2}$

for some sequence ${c_1,\dots,c_J}$ of complex numbers with ${\sum_{j=1}^J |c_j|^2 = 1}$, with the convention that ${|f(n)|^2/\nu(n)}$ vanishes whenever ${f(n), \nu(n)}$ both vanish.

Note by relabeling that we may replace the domain ${{\bf N}}$ here by any other at most countable set, such as the integers ${{\bf Z}}$. (Indeed, one can give an analogue of this lemma on arbitrary measure spaces, but we will not do so here.) This result first appears in this paper of Boas.

Proof: We use the method of duality to replace the role of the function ${f}$ by a dual sequence ${c_1,\dots,c_J}$. By the converse to Cauchy-Schwarz, we may write the left-hand side of (6) as

$\displaystyle \sum_{j=1}^J \overline{c_j} \sum_{n} f(n) \overline{g_j(n)}$

for some complex numbers ${c_1,\dots,c_J}$ with ${\sum_{j=1}^J |c_j|^2 = 1}$. Indeed, if all of the ${\sum_{n} f(n) \overline{g_j(n)}}$ vanish, we can set the ${c_j}$ arbitrarily, otherwise we set ${(c_1,\dots,c_J)}$ to be the unit vector formed by dividing ${(\sum_{n} f(n) \overline{g_j(n)})_{j=1}^J}$ by its length. We can then rearrange this expression as

$\displaystyle \sum_n f(n) \overline{\sum_{j=1}^J c_j g_j(n)}.$

Applying Cauchy-Schwarz (dividing the first factor by ${\nu(n)^{1/2}}$ and multiplying the second by ${\nu(n)^{1/2}}$, after first removing those ${n}$ for which ${\nu(n)}$ vanish), this is bounded by

$\displaystyle (\sum_n |f(n)|^2 / \nu(n))^{1/2} (\sum_n \nu(n) |\sum_{j=1}^J c_j g_j(n)|^2)^{1/2},$

and the claim follows by expanding out the second factor. $\Box$

Observe that Lemma 1 is a special case of Proposition 2 when ${\nu=1}$ and the ${g_j}$ are orthonormal. In general, one can expect Proposition 2 to be useful when the ${g_j}$ are almost orthogonal relative to ${\nu}$, in that the correlations ${\sum_n \nu(n) g_j(n) \overline{g_{j'}(n)}}$ tend to be small when ${j,j'}$ are distinct. In that case, one can hope for the diagonal term ${j=j'}$ in the right-hand side of (6) to dominate, in which case one can obtain estimates of comparable strength to the classical Bessel inequality. The flexibility to choose different weights ${\nu}$ in the above proposition has some technical advantages; for instance, if ${f}$ is concentrated in a sparse set (such as the primes), it is sometimes useful to tailor ${\nu}$ to a comparable set (e.g. the almost primes) in order not to lose too much in the first factor ${\sum_n |f(n)|^2 / \nu(n)}$. Also, it can be useful to choose a fairly “smooth” weight ${\nu}$, in order to make the weighted correlations ${\sum_n \nu(n) g_j(n) \overline{g_{j'}(n)}}$ small.

Remark 3 In harmonic analysis, the use of tools such as Proposition 2 is known as the method of almost orthogonality, or the ${TT^*}$ method. The explanation for the latter name is as follows. For sake of exposition, suppose that ${\nu}$ is never zero (or we remove all ${n}$ from the domain for which ${\nu(n)}$ vanishes). Given a family of finitely supported functions ${g_1,\dots,g_J: {\bf N} \rightarrow {\bf C}}$, consider the linear operator ${T: \ell^2(\nu^{-1}) \rightarrow \ell^2(\{1,\dots,J\})}$ defined by the formula

$\displaystyle T f := ( \sum_{n} f(n) \overline{g_j(n)} )_{j=1}^J.$

This is a bounded linear operator, and the left-hand side of (6) is nothing other than the ${\ell^2(\{1,\dots,J\})}$ norm of ${Tf}$. Without any further information on the function ${f}$ other than its ${\ell^2(\nu^{-1})}$ norm ${(\sum_n |f(n)|^2 / \nu(n))^{1/2}}$, the best estimate one can obtain on (6) here is clearly

$\displaystyle (\sum_n |f(n)|^2 / \nu(n))^{1/2} \times \|T\|_{op},$

where ${\|T\|_{op}}$ denotes the operator norm of ${T}$.

The adjoint ${T^*: \ell^2(\{1,\dots,J\}) \rightarrow \ell^2(\nu^{-1})}$ is easily computed to be

$\displaystyle T^* (c_j)_{j=1}^J := (\sum_{j=1}^J c_j \nu(n) g_j(n) )_{n \in {\bf N}}.$

The composition ${TT^*: \ell^2(\{1,\dots,J\}) \rightarrow \ell^2(\{1,\dots,J\})}$ of ${T}$ and its adjoint is then given by

$\displaystyle TT^* (c_j)_{j=1}^J := (\sum_{j=1}^J c_j \sum_n \nu(n) g_j(n) \overline{g_{j'}}(n) )_{j=1}^J.$

From the spectral theorem (or singular value decomposition), one sees that the operator norms of ${T}$ and ${TT^*}$ are related by the identity

$\displaystyle \|T\|_{op} = \|TT^*\|_{op}^{1/2},$

and as ${TT^*}$ is a self-adjoint, positive semi-definite operator, the operator norm ${\|TT^*\|_{op}}$ is also the supremum of the quantity

$\displaystyle \langle TT^* (c_j)_{j=1}^J, (c_j)_{j=1}^J \rangle_{\ell^2(\{1,\dots,J\})} = \sum_{j=1}^J \sum_{j'=1}^J c_j \overline{c_{j'}} \sum_n \nu(n) g_j(n) \overline{g_{j'}(n)}$

where ${(c_j)_{j=1}^J}$ ranges over unit vectors in ${\ell^2(\{1,\dots,J\})}$. Putting these facts together, we obtain Proposition 2; furthermore, we see from this analysis that the bound here is essentially optimal if the only information one is allowed to use about ${f}$ is its ${\ell^2(\nu^{-1})}$ norm.

For further discussion of almost orthogonality methods from a harmonic analysis perspective, see Chapter VII of this text of Stein.

Exercise 4 Under the same hypotheses as Proposition 2, show that

$\displaystyle \sum_{j=1}^J |\sum_{n} f(n) \overline{g_j(n)}| \leq (\sum_n |f(n)|^2 / \nu(n))^{1/2}$

$\displaystyle \times ( \sum_{j=1}^J \sum_{j'=1}^J |\sum_n \nu(n) g_j(n) \overline{g_{j'}(n)}| )^{1/2}$

as well as the variant inequality

$\displaystyle |\sum_{j=1}^J \sum_{n} f(n) \overline{g_j(n)}| \leq (\sum_n |f(n)|^2 / \nu(n))^{1/2}$

$\displaystyle \times | \sum_{j=1}^J \sum_{j'=1}^J \sum_n \nu(n) g_j(n) \overline{g_{j'}(n)}|^{1/2}.$

Proposition 2 has many applications in analytic number theory; for instance, we will use it in later notes to control the large value of Dirichlet series such as the Riemann zeta function. One of the key benefits is that it largely eliminates the need to consider further correlations of the function ${f}$ (other than its self-correlation ${\sum_n |f(n)|^2 / \nu(n)}$ relative to ${\nu^{-1}}$, which is usually fairly easy to compute or estimate as ${\nu}$ is usually chosen to be relatively simple); this is particularly useful if ${f}$ is a function which is significantly more complicated to analyse than the functions ${g_j}$. Of course, the tradeoff for this is that one now has to deal with the coefficients ${c_j}$, which if anything are even less understood than ${f}$, since literally the only thing we know about these coefficients is their square sum ${\sum_{j=1}^J |c_j|^2}$. However, as long as there is enough almost orthogonality between the ${g_j}$, one can estimate the ${c_j}$ by fairly crude estimates (e.g. triangle inequality or Cauchy-Schwarz) and still get reasonably good estimates.

In this set of notes, we will use Proposition 2 to prove some versions of the large sieve inequality, which controls a square-sum of correlations

$\displaystyle \sum_n f(n) e( -\xi_j n )$

of an arbitrary finitely supported function ${f: {\bf Z} \rightarrow {\bf C}}$ with various additive characters ${n \mapsto e( \xi_j n)}$ (where ${e(x) := e^{2\pi i x}}$), or alternatively a square-sum of correlations

$\displaystyle \sum_n f(n) \overline{\chi_j(n)}$

of ${f}$ with various primitive Dirichlet characters ${\chi_j}$; it turns out that one can prove a (slightly sub-optimal) version of this inequality quite quickly from Proposition 2 if one first prepares the sum by inserting a smooth cutoff with well-behaved Fourier transform. The large sieve inequality has many applications (as the name suggests, it has particular utility within sieve theory). For the purposes of this set of notes, though, the main application we will need it for is the Bombieri-Vinogradov theorem, which in a very rough sense gives a prime number theorem in arithmetic progressions, which, “on average”, is of strength comparable to the results provided by the Generalised Riemann Hypothesis (GRH), but has the great advantage of being unconditional (it does not require any unproven hypotheses such as GRH); it can be viewed as a significant extension of the Siegel-Walfisz theorem from Notes 2. As we shall see in later notes, the Bombieri-Vinogradov theorem is a very useful ingredient in sieve-theoretic problems involving the primes.

There is however one additional important trick, beyond the large sieve, which we will need in order to establish the Bombieri-Vinogradov theorem. As it turns out, after some basic manipulations (and the deployment of some multiplicative number theory, and specifically the Siegel-Walfisz theorem), the task of proving the Bombieri-Vinogradov theorem is reduced to that of getting a good estimate on sums that are roughly of the form

$\displaystyle \sum_{j=1}^J |\sum_n \Lambda(n) \overline{\chi_j}(n)| \ \ \ \ \ (7)$

for some primitive Dirichlet characters ${\chi_j}$. This looks like the type of sum that can be controlled by the large sieve (or by Proposition 2), except that this is an ordinary sum rather than a square sum (i.e., an ${\ell^1}$ norm rather than an ${\ell^2}$ norm). One could of course try to control such a sum in terms of the associated square-sum through the Cauchy-Schwarz inequality, but this turns out to be very wasteful (it loses a factor of about ${J^{1/2}}$). Instead, one should try to exploit the special structure of the von Mangoldt function ${\Lambda}$, in particular the fact that it can be expressible as a Dirichlet convolution ${\alpha * \beta}$ of two further arithmetic sequences ${\alpha,\beta}$ (or as a finite linear combination of such Dirichlet convolutions). The reason for introducing this convolution structure is through the basic identity

$\displaystyle (\sum_n \alpha*\beta(n) \overline{\chi_j}(n)) = (\sum_n \alpha(n) \overline{\chi_j}(n)) (\sum_n \beta(n) \overline{\chi_j}(n)) \ \ \ \ \ (8)$

for any finitely supported sequences ${\alpha,\beta: {\bf N} \rightarrow {\bf C}}$, as can be easily seen by multiplying everything out and using the completely multiplicative nature of ${\chi_j}$. (This is the multiplicative analogue of the well-known relationship ${\widehat{f*g}(\xi) = \hat f(\xi) \hat g(\xi)}$ between ordinary convolution and Fourier coefficients.) This factorisation, together with yet another application of the Cauchy-Schwarz inequality, lets one control (7) by square-sums of the sort that can be handled by the large sieve inequality.

As we have seen in Notes 1, the von Mangoldt function ${\Lambda}$ does indeed admit several factorisations into Dirichlet convolution type, such as the factorisation ${\Lambda = \mu * L}$. One can try directly inserting this factorisation into the above strategy; it almost works, however there turns out to be a problem when considering the contribution of the portion of ${\mu}$ or ${L}$ that is supported at very small natural numbers, as the large sieve loses any gain over the trivial bound in such settings. Because of this, there is a need for a more sophisticated decomposition of ${\Lambda}$ into Dirichlet convolutions ${\alpha * \beta}$ which are non-degenerate in the sense that ${\alpha,\beta}$ are supported away from small values. (As a non-example, the trivial factorisation ${\Lambda = \Lambda * \delta}$ would be a totally inappropriate factorisation for this purpose.) Fortunately, it turns out that through some elementary combinatorial manipulations, some satisfactory decompositions of this type are available, such as the Vaughan identity and the Heath-Brown identity. By using one of these identities we will be able to complete the proof of the Bombieri-Vinogradov theorem. (These identities are also useful for other applications in which one wishes to control correlations between the von Mangoldt function ${\Lambda}$ and some other sequence; we will see some examples of this in later notes.)

For further reading on these topics, including a significantly larger number of examples of the large sieve inequality, see Chapters 7 and 17 of Iwaniec and Kowalski.

Remark 5 We caution that the presentation given in this set of notes is highly ahistorical; we are using modern streamlined proofs of results that were first obtained by more complicated arguments.

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.
Read the rest of this entry »