2 Ring of Integers of a Quadratic Field (Human-Oriented)

Let \(d\) be an integer different from \(0\) and \(1\). We also assume that \(d\) is squarefree. We write \(K\) for the subring of \(\mathbb {C}\) generated by \(\sqrt{d}\). By our assumptions on \(d\), we have that \(K = \mathbb {Q}(\sqrt{d})\) is a field.

Lemma 27

We have that \(d = \pm 1 \bmod 4\) or \(d = 2 \bmod 4\).

Proof ▶

If \(d = 0 \bmod 4\) then \(d\) would not be squarefree.

Lemma 28

We have that \(\sqrt{d} \in \mathcal{O}_K\).

Proof ▶

Clear since \(\sqrt{d}\) is a root of \(x^2-d\).

Lemma 29

If \(d = 1 \bmod {4}\) then \(\frac{1+\sqrt{d}}{2} \in \mathcal{O}_K\).

Proof ▶

Write \(d = 4a + 1\), with \(a \in \mathbb {Z}\). Then \(\frac{1+\sqrt{d}}{2}\) is a root of \(x^2 - x - a \in \mathbb {Z}[x]\).

Let \(t \in K\), so \(t = a + b \sqrt{d}\) for some \(a, b \in \mathbb {Q}\). We assume that \(t \notin \mathbb {Q}\), i.e. that \(b \neq 0\).

Lemma 30

The minimal polynomial of \(t\) over \(\mathbb {Q}\) is

\[ P(x) = x^2-2at+(a^2-db^2) \]

Proof ▶

It’s clear that \(t\) is a root of \(P\) and that \(P \in \mathbb {Q}[x]\) is monic.

Irreducibility follows by the fact that \(P\) has a root that is not rational.

Lemma 31

We have that the trace of \(t\) is \(2a\).

Proof ▶

Clear by Lemma 30.

Lemma 32

We have that the norm of \(t\) is \(a^2-db^2\).

Proof ▶

Clear by Lemma 30.

We suppose now that \(t \in \mathcal{O}_K\).

Lemma 33

We have that \(2a \in \mathbb {Z}\)

Proof ▶

Since the trace of an algebraic integer is an integers, this follows by Lemma 31.

Lemma 34

We have that \(a^2-db^2 \in \mathbb {Z}\)

Proof ▶

Since the norm of an algebraic integer is an integers, this follows by Lemma 32.

Lemma 35

We have that \((2a)^2 - d(2b)^2\) is an integer divisible by \(4\).

Proof ▶

Clear since \((2a)^2 - d(2b)^2 = 4(a^2-db^2)\) and \(a^2-db^2 \in \mathbb {Z}\) by Lemma 34.

Lemma 36

We have that \(2b \in \mathbb {Z}\).

Proof ▶

By Lemma 35, \((2a)^2 - d(2b)^2\) is an integer and so, by Lemma 33, we know that \(d(2b)^2 \in \mathbb {Z}\). Since \(d\) is squarefree, we conclude that \(2b \in \mathbb {Z}\).

Lemma 37

If \(a \in \mathbb {Z}\) then \(b \in \mathbb {Z}\).

Proof ▶

By Lemma 35 and our assumption, both \((2a)^2\) and \((2a)^2 - d(2b)^2\) are integers divisible by \(4\), so the same holds for \(d(2b)^2\). In particular \(db^2 \in \mathbb {Z}\) and \(b \in \mathbb {Z}\) since \(d\) is squarefree.

Lemma 38

If \(a \not\in \mathbb {Z}\) then \(d = 1 \bmod {4}\).

Proof ▶

We have that \(2a\), that is an integer, must be odd. By Lemmas 35 and 36, we have \((2a)^2 = d(2b)^2 \bmod {4}\), so \(2b\) must be odd and \(d = 1 \bmod {4}\) as required.

Theorem 1

Assume that \(d = 2 \bmod {4}\) or \(d = 3 \bmod {4}\). Then

\[ \mathcal{O}_K = \mathbb {Z}[\sqrt{d}] \]

Proof ▶

By Lemma 28 we know that \(\mathbb {Z}[\sqrt{d}] \subseteq \mathcal{O}_K\). Let \(t = a + b \sqrt{d} \in \mathcal{O}_K\), with \(a, b \in \mathbb {Q}\). By Lemma 38 we have that \(a \in \mathbb {Z}\) (since by Lemma 27 we cannot have \(d = 1 \bmod {4}\)), and so by Lemma 37 we have \(b \in \mathbb {Z}\), so \(t \in \mathbb {Z}[\sqrt{d}]\).

Lemma 39

Assume that \(d = 1 \bmod {4}\) and take \(t = a + b \sqrt{d} \in \mathcal{O}_K\) with \(a, b \in \mathbb {Q}\). If \(a \in \mathbb {Z}\) then \(t \in \mathbb {Z}\left[ \frac{1+\sqrt{d}}{2} \right]\).

Proof ▶

By Lemma 37 we have that \(b \in \mathbb {Z}\) and so \(t \in \mathbb {Z}[\sqrt{d}] \subseteq \mathbb {Z}\left[ \frac{1+\sqrt{d}}{2} \right]\).

Theorem 2

Assume that \(d = 1 \bmod {4}\). Then

\[ \mathcal{O}_K = \mathbb {Z}\left[ \frac{1+\sqrt{d}}{2} \right] \]

Proof ▶

By Lemma 29 we know that \(\mathbb {Z}\left[ \frac{1+\sqrt{d}}{2} \right] \subseteq \mathcal{O}_K\). Let \(t = a + b \sqrt{d} \in \mathcal{O}_K\), with \(a, b \in \mathbb {Q}\).

If \( a \in \mathbb {Z}\) we conclude by Lemma 39.
If \(a \notin \mathbb {Z}\), let us consider

\[ t' = t - \frac{1+\sqrt{d}}{2} = a - \frac{1}{2} + \left( b - \frac{1}{2} \right) \sqrt{d} \in \mathcal{O}_K \]

Since \(2a \in \mathbb {Z}\) and \(a \notin \mathbb {Z}\), we have that \(a - \frac{1}{2} \in \mathbb {Z}\), so by Lemma 39, we have that \(t' \in \mathbb {Z}\left[ \frac{1+\sqrt{d}}{2} \right]\) and so \(t \in \mathbb {Z}\left[ \frac{1+\sqrt{d}}{2} \right]\).