Invertible elements in ℤ[√d]

We shall study the invertible elements of ℤ[√d], with d is a square-free integer. E.g. if a = √−3:

To each element of ℤ[√d] we associate a norm defined as follows

This definition arises from the observation that defining, the conjugate of x = a + b√d the element x̄ = a − b√d, the the norm just introduced is equal to N(x) = xx̄. In the case of negative d, the number a + b√d is complex and this norm coincides with the complex norm.
The case where d = −1, coincides with the ring of Gaussian integers and is usually written ℤ[i].

The norm we've introduce is multiplicative, that is

Indeed, we have

We premise to our analysis, the following statements:

4.14.1 Proposition. The invertible elements of ℤ[√d], with d not a square in ℤ, are only those elements a + b√d such that N(a + b√d) = a² − db² = ±1, that is

Proof. Let x invertible, and x^-1 its inverse, then

from which N(x) = ±1. Conversely, if x is invertible, and x⁻¹ its inverse, then

from which x is invertible (if N(x) = 1, the inverse of x is x̄, if N(x) = -1, the inverse of x is -x̄.□

We observe now that if z is a unit then also −z and 1/z are units. Moreover if |z| > 1 then |1/ z| < 1. Thus the unit of ℤ[√d] are of the form ±z, z⁻¹.

4.14.2 Theorem. Let u ∈ ℤ[√d], d > 0, the smallest unit greater that 1, then all the units greater than 1 of ℤ[√d] are of the form uⁿ, n ∈ ℕ.

Proof. Suppose first there exists a unit x > 1, which is not a power of u. Then x lies between two successive powers of u, that is there exists a positive integer m, such that

If we multiply by u^-m, we obtain

Since u was chosen as the smallest unit, we cannot have another unit between 1 and u. Thus every unit greater that 1 is a power with positive integer of u.□

Let’s now turn to the case ℤ[√d], with d not a square in ℤ. We must solve in ℤ, the following equations

1. d = −1 (case of the Gauss integers): we have to solve the following equations

   a² + b² = 1   and   a² + b² = −1

   Only the first equation is solvable: a = ± 1, b = 0, or a = 0 and b = ±1. Thus the invertible elements of ℤ[i] are ±1 and ±i.

2. d < −1 (e.g. ℤ[√−3]): we have a² − db² = a² + |d|b² hence a² − db² = −1 does not admit solutions. Conversely the solutions of a² + b² = 1 are b = 0 and a = ±1. Thus the only invertible elements are ±1.

3. d > 0: Consider the equation

   a² − db² = 1,   d > 0

   Such Diophantine equations have a long history and are known as Pell’s equations (they were also studied by Fermat). It can be proven that such an equations admits always a not-trivial solution i.e different from ±1. Let u the smallest unit greater than 1: we can find a solution imposing b = 1,2, ... till 1 + db² becomes a perfect square. The description of how Pell’s equations are solved in general is a more sophisticated issue that is unfortunately beyond the scope of our text. We have that also uⁿ, n ∈ ℕ, are units greater than 1.

   Pell’s equation admits infinte solutions, instead the equation a² − db² = −1, can be unsolvable.

4.14.3 Example. In ℤ[√2] the smallest unit greater than 1 is 1 + √2 as it can be easily noticed. Then for what we discussed earlier we can obtain the following list of units of ℤ[√2]

and

This infinite list of units means that detecting whether or not two elements in ℤ[√2] are associates is not the trivial matter it is in ℤ (or even in ℤ[i]). For example, 4 + √2 and 8 − 5√2 are associates, because (4 + √2)(3 − 2√2) = 8 −5 √2, and 3 − 2√2 is a unit.

4.14.4 Example. Which of the following elements of ℤ[√8] are associates.

We calculate the norms N(4 +3√8) = 4² + 8 · 3² = −56; N(2 + √8) = 4 + 8 = 12; N(36 + 13√8) = 36² + 8 · 13² = -56. Thus 4 +3√8, and 36 + 13√8 are associates.

4.14.5 Example. In ℤ[√8] the smallest unit greater than 1 is 3 + √8, as it can be easily verified. Then also 17 + 6√8 = (3 + √8)³ is a unit.  ■