Quadratic Fields

In this post, I will connect the two sections of my previous post, and show how to produce many more special value formulas like the ones for $L(\chi_{-4}, 1)$ and $L(\chi_8, 1)$. Recall that these two special value formulas are

\begin{equation} L(\chi_{-4}, 1) = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \frac{1}{9} - \frac{1}{11} + \frac{1}{13} - \frac{1}{15} + \cdots = \frac{\pi}{4} \end{equation}

and

\begin{equation}L(\chi_8, 1) = 1 - \frac{1}{3} - \frac{1}{5} + \frac{1}{7} + \frac{1}{9} - \frac{1}{11} - \frac{1}{13} + \frac{1}{15} + \cdots = \frac{\log(1 + \sqrt{2})}{\sqrt{2}}.\end{equation}

I will also explain where the sequences defining $L(\chi_{-4}, s)$ and $L(\chi_8, s)$ come from as well as where the numbers $\frac{\pi}{4}$ and $\frac{\log(1 + \sqrt{2})}{\sqrt{2}}$ come from. To do this, the most simple number systems that are not the rational numbers will be introduced and studied.

Quadratic Fields

Quadratic fields are the first fields after the rational numbers to study. As a $\mathbb Q$-vector space, they have dimension $2$, and are obtained by adjoining the square root of an integer to $\mathbb Q$.

Let $d$ be a square free integer, which means that no square numbers in $\mathbb Z$ divide $d$. The quadratic field associated to $d$ is

\[\mathbb Q(\sqrt{d}) = \{ \alpha + \beta\sqrt{d} : \alpha, \beta \in \mathbb{Q}\}.\]

Let $K = \mathbb Q(\sqrt{d})$. In $K$ one can add, subtract, multiply and divide:

\[(\alpha + \beta\sqrt{d}) \pm (\gamma + \delta\sqrt{d}) = \alpha + \gamma \pm (\beta + \delta)\sqrt{d}\] \[(\alpha + \beta\sqrt{d})(\gamma + \delta\sqrt{d}) = \alpha\gamma + \beta\delta d + (\alpha\delta + \beta\gamma)\sqrt{d}\] \[\frac{\alpha + \beta \sqrt{d}}{\gamma + \delta\sqrt{d}} = \frac{\alpha\gamma - \delta\beta d}{\gamma^2 - d\delta^2} + \frac{\beta\gamma - \alpha\delta}{\gamma^2 - d\delta^2}\sqrt{d},\]

where to divide, we rationalized $\frac{\alpha + \beta\sqrt{d}}{\gamma + \delta\sqrt{d}}$ by multiplying by $\frac{\gamma - \delta\sqrt{d}}{\gamma - \delta\sqrt{d}}$.

The ring of integers of $K$ comes in two different types depending on what $d$ is modulo $4$. If $d\equiv 2$ or $3\bmod 4$, then

\[\mathscr O_K = \{\alpha + \beta\sqrt{d} : \alpha, \beta \in \mathbb Z\},\]

and if $d\equiv 1\bmod 4$, then

\[\mathscr O_K = \left\{\alpha + \beta\frac{1 + \sqrt{d}}{2} : \alpha, \beta \in \mathbb Z\right\}.\]

If one were to apriori guess the ring of integers of $\mathscr O_K$, the natural candidate is the representation written for when $d\equiv 2$ or $3\bmod 4$. To see that $\frac{1 + \sqrt{d}}{2}$ is an algebraic integer when $d\equiv 1\bmod 4$, consider the polynomial

\[\left(x - \frac{1 + \sqrt{d}}{2}\right)\left(x - \frac{1 - \sqrt{d}}{2}\right) = x^2 - x + \frac{1-d}{4},\]

which is a monic polynomial that has integer coefficients if and only if $d\equiv 1\bmod 4$ since $\frac{1-d}{4}\in \mathbb Z$ if and only if $d\equiv 1\bmod 4$. It turns out that you cannot get more than this denominator of $2$ showing up in the ring of integers of these quadratic fields.

The first invariant of $\mathscr O_K$ I’ll discuss is the discriminant of $\mathscr O_K$. Recall that the discriminant measures the covolume of $\mathscr O_K$ in $K$. since the definition of $\mathscr O_K$ differs whether or not $d$ is congruent to $1\bmod 4$, the discriminant will differ as well. In general we have the inclusion of sets

\[\{\alpha + \beta\sqrt{d} : \alpha,\beta\in\mathbb Z\}\subset\left\{\alpha + \beta \frac{1 + \sqrt{d}}{2} : \alpha,\beta\in\mathbb Z\right\}\]

since $\sqrt{d} = -1 + 2\frac{1 + \sqrt{d}}{2}$. This inclusion shows that when $d\equiv 1\bmod 4$, the size of $\mathscr O_K$ relative to $K$ is larger than when $d\equiv 2$ and $3=\bmod 4$, so the covolume will be smaller.

To precisely calculate the discriminant, we need to consider the embeddings of $K$ into $\mathbb C$. An embedding of $K$ into $\mathbb C$ is a function from $K$ to $\mathbb C$ that preserves the arithmetic operations of $K$. Let $\varphi :K\rightarrow \mathbb C$ be an embedding of $K$ into $\mathbb C$. Then it turns out that $\varphi$ must be the identity on $\mathbb Q$, and $\varphi$ is determined by where it sends $\sqrt{d}$. The two possibilities for $\varphi(\sqrt{d})$ are $\sqrt{d}$ and $-\sqrt{d}$. I’m abusing notation a bit here by apriori not thinking of $K$ as a subset of $\mathbb C$ and saying that the two ways to make $K$ a subset of $\mathbb C$ such that the arithmetic operations in $K$ agree with those in $\mathbb C$ are to send $\sqrt{d}\in K$ (thought of abstractly as an element of $K$) to either $\sqrt{d}\in \mathbb C$ or $-\sqrt{d}\in \mathbb C$ (now thought of as concrete complex numbers).

Before calculating the discriminant of $\mathscr O_K$, I want to mention an important dichotomy that happens with quadratic fields. When $d >0$, $\sqrt{d}$ and $-\sqrt{d}$ are real numbers, so both of the embeddings of $K$ into $\mathbb C$ land in $\mathbb R$. In this case $K$ is called a real quadratic field and the invariants $r_1$ and $r_2$ are $r_1 = 2, r_2 = 0$. When $d < 0$, $\sqrt{d}$ and $-\sqrt{d}$ are purely imaginary complex numbers, so both the embeddings of $K$ into $\mathbb C$ do not land in $\mathbb R$. Further, $-\sqrt{d}$ is the complex conjugate of $\sqrt{d}$. In this case $K$ is called an imaginary quadratic field and the invariants $r_1$ and $r_2$ are $r_1 = 0, r_2 = 1$.

Now to calculate the discriminant of $\mathscr O_K$. If $d\equiv 2$ or $3\bmod 4$, then $1$, $\sqrt{d}$ is a basis for $\mathscr O_K$ over $\mathbb Z$ and the embeddings of $K$ into $\mathbb C$ either send $1$ to $1$ and $\sqrt{d}$ to $\sqrt{d}$ or send $1$ to $1$ and $\sqrt{d}$ to $-\sqrt{d}$. Then the discriminant of $\mathscr O_K$ is

\[D_K = \det\begin{pmatrix} 1 & 1\\ \sqrt{d} &-\sqrt{d}\end{pmatrix}^2 = 4d.\]

If $d\equiv 1\bmod 4$, then $1$, $\frac{1 + \sqrt{d}}{2}$ is a basis for $\mathscr O_K$ over $\mathbb Z$ and the embeddings of $K$ into $\mathbb C$ either send $1$ to $1$ and $\frac{1 + \sqrt{d}}{2}$ to $\frac{1 + \sqrt{d}}{2}$ or send $1$ to $1$ and $\frac{1 + \sqrt{d}}{2}$ to $\frac{1 - \sqrt{d}}{2}$. Then the discriminant is

\[D_K = \det\begin{pmatrix} 1 & 1\\ \frac{1 + \sqrt{d}}{2} & \frac{1 - \sqrt{d}}{2}\end{pmatrix}^2 = d.\]

The covolume of $\mathscr O_K$ in $K$ is the square root of the absolute value of the discriminant, and these calculations match the observation that the covolume of $\mathscr O_K$ in $K$ should be smaller when $d\equiv 1\bmod 4$.

Associated $L$-series

Before I discuss the other terms in the class number formula for $K$, I will associate an $L$-series to $K$. Let $D = D_K$ be the discriminant of $K$. I will define a function

\[\chi_D:\mathbb N\longrightarrow \{-1, 0, 1\}\]

which determines a sequence $a_n = \chi_D(n)$. The $L$-series associated to $K$ is defined as

\[L(\chi_D, s) = \sum_{n = 1}^\infty\frac{\chi_D(n)}{n^s}.\]

This is where the functions $L(\chi_{-4}, s)$ and $L(\chi_8, s)$ from the previous post come from.

Let $n\in\mathbb N$. If $\gcd(n, D) > 1$, define $\chi_D(n) = 0$. If $p$ is a prime, and $p\nmid 2d$, define

\[\chi_D(p) = \begin{cases} -1 & \text{ if } d \text{ is not a square} \bmod p\\ 1 & \text{ if } d \text{ is a square} \bmod p.\end{cases}\]

If $d\equiv 1\bmod 4$, define $\chi_D(2)$ as

\[\chi_D(2) = \begin{cases} 1 &\text{ if } d\equiv 1\bmod 8\\ -1 &\text{ if } d\equiv 5\bmod 8.\end{cases}\]

$\chi_D$ is now defined on all prime numbers, and we extend $\chi_D$ to all of $\mathbb N$ multiplicatively. That is, if $n$ has prime factorization $n = p_1^{a_1}p_2^{a_2}\cdots p_m^{a_m}$, then define

\[\chi_D(n) = \prod_{i = 1}^m\chi_D(p_i)^{a_i}.\]

The function $\chi_D$ is known as the Kronecker symbol and it is a generalization of the Legendre symbol. It is related to the arithmetic of $\mathscr O_K$ in the following way. Let $p$ be a prime number that does not divide $D$. $\chi_D(p) = -1$ if the ideal $p\mathscr O_K$ is a prime ideal of $\mathscr O_K$. In this case the prime number $p$ continues to be a prime number in $\mathscr O_K$. $\chi_D(p) = 1$ if the ideal $p\mathscr O_K$ is the product of two distinct prime ideals of $\mathscr O_K$. In this case the prime number $p$ factors into two different prime numbers in $\mathscr O_K$, so $p$ is not a prime number in $\mathscr O_K$, but a product of two new prime numbers. If $p$ divides $D$, then $\chi_D(p) = 0$ and the prime ideal $p\mathscr O_K$ is the product of a prime ideal of $\mathscr O_K$ with itself.

These three arithmetic behaviors of prime numbers can be seen in the canonical example $K = \mathbb Q(\sqrt{-1})$. Let $i = \sqrt{-1}$. In this case $d = -1 \equiv 3\bmod 4$, so

\[\mathscr O_K = \{\alpha + \beta i : \alpha, \beta \in \mathbb Z\},\]

and $D = 4$. The units, $\mathscr O_K^\times$, of $\mathscr O_K$ are $1,-1,i$, and $-i$. These play the role of $1$ and $-1$ in $\mathbb Z$. The “positive” numbers in $\mathscr O_K$ are $\alpha + \beta i$ with $\alpha > 0$ and $\beta\geq 0$. Any $\alpha + \beta i\in\mathscr O_K$ can be multiplied by $1,-1, i,$ or $-i$ to be made positive (just as any integer can be multiplied by $1$ or $-1$ to be made positive). A prime number in $\mathscr O_K$ is a “positive number” in $\mathscr O_K$ that cannot be written as a product of more than one other positive number in $\mathscr O_K$, where we disregard the use of $1,-1,i,$ and $-i$ in any expressions because they are units (just like we disregard $-1$ in the factorization of a negative integer).

In $\mathscr O_K$, the prime number $2$ factors in $\mathscr O_K$ as $2 = -i(1 + i)^2$. Here $1 + i$ is the new prime in $\mathscr O_K$ and we disregard $-i$ in the prime factorization. Because of how one multiplies complex numbers, the product of two “positive” numbers may not be “positive” anymore. This happens with $(1 + i)^2 = 2i$, which we then multiply by $-i$ to make positive and equal to $2$. These three examples of $2,3$, and $5$ show the three possible behaviors of prime numbers in $K$.

The prime number $3$ stays prime in $\mathscr O_K$. Try writing $3$ as a product of two or more “positive” elements of $\mathscr O_K$, disregarding any $1,-1,i$ or $-i$ that you use. The prime number $5$ factors in $\mathscr O_K$ as $5 = -i(1+2i)(2 + i)$. Here $1 + 2i$ and $2 + i$ are the new prime numbers that $5$ is written as a product of.

You may wonder why the subscript on $\chi_D$ is $D$ and not $d$. Using some language from abstract algebra I can explain one reason for this. Using $\chi_D$, we can define a function

\[\widetilde{\chi}_D:(\mathbb Z/|D|\mathbb Z)^\times \longrightarrow \{\pm 1\}\] \[\widetilde{\chi}_D(a + |D|\mathbb Z) = \chi_D(a),\]

which is a well defined group homomorphism ($\left\lvert D\right\rvert$ is the absolute value of $D$). The number $\left\lvert D\right\rvert$ has the following property: among all positive integers $N$, such that the function

\[\widetilde{\chi}:(\mathbb Z/N\mathbb Z)^\times\longrightarrow \{\pm 1\}\] \[\widetilde{\chi}(a + N\mathbb Z) = \chi_D(a)\]

is a well-defined group homomorphism, $\left\lvert D\right\rvert$ is the smallest one.

Relation to Class Number Formula

Since $\chi_D$ is defined to be multiplicative, which means that for all $a,b$, $\chi_D(ab) = \chi_D(a)\chi_D(b)$, $L(\chi_D,s)$ may be written like the Riemann zeta function as a product over all the prime numbers:

\[L(\chi_D,s) = \sum_{n=1}^\infty\frac{\chi_D(n)}{n^s} = \prod_{p-\text{prime}}\left(1 - \frac{\chi_D(p)}{p^s}\right)^{-1}.\]

We’ve considered three $L$-functions related to the two fields $K = \mathbb Q(\sqrt{d})$ and $\mathbb Q$: $L(\chi_D, s)$, the Riemann zeta function, $\zeta(s)$, and the Dedekind zeta function associated to $K$, $\zeta_K(s)$. It turns out that among these three functions the following relation holds:

\begin{equation}\zeta_K(s) = \zeta(s)L(\chi_D,s).\end{equation}

If you know some algebraic number theory and/or commutative algebra, then you can see why this relation holds by looking at the product representation of each function, and using the property of $\chi_D$ that for a prime $p$, $\chi_D(p)$ is related to the behavior of the ideal $p\mathscr O_K$. (If you do this, it is important to note that $\zeta(s)$ and $L(\chi_D,s)$ are products over primes of $\mathbb Z$, and $\zeta_K(s)$ is a product over prime ideals of $\mathscr O_K$.) Using (3), the class number formula for $\mathbb Q$ and $K$ tells us what $L(\chi_D, 1)$ is.

Let us quickly determine the class number formula for $\mathbb Q$. The class number of $\mathbb Z$ is $1$ since there is unique factorization in $\mathbb Z$. The units of $\mathbb Z$ are $1$ and $-1$, so the regulator term in the class number formula for $\mathbb Q$ is $1$ and $\mu_\mathbb Q = 2$. There is only one embedding of $\mathbb Q$ into $\mathbb C$, and it is the identity on $\mathbb Q$, so it lands in $\mathbb R$. Hence $r_1 = 1, r_2=0$. Finally, using the one embedding and that $1$ is a $\mathbb Z$-basis for $\mathbb Z$ over $\mathbb Z$ gives use that the discriminant of $\mathbb Z$ is $1$. Putting this all together, the class number formula for $\mathbb Q$ says

\[\lim_{s\to1}(s-1)\zeta(s) = 1.\]

Then using the class number formula for $K$, the equality (3) gives us that

\begin{equation}L(\chi_D, 1) = \lim_{s\to 1}(s-1)\zeta_K(s) = \frac{2^{r_1}(2\pi)^{r_2}h_K\text{Reg}_K}{\sqrt{\left\lvert D\right\rvert}\mu_K},\end{equation}

where $r_1$ and $r_2$ are the invariants of $K$. To further analyze this formula, we need to break into the two cases of $d > 0$ and $d < 0$.

When $d < 0$, $r_1 = 0$ and $r_2 = 1$, so by Dirichlet’s unit theorem $\mathscr O_K^\times/\mu(\mathscr O_K)$ is a free $\mathbb Z$-module of rank 0, so $\mu(\mathscr O_K) = \mathscr O_K^\times$ and $\text{Reg}_K = 1$. It turns out, and I will say more about this in my next post, that $\mu(\mathscr O_K) = {\pm 1}$ unless $d = -1$ or $-3$. When $d = -1$, \(\mu(\mathscr O_K) = \{\pm 1, \pm\sqrt{-1}=\pm i\}\) and when $d = -3$, \(\mu(\mathscr O_K) = \{\pm 1, \pm e^{2\pi i/3}, \pm e^{4\pi i/3}\}\). Equation (4), then says the following: if $d = -1$

\[L(\chi_{-4}, 1) = h_K\frac{\pi}{4},\]

if $d = -3$

\[L(\chi_{-3}, 1) = h_K\frac{\pi}{3\sqrt{3}},\]

and if $d \not= -1, -3$

\[L(\chi_D, 1) = h_K\frac{\pi}{\sqrt{|D|}}.\]

These are remarkable formulas and explain where the formula (1) for $L(\chi_{-4}, 1)$ comes from (using in addition the fact that $h_K = 1$ in this case).

Now we consider when $d>0$. In this case $r_1 = 2$, $r_2 = 0$, $\mu(\mathscr O_K) = {\pm 1}$, and $\mathscr O_K^\times/\mu(\mathscr O_K)$ is a free $\mathbb Z$-module of rank $1$. Let $u_K\in\mathscr O_K^\times$ be an element of $\mathscr O_K^\times$ that generates $\mathscr O_K^\times/\mu(\mathscr O_K)$ as a $\mathbb Z$-module and such that $\left\lvert u_K\right\rvert > 1$. The element $u_K$ is called the fundamental unit of $K$. More about these will be said in my following posts. In this case, equation (4) says

\[L(\chi_D, 1) = \frac{2h_K\log(u_K)}{\sqrt{D}}.\]

This recovers the formula (2) for $L(\chi_8, 1)$, if we assume that we may take $u_K = 1 + \sqrt{2}$ for $K = \mathbb Q(\sqrt{2})$.

These remarkable formulas give tons of special value formulas resulting in identities for the infinite series $L(\chi_D, 1)$. Further, they give use a way to calculate $h_K$ (if in the real quadratic case we can determine $u_K$). Dirichlet was the first to prove these formulas, and that is why the class number formula is known as Dirichlet’s class number formula. Further, functions such as $L(\chi_D, s)$ are known as Dirichlet $L$-functions. In my following posts, I will use these formulas and write code to calculate the class numbers, $h_K$, for real and imaginary quadratic fields with $\left\lvert d\right\rvert < 100$.

Below is a reference that goes through all of this material in gory detail from a theoretical perspective.

Reference

H. Davenport, Multiplicative Number Theory. Graduate Texts in Mathematics, Springer New York, 2000.