1. | General properties |

2. | Arithmetics in Gaussian integers |

3. | Arithmetics in \(\mathbb Z[\omega]\) |

4. | Arithmetics in other quadratic rings |

What makes work with rational numbers and integers comfortable are
the essential properties they have, especially the unique
factorization property (the Fundamental Theorem of Arithmetic). However,
the might of the arithmetic in \(\mathbb{Q}\) is bounded. Thus, some
polynomials, although they have zeros, cannot be factorized into
polynomials with rational coefficients. Nevertheless, such
polynomials can always be factorized in a wider field. For instance,
the polynomial \(x^2+1\) is irreducible over \(\mathbb{Z}\) or
\(\mathbb{Q}\), but over the ring of the so called *Gaussian
integers* \(\mathbb Z [i]=\{a+b i \mid a,b\in\mathbb Z \}\) it can be
factorized as \((x+i)(x-i)\). Sometimes the wider field retains many
properties of the rational numbers. In particular, it will turn out
that the Gaussian integers are a unique factorization domain, just
like the (rational) integers \(\mathbb{Z}\). We shall first discuss
some basics of higher algebra.

A number \(\alpha\in\mathbb{C}\) is
* algebraic* if there is a polynomial \(p(x)=a_nx^n+a_{n-1}
x^{n-1}+\dots+a_0\) with integer coefficients such that
\(p(\alpha)=0\). If \(a_n=1\), then \(\alpha\) is an * algebraic
integer*.

Further, \(p(x)\) is the * minimal polynomial* of \(\alpha\) if it is
irreducible over \(\mathbb{Z}[x]\) (i.e. it cannot be written as a
product of nonconstant polynomials with integer coefficients).

The theme of this text are extensions of the ring \(\mathbb{Z}\) of
degree 2, so called * quadratic extensions*. Thus, for example,
the polynomials \(x^2+1\) and \(x^2+x+1\) determine the extensions
\(\mathbb{Z}[i]\) and \(\mathbb{Z}[\omega]\), where
\(\omega=\frac{-1+i\sqrt3}2\) (this notation will be used later).

All elements of a quadratic extension of \(\mathbb{Z}\) are algebraic
integers with the minimal polynomial of second degree. Two elements
having the same minimal polynomials are said to be * conjugates*.
Each nonrational element \(z\) of the quadratic extension has exactly
one conjugate, called the conjugate of \(z\) and denoted
\(\overline{z}\). For a rational integer \(z\) we define
\(\overline{z}=z\).

The norm is always an integer. Roughly speaking, it is a kind of equivalent of the absolute value in the set of integers \(\mathbb{Z}\).

If \(z\in\mathbb{Z}[\sqrt{d}]\), \(z=a+b\sqrt{d}\) (\(a,b\in\mathbb{Z}\)), then \(\overline{z}=a-b\sqrt{d}\) and \(N(z)=a^2-db^2\). In particular, in \(\mathbb{Z}[i]\) the norm of element \(a+bi\) (\(a,b\in\mathbb{N}\)) is \(N(a+bi)=a^2+b^2\).

If \(z=a+b\omega\in\mathbb{Z}[\omega]\) (\(a,b\in\mathbb{Z}\)), then \(\overline{z}=a-b-b\omega\) and \(N(z)=a^2-ab+b^2\).

In every complex quadratic field the conjugation corresponds to the complex conjugation.

The following two propositions follow directly from definition.

The norm is multiplicative, i.e. for arbitrary elements \(z_1,z_2\) of a quadratic extension of \(\mathbb{Z}\) it holds that \(N(z_1z_2)=N(z_1)N(z_2)\).

An element \(\epsilon\in\mathbb{Z}[\alpha]\) is called a * unit* if
there exists \(\epsilon^{\prime}\in\mathbb{Z}[\alpha]\) such that
\(\epsilon\epsilon^{\prime}=1\). In that case
\(N(\epsilon)N(\epsilon^{\prime})=N(1)=1\), so \(N(\epsilon)=\pm1\). In fact,
\(\epsilon\) is a unit if and only if its norm is \(\pm1\): indeed, if
\(N(\epsilon)=\pm1\) then \(\epsilon\overline{\epsilon}=\pm1\) by
definition.

An element \(\epsilon\in\mathbb{Z}[\alpha]\) is called a * unit* if
there exists \(\epsilon^{\prime}\in\mathbb{Z}[\alpha]\) such that
\(\epsilon\epsilon^{\prime}=1\). In that case
\(N(\epsilon)N(\epsilon^{\prime})=N(1)=1\), so \(N(\epsilon)=\pm1\). In fact,
\(\epsilon\) is a unit if and only if its norm is \(\pm1\): indeed, if
\(N(\epsilon)=\pm1\) then \(\epsilon\overline{\epsilon}=\pm1\) by
definition.

The only units in \(\mathbb{Z}\) are \(\pm1\).

If \(a+bi\) (\(a,b\in\mathbb{Z}\)) is a unit, then \(N(a+bi)=a^2+b^2=\pm1\), which implies \(a+bi\in\{\pm1,\pm i\}\).

All units in \(\mathbb{Z}[\omega]\) are \(\pm1,\pm\omega, \pm(1+\omega)\). Indeed, if \(a+b\omega\) is a unit then \(a^2-ab+b^2=1\), i.e. \((2a-b)^2+3b^2=4\) and he result follows. Note that \(\omega^2\) equals \(-(1+\omega)\).

Let \(p\) be a prime number and \( N=\prod _{k=1}^{p-1}(k^2+1)\). Determine the remainder of \(N\) upon division by \(p\).

The divisibility and congruences in an extension \(K\) of the ring \(\mathbb{Z}\) is defined in the usual way: \(x\in K\) is divisible by \(y\in K\) (denoted \(y\mid x\)) if there exists \(z\in K\) such that \(x=yz\), and \(x\equiv y\) (mod \(z\)) if \(z\mid x-y\).

Since every element of a quadratic ring is divisible by every unit, the definition of the notion of a prime must be adjusted to the new circumstances.

We have the following simple proposition.

The converse does not hold, as \(3\) is a prime in \(\mathbb{Z}[i]\), but \(N(3)=9\) is composite.

Of course, the elements conjugate or adjoint to a prime are also primes. Therefore the smallest positive rational integer divisible by a prime \(z\) equals \(z\overline{z}=N(z)\).

Consider an arbitrary nonzero and nonunit element \(x\in K\). If \(x\) is not prime then there are nonunit elements \(y,z\in K\) such that \(yz=x\). Hereby \(N(y)N(z)=N(x)\) and \(N(y),N(z) > 1\). Hence \(N(y),N(z) < N(x)\). Continuing this procedure we end up with a factorization \(x=x_1x_2\cdots x_k\) in which all elements are prime. Therefore, we have

Naturally, we would like to know when the factorization into primes is unique, i.e. when the Fundamental Theorem of Arithmetic holds. But let us first note that, by the above definition, the primes of \(\mathbb{Z}\) are \(\pm2,\pm3,\pm5\), etc, so the factorization into primes is not exactly unique, as e.g. \(2\cdot3=(-2)(-3)\). Actually, in this case the uniqueness of factorization is true in the following wording.

The division with remainder in a quadratic extension \(K\) of \(\mathbb{Z}\) can be formulated as follows:

Obviously, such a division, if it exists, is not necessarily unique - it is not so even in \(\mathbb{Z}\) itself. Moreover, it does not exist in some quadratic extensions, as we shall see later. The significance of the existence of a division with remainder, however, lies in the fact that it implies the uniqueness of factorization:

There are quadratic rings in which FTA holds despite the nonexistence of a division with remainder. However, FTA is an exception rather than a rule.