2 Third Cyclotomic Extensions

Theorem 2.1

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Let \(K = \mathbb {Q}(\zeta _3)\) be the third cyclotomic field.
Let \(\mathcal{O}_K = \mathbb {Z}[\zeta _3]\) be the ring of integers of \(K\).
Let \(\mathcal{O}^\times _K\) be the group of units of \(\mathcal{O}_K\).
Let \(\zeta _3 \in K\) be any primitive third root of unity.
Let \(\eta \in \mathcal{O}_K\) be the element corresponding to \(\zeta _3 \in K\).
Let \(\lambda \in \mathcal{O}_K\) be such that \(\lambda = \eta -1\).
Let \(u \in \mathcal{O}^\times _K\) be a unit.

Then \(u \in \left\{ 1, -1, \eta , -\eta , \eta ^2, -\eta ^2\right\} \).

Proof ▶

Let \(\mathcal{F}\) be the fundamental system of \(K\).
By properties of cyclotomic fields, we know that \(\mathrm{rank}\left(K\right) = 0\) (see this lemma, this lemma and this lemma which have already been formalised and included in Mathlib). By the Dirichlet Unit Theorem (see Mathlib), we know that

\[ \exists x \in K \text{ with finite order, such that } u = x \prod _{v\in \mathcal{F}} v, \]

but since \(\mathrm{rank}\left(K\right) = 0\), then \(\mathcal{F} = \emptyset \), which implies that \(u = x\).
Since \(u = x\) has finite order, by properties of primitive roots (see this lemma that has already been formalised and included in Mathlib), we can deduce that

\[ \exists r {\lt} 3 \text{ such that } u = \eta ^r \lor u = -\eta ^r. \]

Therefore, we can conclude

\[ u \in \left\{ \pm \eta ^r\mid r \in \left\{ 0,1,2\right\} \right\} = \left\{ 1, -1, \eta , -\eta , \eta ^2, -\eta ^2\right\} . \]

Theorem 2.2

✓

Proof ▶

By properties of cyclotomic fields, we know that \(\left\{ 1,\eta \right\} \) is an integral power basis of \(\mathcal{O}_K\) (see this lemma, this lemma and this lemma which have already been formalised and included in Mathlib).
For every \(\xi \in \mathcal{O}_K\), we define \(\pi _1(\xi )\) and \(\pi _2(\xi )\) to be the first and second coordinates of \(\xi \) with respect to the basis \(\left\{ 1,\eta \right\} \in \mathcal{O}_K\), i.e.

\[ \xi = \pi _1(\xi ) + \pi _2(\xi )\eta . \]

By contradiction we assume that

\[ \exists m \in \mathbb {Z}\text{ such that } 3 \mid \eta - m, \]

which implies that

\[ \exists x \in \mathcal{O}_K \text{ such that } \eta - m = 3 x. \]

By linearity of \(\pi _2\),

\[ \pi _2(\eta ) = \pi _2(3x + m) = 3\pi _2(x) + \pi _2(m). \]

Since \(\pi _2(\eta ) = 1\) and \(\pi _2(m) = 0\), then we have that \(3 \mid 1\), which is a contradiction.

Lemma 2.3

✓

Proof ▶

By definition we have that \(\lambda = \eta -1\), which implies that

\[ \lambda ^2 = (\eta - 1)^2 = \eta ^2 - 2\eta + 1. \]

Since \(\eta \) corresponds to a root of the equation \(x^2 + x + 1 = 0\), then \(\eta ^2 = -1 - \eta \). Substituting back, we can conclude that

\[ \lambda ^2 = (-1 - \eta ) - 2\eta + 1 = -3\eta . \]

Theorem 2.4

✓

Let \(K = \mathbb {Q}(\zeta _3)\) be the third cyclotomic field.
Let \(\mathcal{O}_K = \mathbb {Z}[\zeta _3]\) be the ring of integers of \(K\).
Let \(\mathcal{O}^\times _K\) be the group of units of \(\mathcal{O}_K\).
Let \(\zeta _3 \in K\) be any primitive third root of unity.
Let \(\eta \in \mathcal{O}_K\) be the element corresponding to \(\zeta _3 \in K\).
Let \(\lambda \in \mathcal{O}_K\) be such that \(\lambda = \eta -1\).
Let \(u \in \mathcal{O}^\times _K\) be a unit.

If \(\exists m \in \mathbb {Z}\) such that \(\lambda ^2 \mid u - m\), then \(u = 1 \lor u = -1\).
This is a special case of the Kummer’s Lemma.

Proof ▶

By Lemma 2.3, we have that \(-3\eta = \lambda ^2 \mid u - m\), which implies that \(3 \mid u - m\).
By Theorem 2.1, we know that \(u \in \left\{ 1, -1, \eta , -\eta , \eta ^2, -\eta ^2\right\} \).
We proceed by analysing each case:

Case \(u = 1 \lor u = -1\). This finishes the proof.
Case \(u = \eta \).
Since \(3 \mid u - m\), we have that \(3 \mid \eta - m\), which contradicts Theorem 2.2 forcing us to conclude that \(u \neq \eta \).
Case \(u = -\eta \).
Since \(3 \mid u - m\), we have that \(3 \mid - \eta - m\), then by properties of divisibility \(3 \mid \eta + m\), which contradicts Theorem 2.2 forcing us to conclude that \(u \neq -\eta \).
Case \(u = \eta ^2\).
Since \(3 \mid u - m\), we have that \(3 \mid \eta ^2 - m\), which contradicts Theorem 2.2 since \(\eta ^2\) is a third root of unity (see Mathlib), forcing us to conclude that \(u \neq \eta ^2\).
Case \(u = -\eta ^2\).
Since \(3 \mid u - m\), we have that \(3 \mid - \eta ^2 - m\), then by properties of divisibility \(3 \mid \eta ^2 + m\), which contradicts Theorem 2.2 since \(\eta ^2\) is a third root of unity (see Mathlib), forcing us to conclude that \(u \neq -\eta ^2\).

Therefore, \(u = 1 \lor u = -1\).

Lemma 2.5

✓

Proof ▶

Since the third cyclotomic polynomial over \(\mathbb {Q}\) is irreducible, then the norm of \(\lambda \) is \(3\) by properties of primitive roots (see this lemma that has already been formalised and included in Mathlib).

Lemma 2.6

✓

Proof ▶

It directly follows from Lemma 2.5 since \(3\) is a prime number.

Lemma 2.7

✓

Proof ▶

By properties of norms and divisibility, if the norm of an element in the ring of integers divides a number, then the element itself must divide that number. In this case, by Lemma 2.5 we know that the norm of \(\lambda \) is \(3\), that divides \(3\), which implies that \(\lambda \mid 3\).

Lemma 2.8

✓

Proof ▶

Since \(3\) is prime and \(\zeta _3\) is a primitive third root of unity, then \(\lambda \) is prime by Theorem 1.29.

Lemma 2.9

✓

Proof ▶

It directly follows from Lemma 2.8 since zero is not prime.

Lemma 2.10

✓

Proof ▶

It directly follows from Lemma 2.8 since prime numbers are not units.

Lemma 2.11

✓

Let \(K = \mathbb {Q}(\zeta _3)\) be the third cyclotomic field.
Let \(\mathcal{O}_K = \mathbb {Z}[\zeta _3]\) be the ring of integers of \(K\).
Let \(\mathcal{O}^\times _K\) be the group of units of \(\mathcal{O}_K\).
Let \(\zeta _3 \in K\) be any primitive third root of unity.
Let \(\eta \in \mathcal{O}_K\) be the element corresponding to \(\zeta _3 \in K\).
Let \(\lambda \in \mathcal{O}_K\) be such that \(\lambda = \eta -1\).
Let \(I\) be the ideal generated by \(\lambda \).

Then \(\mathcal{O}_K / I\) has cardinality \(3\).

Proof ▶

It directly follows from Lemma 2.5 by the fundamental properties of ideals.

Lemma 2.12

✓

Let \(K = \mathbb {Q}(\zeta _3)\) be the third cyclotomic field.
Let \(\mathcal{O}_K = \mathbb {Z}[\zeta _3]\) be the ring of integers of \(K\).
Let \(\mathcal{O}^\times _K\) be the group of units of \(\mathcal{O}_K\).
Let \(\zeta _3 \in K\) be any primitive third root of unity.
Let \(\eta \in \mathcal{O}_K\) be the element corresponding to \(\zeta _3 \in K\).
Let \(\lambda \in \mathcal{O}_K\) be such that \(\lambda = \eta -1\).
Let \(I\) be the ideal generated by \(\lambda \).
Let \(2 \in \mathcal{O}_K / I\).

Then \(2 \neq 0\).

Proof ▶

By contradiction we assume that \(2 \in I\), then, by definition, \(\lambda \) would divide \(2 \in \mathcal{O}_K\). Recall from Lemma 2.5 that the norm of \(\lambda \) is \(3\). If \(\lambda \) divided \(2\), then by properties of divisibility in number fields, the norm of \(\lambda \) would also divide \(2\). However \(3 \nmid 2\) showing a contradiction. Therefore, \(\lambda \nmid 2\), then \(2 \notin I\), which implies that \(2 \in \mathcal{O}_K / I\) is non-zero.

Lemma 2.13

✓

Proof ▶

By contradiction we assume that \(\lambda \mid 2\), that implies that \(2 \in I\) from which it follows that \(2 = 0\) contradicting Lemma 2.12 forcing us to conclude that \(\lambda \nmid 2\).

Lemma 2.14

✓

Let \(K = \mathbb {Q}(\zeta _3)\) be the third cyclotomic field.
Let \(\mathcal{O}_K = \mathbb {Z}[\zeta _3]\) be the ring of integers of \(K\).
Let \(\mathcal{O}^\times _K\) be the group of units of \(\mathcal{O}_K\).
Let \(\zeta _3 \in K\) be any primitive third root of unity.
Let \(\eta \in \mathcal{O}_K\) be the element corresponding to \(\zeta _3 \in K\).
Let \(\lambda \in \mathcal{O}_K\) be such that \(\lambda = \eta -1\).
Let \(I\) be the ideal generated by \(\lambda \).

Then \(\mathcal{O}_K / I = \left\{ 0, 1, -1\right\} \).

Proof ▶

By Lemma 2.11, the cardinality of \(\mathcal{O}_K / I\) is \(3\), so it suffices to prove that \(1,-1\) and \(0\) are distinct.
We proceed by contradiction analysing each case:

Case \(1 = -1\). By basic algebraic properties, \(1 = -1\) implies that \(2 = 0\), which contradicts Lemma 2.12 forcing us to conclude that \(1 \neq -1\).
Case \(1 = 0\). Trivial contradiction.
Case \(-1 = 0\). It implies that \(1 = 0\), which is a contradiction.

Lemma 2.15

✓

Let \(K = \mathbb {Q}(\zeta _3)\) be the third cyclotomic field.
Let \(\mathcal{O}_K = \mathbb {Z}[\zeta _3]\) be the ring of integers of \(K\).
Let \(\mathcal{O}^\times _K\) be the group of units of \(\mathcal{O}_K\).
Let \(\zeta _3 \in K\) be any primitive third root of unity.
Let \(\eta \in \mathcal{O}_K\) be the element corresponding to \(\zeta _3 \in K\).
Let \(\lambda \in \mathcal{O}_K\) be such that \(\lambda = \eta -1\).
Let \(x \in \mathcal{O}_K\).

Then \((\lambda \mid x) \lor (\lambda \mid x-1) \lor (\lambda \mid x+1)\).

Proof ▶

Let \(I\) be the ideal generated by \(\lambda \). Let \(\pi : \mathcal{O}_K \to \mathcal{O}_K / I\).
By Lemma 2.14, we have that \(\pi (x) \in \mathcal{O}_K / I = \left\{ 0, 1, -1\right\} \).
We proceed by analysing each case:

Case \(\pi (x) = 0\). By properties of ideals, \(\lambda \mid x\).
Case \(\pi (x) = 1\). Then \(0=\pi (x)-1=\pi (x-1)\), which, by properties of ideals, implies that \(\lambda \mid x-1\).
Case \(\pi (x) = -1\). Then \(0=\pi (x)+1=\pi (x+1)\), which, by properties of ideals, implies that \(\lambda \mid x+1\).

Lemma 2.16

✓

Proof ▶

Since \(\zeta _3 \in K\) is a primitive third root of unity, then \(\zeta _3^3 = 1\). Given that \(\eta \in \mathcal{O}_K\) is the element corresponding to \(\zeta _3 \in K\), then \(\eta ^3 = 1\) by the extension of the field properties into the ring of integers.

Lemma 2.17

✓

Proof ▶

It directly follows from Lemma 2.16.

Lemma 2.18

✓

Proof ▶

Since \(\eta \) corresponds to a root of the equation \(x^2 + x + 1 = 0\), then \(\eta ^2 + \eta + 1 = 0\).

Lemma 2.19

✓

Let \(K = \mathbb {Q}(\zeta _3)\) be the third cyclotomic field.
Let \(\mathcal{O}_K = \mathbb {Z}[\zeta _3]\) be the ring of integers of \(K\).
Let \(\mathcal{O}^\times _K\) be the group of units of \(\mathcal{O}_K\).
Let \(\zeta _3 \in K\) be any primitive third root of unity.
Let \(\eta \in \mathcal{O}_K\) be the element corresponding to \(\zeta _3 \in K\).
Let \(x \in \mathcal{O}_K\).

Then \(x^3 - 1 = (x - 1)(x - \eta )(x - \eta ^2)\).

Proof ▶

Applying Lemma 2.16 and Lemma 2.18, we have that

\begin{align*} (x - 1)(x - \eta )(x - \eta ^2) & = x^3 - x^2 (\eta ^2 + \eta + 1) + x (\eta ^2 + \eta + \eta ^3) - \eta ^3 \\ & = x^3 - x^2 (\eta ^2 + \eta + 1) + x (\eta ^2 + \eta + 1) - 1 \\ & = x^3 - 1. \end{align*}

Lemma 2.20

✓

Let \(K = \mathbb {Q}(\zeta _3)\) be the third cyclotomic field.
Let \(\mathcal{O}_K = \mathbb {Z}[\zeta _3]\) be the ring of integers of \(K\).
Let \(\mathcal{O}^\times _K\) be the group of units of \(\mathcal{O}_K\).
Let \(\zeta _3 \in K\) be any primitive third root of unity.
Let \(\eta \in \mathcal{O}_K\) be the element corresponding to \(\zeta _3 \in K\).
Let \(\lambda \in \mathcal{O}_K\) be such that \(\lambda = \eta -1\).
Let \(x \in \mathcal{O}_K\).

Then \(\lambda \mid x(x - 1)(x - (\eta + 1))\).

Proof ▶

By Lemma 2.15, we have that

\[ (\lambda \mid x) \lor (\lambda \mid x-1) \lor (\lambda \mid x+1). \]

We proceed by analysing each case:

Case \(\lambda \mid x\).
By properties of divisibility, we have that \(\lambda \mid x(x - 1)(x - (\eta + 1))\).
Case \(\lambda \mid x-1\).
By properties of divisibility, we have that \(\lambda \mid x(x - 1)(x - (\eta + 1))\).
Case \(\lambda \mid x+1\).
By properties of divisibility, it suffices to prove that

\[ \lambda \mid x - (\eta + 1) = x + 1 - (\eta - 1 + 3). \]

By definition of \(\lambda \), we have that

\[ x + 1 - (\eta - 1 + 3) = x + 1 - (\lambda + 3). \]

By properties of divisibility and Lemma 2.7, we can deduce that \(\lambda \mid \lambda + 3\).
Therefore, by properties of divisibility, we can conclude that

\[ \lambda \mid x(x - 1)(x - (\eta + 1)). \]

Lemma 2.21

✓

Let \(K = \mathbb {Q}(\zeta _3)\) be the third cyclotomic field.
Let \(\mathcal{O}_K = \mathbb {Z}[\zeta _3]\) be the ring of integers of \(K\).
Let \(\mathcal{O}^\times _K\) be the group of units of \(\mathcal{O}_K\).
Let \(\zeta _3 \in K\) be any primitive third root of unity.
Let \(\eta \in \mathcal{O}_K\) be the element corresponding to \(\zeta _3 \in K\).
Let \(\lambda \in \mathcal{O}_K\) be such that \(\lambda = \eta -1\).
Let \(x \in \mathcal{O}_K\).

If \(\lambda \mid x - 1\), then \(\lambda ^4 \mid x^3 - 1\).

Proof ▶

Let \(\lambda \mid x - 1\), which is equivalent to say that

\[ \exists y\in \mathcal{O}_K \text{ such that } x - 1 = \lambda y. \]

By ring properties and Lemma 2.19, we have that

\[ x^3 - 1 = \lambda ^3 (y (y - 1) (y - (\eta + 1))). \]

By properties of divisibility and Lemma 2.20, we can conclude that

\[ \lambda ^4 \mid x^3 - 1. \]

Lemma 2.22

✓

Let \(K = \mathbb {Q}(\zeta _3)\) be the third cyclotomic field.
Let \(\mathcal{O}_K = \mathbb {Z}[\zeta _3]\) be the ring of integers of \(K\).
Let \(\mathcal{O}^\times _K\) be the group of units of \(\mathcal{O}_K\).
Let \(\zeta _3 \in K\) be any primitive third root of unity.
Let \(\eta \in \mathcal{O}_K\) be the element corresponding to \(\zeta _3 \in K\).
Let \(\lambda \in \mathcal{O}_K\) be such that \(\lambda = \eta -1\).
Let \(x \in \mathcal{O}_K\).

If \(\lambda \mid x + 1\), then \(\lambda ^4 \mid x ^3 + 1\).

Proof ▶

By properties of divisibility, if \(\lambda \mid x + 1\) then

\[ \lambda \mid - (x + 1) = (- x) - 1. \]

By Lemma 2.20, we can deduce that

\[ \lambda ^4 \mid (-x)^3 - 1. \]

By divisibility and ring properties we can conclude that

\[ \lambda ^4 \mid x^3 + 1. \]

Lemma 2.23

✓

Let \(K = \mathbb {Q}(\zeta _3)\) be the third cyclotomic field.
Let \(\mathcal{O}_K = \mathbb {Z}[\zeta _3]\) be the ring of integers of \(K\).
Let \(\mathcal{O}^\times _K\) be the group of units of \(\mathcal{O}_K\).
Let \(\zeta _3 \in K\) be any primitive third root of unity.
Let \(\eta \in \mathcal{O}_K\) be the element corresponding to \(\zeta _3 \in K\).
Let \(\lambda \in \mathcal{O}_K\) be such that \(\lambda = \eta -1\).
Let \(x \in \mathcal{O}_K\).

If \(\lambda \nmid x\), then \((\lambda ^4 \mid x^3 - 1) \lor (\lambda ^4 \mid x^3 + 1)\).

Proof ▶

By Lemma 2.15, we have that

\[ (\lambda \mid x) \lor (\lambda \mid x-1) \lor (\lambda \mid x+1). \]

We proceed by analysing each case:

Case \(\lambda \mid x\). From trivially contradictory hypotheses we can conclude that

\[ (\lambda ^4 \mid x^3 - 1) \lor (\lambda ^4 \mid x^3 + 1). \]
Case \(\lambda \mid x-1\). By Lemma 2.21, we have that \(\lambda ^4 \mid x^3 - 1\), which implies that

\[ (\lambda ^4 \mid x^3 - 1) \lor (\lambda ^4 \mid x^3 + 1). \]
Case \(\lambda \mid x+1\). By Lemma 2.22, we have that \(\lambda ^4 \mid x^3 + 1\), which implies that

\[ (\lambda ^4 \mid x^3 - 1) \lor (\lambda ^4 \mid x^3 + 1). \]